{"id":12612,"date":"2025-09-19T10:17:01","date_gmt":"2025-09-19T09:17:01","guid":{"rendered":"https:\/\/mcqsadda.com\/?p=12612"},"modified":"2025-10-21T07:54:35","modified_gmt":"2025-10-21T06:54:35","slug":"probability-top-100-mcqs-with-answer-and-explanation","status":"publish","type":"post","link":"https:\/\/mcqsadda.com\/index.php\/2025\/09\/19\/probability-top-100-mcqs-with-answer-and-explanation\/","title":{"rendered":"Probability Top 100 MCQs With Answer and Explanation"},"content":{"rendered":"\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">1. What is the probability of getting a head when a fair coin is tossed?<\/mark><\/strong><br>A) 0<br>B) 1\/2<br>C) 1<br>D) 2\/3<br><strong>Answer:<\/strong> B) 1\/2<br><strong>Explanation:<\/strong> Two outcomes \u2192 {Head, Tail}. One favorable \u2192 Probability = 1\/2.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\"><strong>2. Probability of an impossible event is:<br><\/strong><\/mark>A) 1<br>B) 0<br>C) 1\/2<br>D) Undefined<br><strong>Answer:<\/strong> B) 0<br><strong>Explanation:<\/strong> Impossible event has no favorable outcomes.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">3. Probability of a sure event is:<\/mark><\/strong><br>A) 0<br>B) 1\/2<br>C) 1<br>D) Cannot be determined<br><strong>Answer:<\/strong> C) 1<br><strong>Explanation:<\/strong> A certain event always occurs \u2192 probability = 1.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">4. If P(E) = 0.7, then probability of not E is:<\/mark><\/strong><br>A) 0.3<br>B) 0.5<br>C) 0.7<br>D) 1.7<br><strong>Answer:<\/strong> A) 0.3<br><strong>Explanation:<\/strong> P(not E) = 1 \u2013 P(E) = 1 \u2013 0.7 = 0.3.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">5. Which of the following is not a valid probability value?<\/mark><\/strong><br>A) 0<br>B) 1\/2<br>C) 3\/4<br>D) 1.2<br><strong>Answer:<\/strong> D) 1.2<br><strong>Explanation:<\/strong> Probability \u2208 [0, 1].<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">6. A dice is thrown. Probability of getting a number greater than 4?<\/mark><\/strong><br>A) 1\/6<br>B) 2\/6<br>C) 1\/3<br>D) 1\/2<br><strong>Answer:<\/strong> C) 1\/3<br><strong>Explanation:<\/strong> Outcomes > 4 \u2192 {5, 6} \u2192 2\/6 = 1\/3.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">7. A card is drawn from a pack of 52. Probability of drawing a red card?<\/mark><\/strong><br>A) 1\/4<br>B) 1\/2<br>C) 1\/3<br>D) 2\/3<br><strong>Answer:<\/strong> B) 1\/2<br><strong>Explanation:<\/strong> Red cards = 26\/52 = 1\/2.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">8. Probability of getting an odd number when a dice is thrown?<\/mark><\/strong><br>A) 1\/6<br>B) 1\/2<br>C) 2\/3<br>D) 5\/6<br><strong>Answer:<\/strong> B) 1\/2<br><strong>Explanation:<\/strong> Odd numbers {1,3,5} \u2192 3\/6 = 1\/2.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">9. If a coin is tossed twice, total possible outcomes = ?<\/mark><\/strong><br>A) 2<br>B) 3<br>C) 4<br>D) 6<br><strong>Answer:<\/strong> C) 4<br><strong>Explanation:<\/strong> Sample space {HH, HT, TH, TT}.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">10. Probability of getting at least one head when two coins are tossed?<\/mark><\/strong><br>A) 1\/4<br>B) 1\/2<br>C) 3\/4<br>D) 1<br><strong>Answer:<\/strong> C) 3\/4<br><strong>Explanation:<\/strong> Outcomes = 4. Favourable = {HH, HT, TH} = 3. \u2192 3\/4.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">11. A number is chosen at random from 1 to 10. Probability it is prime?<\/mark><\/strong><br>A) 2\/5<br>B) 3\/5<br>C) 4\/5<br>D) 1\/2<br><strong>Answer:<\/strong> A) 2\/5<br><strong>Explanation:<\/strong> Primes \u2264 10: {2,3,5,7} \u2192 4\/10 = 2\/5.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">12. If a card is drawn, probability it is a King?<\/mark><\/strong><br>A) 1\/52<br>B) 1\/26<br>C) 1\/13<br>D) 1\/12<br><strong>Answer:<\/strong> C) 1\/13<br><strong>Explanation:<\/strong> Total Kings = 4. Probability = 4\/52 = 1\/13.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">13. What is the probability of getting number 7 on a dice?<\/mark><\/strong><br>A) 0<br>B) 1\/6<br>C) 1\/12<br>D) 1\/7<br><strong>Answer:<\/strong> A) 0<br><strong>Explanation:<\/strong> Dice outcomes 1\u20136. 7 is impossible \u2192 probability 0.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">14. In tossing 3 coins, probability of getting all heads?<\/mark><\/strong><br>A) 1\/8<br>B) 1\/4<br>C) 1\/2<br>D) 3\/8<br><strong>Answer:<\/strong> A) 1\/8<br><strong>Explanation:<\/strong> Outcomes = 8. Only 1 favorable (HHH). \u2192 1\/8.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">15. If two dice are thrown, total sample space size = ?<\/mark><\/strong><br>A) 6<br>B) 12<br>C) 18<br>D) 36<br><strong>Answer:<\/strong> D) 36<br><strong>Explanation:<\/strong> Each die 6 outcomes. 6\u00d76 = 36.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">16. Probability of getting a sum of 7 on two dice?<\/mark><\/strong><br>A) 1\/6<br>B) 1\/12<br>C) 1\/18<br>D) 1\/36<br><strong>Answer:<\/strong> A) 1\/6<br><strong>Explanation:<\/strong> Pairs giving 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) \u2192 6\/36 = 1\/6.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">17. Probability of drawing a black card?<\/mark><\/strong><br>A) 1\/2<br>B) 1\/4<br>C) 2\/3<br>D) 1\/3<br><strong>Answer:<\/strong> A) 1\/2<br><strong>Explanation:<\/strong> Black cards = 26. Probability = 26\/52 = 1\/2.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">18. A die is rolled. Probability of a multiple of 3?<\/mark><\/strong><br>A) 1\/6<br>B) 1\/3<br>C) 1\/2<br>D) 2\/3<br><strong>Answer:<\/strong> B) 1\/3<br><strong>Explanation:<\/strong> Multiples of 3 = {3,6} \u2192 2\/6 = 1\/3.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">19. If P(E) = 0.35, then P(not E) = ?<\/mark><\/strong><br>A) 0.45<br>B) 0.55<br>C) 0.65<br>D) 0.75<br><strong>Answer:<\/strong> C) 0.65<br><strong>Explanation:<\/strong> 1 \u2013 0.35 = 0.65.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">20. Probability of selecting a vowel from English alphabets (A\u2013Z)?<\/mark><\/strong><br>A) 5\/26<br>B) 21\/26<br>C) 1\/5<br>D) 5\/21<br><strong>Answer:<\/strong> A) 5\/26<br><strong>Explanation:<\/strong> Vowels = {A,E,I,O,U} \u2192 5\/26.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">21. A card is drawn. Probability of drawing an ace?<\/mark><\/strong><br>A) 1\/52<br>B) 1\/13<br>C) 1\/26<br>D) 4\/13<br><strong>Answer:<\/strong> B) 1\/13<br><strong>Explanation:<\/strong> 4 aces in 52 cards \u2192 4\/52 = 1\/13.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">22. In a lottery of 1000 tickets, one ticket wins. If you buy 1 ticket, probability of winning = ?<\/mark><\/strong><br>A) 1\/1000<br>B) 1\/100<br>C) 1\/10<br>D) 0<br><strong>Answer:<\/strong> A) 1\/1000<br><strong>Explanation:<\/strong> Favourable = 1, total = 1000 \u2192 1\/1000.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">23. A bag has 3 red and 2 black balls. Probability of drawing a red ball?<\/mark><\/strong><br>A) 2\/5<br>B) 3\/5<br>C) 1\/2<br>D) 1\/3<br><strong>Answer:<\/strong> B) 3\/5<br><strong>Explanation:<\/strong> Total = 5, Red = 3 \u2192 3\/5.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">24. A die is thrown. Probability of getting even number?<\/mark><\/strong><br>A) 1\/2<br>B) 1\/3<br>C) 2\/3<br>D) 5\/6<br><strong>Answer:<\/strong> A) 1\/2<br><strong>Explanation:<\/strong> Even {2,4,6} \u2192 3\/6 = 1\/2.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">25. Which is correct for any event E?<\/mark><\/strong><br>A) 0 \u2264 P(E) \u2264 1<br>B) \u20131 \u2264 P(E) \u2264 1<br>C) P(E) \u2265 0 always<br>D) P(E) \u2265 1 always<br><strong>Answer:<\/strong> A) 0 \u2264 P(E) \u2264 1<br><strong>Explanation:<\/strong> Probability always lies between 0 and 1.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">26. Two dice are thrown. Probability of getting doublets (same number on both dice)?<\/mark><\/strong><br>A) 1\/6<br>B) 1\/12<br>C) 1\/18<br>D) 1\/36<br><strong>Answer:<\/strong> A) 1\/6<br><strong>Explanation:<\/strong> Doublets = (1,1),(2,2)\u2026(6,6) \u2192 6 outcomes. Probability = 6\/36 = 1\/6.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">27. Probability of getting a sum of 11 on two dice?<\/mark><\/strong><br>A) 1\/18<br>B) 1\/12<br>C) 1\/6<br>D) 1\/36<br><strong>Answer:<\/strong> A) 1\/18<br><strong>Explanation:<\/strong> Pairs = (5,6),(6,5) \u2192 2 outcomes. Probability = 2\/36 = 1\/18.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">28. A card is drawn. Probability it is neither king nor queen?<\/mark><\/strong><br>A) 12\/13<br>B) 25\/26<br>C) 23\/26<br>D) 11\/13<br><strong>Answer:<\/strong> A) 12\/13<br><strong>Explanation:<\/strong> Kings = 4, Queens = 4 \u2192 8 total. Favorable = 52\u20138=44. Probability = 44\/52=11\/13.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">29. A coin is tossed 3 times. Probability of getting exactly 2 heads?<\/mark><\/strong><br>A) 1\/8<br>B) 3\/8<br>C) 3\/4<br>D) 1\/2<br><strong>Answer:<\/strong> B) 3\/8<br><strong>Explanation:<\/strong> Outcomes = 8. Favourable = {HHT, HTH, THH} = 3. Probability = 3\/8.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">30. A card is drawn from 52. Probability it is a spade or a king?<\/mark><\/strong><br>A) 4\/13<br>B) 17\/52<br>C) 16\/52<br>D) 13\/52<br><strong>Answer:<\/strong> B) 17\/52<br><strong>Explanation:<\/strong> Spades = 13, Kings = 4, but overlap (king of spades counted twice). Total favorable = 13+4\u20131=16. Probability = 16\/52 = 4\/13.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">31. Probability of getting at least one 6 in two dice throws?<\/mark><\/strong><br>A) 5\/36<br>B) 11\/36<br>C) 25\/36<br>D) 1\/6<br><strong>Answer:<\/strong> C) 25\/36<br><strong>Explanation:<\/strong> P(no 6) = (5\/6)\u00d7(5\/6)=25\/36. So P(at least one 6)=1\u201325\/36=11\/36.<br>(\u26a0 Correction: Actually \u2192 11\/36.)<\/p>\n\n\n\n<p class=\"has-large-font-size\">\u2714 Correct Answer: <strong>B) 11\/36<\/strong><\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">32. A coin is tossed 4 times. Probability of getting all tails?<\/mark><\/strong><br>A) 1\/8<br>B) 1\/16<br>C) 1\/32<br>D) 1\/4<br><strong>Answer:<\/strong> B) 1\/16<br><strong>Explanation:<\/strong> Total = 16 outcomes. Only TTTT favorable. Probability = 1\/16.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">33. A card is drawn. Probability it is a red face card?<\/mark><\/strong><br>A) 3\/13<br>B) 3\/26<br>C) 6\/52<br>D) 12\/52<br><strong>Answer:<\/strong> B) 3\/26<br><strong>Explanation:<\/strong> Face cards = 12 (J,Q,K). Red face cards = 6 (\u2665\u2666). Probability = 6\/52=3\/26.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">34. Two coins are tossed. Probability of getting at least one tail?<\/mark><\/strong><br>A) 1\/2<br>B) 3\/4<br>C) 1\/4<br>D) 2\/3<br><strong>Answer:<\/strong> B) 3\/4<br><strong>Explanation:<\/strong> Total outcomes = 4. Favourable = {HT,TH,TT}=3. Probability=3\/4.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">35. Probability of getting a sum of 9 on two dice?<\/mark><\/strong><br>A) 1\/9<br>B) 1\/8<br>C) 1\/12<br>D) 1\/18<br><strong>Answer:<\/strong> C) 1\/12<br><strong>Explanation:<\/strong> Outcomes = (3,6),(4,5),(5,4),(6,3) \u2192 4. Probability=4\/36=1\/9.<br>\u2714 Correct Answer: <strong>A) 1\/9<\/strong><\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">36. A card is drawn. Probability it is not a heart?<\/mark><\/strong><br>A) 1\/2<br>B) 3\/4<br>C) 1\/3<br>D) 2\/3<br><strong>Answer:<\/strong> B) 3\/4<br><strong>Explanation:<\/strong> Hearts=13, not hearts=39. Probability=39\/52=3\/4.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">37. Probability of getting exactly one head in 3 coin tosses?<\/mark><\/strong><br>A) 1\/8<br>B) 2\/8<br>C) 3\/8<br>D) 4\/8<br><strong>Answer:<\/strong> C) 3\/8<br><strong>Explanation:<\/strong> Favourable = {HTT, THT, TTH} \u2192 3. Probability=3\/8.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\"><strong>38. If two dice are thrown, probability that both numbers are even?<\/strong><\/mark><br><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-dark-gray-color\">A) 1\/4<\/mark><br>B) 1\/9<br>C) 1\/3<br>D) 1\/6<br><strong>Answer:<\/strong> A) 1\/4<br><strong>Explanation:<\/strong> Even on each die=3\/6=1\/2. Together=(1\/2)\u00d7(1\/2)=1\/4.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">39. Probability of drawing a black queen from 52 cards?<\/mark><\/strong><br>A) 1\/26<br>B) 1\/52<br>C) 1\/13<br>D) 1\/12<br><strong>Answer:<\/strong> A) 1\/26<br><strong>Explanation:<\/strong> Black queens=2. Probability=2\/52=1\/26.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">40. Two dice are thrown. Probability of getting an even sum?<\/mark><\/strong><br>A) 1\/4<br>B) 1\/2<br>C) 2\/3<br>D) 3\/4<br><strong>Answer:<\/strong> B) 1\/2<br><strong>Explanation:<\/strong> Even sum occurs if both even or both odd. Half the outcomes are even \u2192 18\/36=1\/2.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">41. A coin is tossed. Probability of not getting a tail?<\/mark><\/strong><br>A) 0<br>B) 1\/2<br>C) 1<br>D) 1\/3<br><strong>Answer:<\/strong> B) 1\/2<br><strong>Explanation:<\/strong> Not tail= head \u2192 probability 1\/2.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">42. Probability of drawing a king of spades?<\/mark><\/strong><br>A) 1\/52<br>B) 1\/26<br>C) 1\/13<br>D) 1\/12<br><strong>Answer:<\/strong> A) 1\/52<br><strong>Explanation:<\/strong> Only one king of spades. Probability=1\/52.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">43. If 2 dice are thrown, probability of both numbers being prime?<\/mark><\/strong><br>A) 1\/9<br>B) 1\/6<br>C) 1\/4<br>D) 1\/3<br><strong>Answer:<\/strong> A) 1\/9<br><strong>Explanation:<\/strong> Prime numbers on dice= {2,3,5} =3. Probability=3\/6=1\/2 each. Combined=1\/4.<br>Correction: Actually \u2192 (3\/6)\u00d7(3\/6)=9\/36=1\/4.<br>\u2714 Correct Answer: <strong>C) 1\/4<\/strong><\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">44. Probability of getting 2 or 3 on a dice?<\/mark><\/strong><br>A) 1\/6<br>B) 2\/6<br>C) 1\/3<br>D) 1\/2<br><strong>Answer:<\/strong> B) 2\/6 = 1\/3<br><strong>Explanation:<\/strong> Favourable={2,3}. Probability=2\/6=1\/3.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">45. A card is drawn. Probability that it is neither red nor a king?<\/mark><\/strong><br>A) 3\/4<br>B) 11\/13<br>C) 12\/13<br>D) 25\/26<br><strong>Answer:<\/strong> B) 11\/13<br><strong>Explanation:<\/strong> Red cards=26, kings=4, overlap=2 (red kings). Total=28. Favorable=24. Probability=24\/52=6\/13.<br>\u2714 Correct Answer: <strong>6\/13 (not in given options)<\/strong><\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">46. Two dice thrown. Probability of getting sum \u2264 4?<\/mark><\/strong><br>A) 1\/9<br>B) 1\/6<br>C) 1\/12<br>D) 1\/18<br><strong>Answer:<\/strong> B) 1\/6<br><strong>Explanation:<\/strong> Favourable sums=2,3,4. Outcomes=(1,1),(1,2),(2,1),(1,3),(2,2),(3,1) \u21926. Probability=6\/36=1\/6.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">47. Probability of drawing a non-face card?<\/mark><\/strong><br>A) 10\/13<br>B) 11\/13<br>C) 3\/4<br>D) 2\/3<br><strong>Answer:<\/strong> A) 10\/13<br><strong>Explanation:<\/strong> Face cards=12. Non-face=40. Probability=40\/52=10\/13.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">48. A card is drawn. Probability it is a diamond or club?<\/mark><\/strong><br>A) 1\/2<br>B) 1\/4<br>C) 1\/3<br>D) 2\/3<br><strong>Answer:<\/strong> A) 1\/2<br><strong>Explanation:<\/strong> Diamonds=13, clubs=13 \u2192 26. Probability=26\/52=1\/2.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">49. Probability of getting a sum of 5 on two dice?<\/mark><\/strong><br>A) 1\/9<br>B) 1\/12<br>C) 1\/18<br>D) 1\/36<br><strong>Answer:<\/strong> B) 1\/12<br><strong>Explanation:<\/strong> Outcomes: (1,4),(2,3),(3,2),(4,1)=4. Probability=4\/36=1\/9.<br>\u2714 Correct Answer: <strong>A) 1\/9<\/strong><\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">50. If two coins are tossed, probability of getting one head and one tail?<\/mark><\/strong><br>A) 1\/4<br>B) 1\/2<br>C) 3\/4<br>D) 2\/3<br><strong>Answer:<\/strong> B) 1\/2<br><strong>Explanation:<\/strong> Outcomes= {HH,HT,TH,TT}. Favorable={HT,TH}=2. Probability=2\/4=1\/2.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">51. If P(A) = 0.5, P(B) = 0.4 and A &amp; B are independent, find P(A \u2229 B).<\/mark><\/strong><br>A) 0.1<br>B) 0.2<br>C) 0.25<br>D) 0.9<br><strong>Answer:<\/strong> B) 0.2<br><strong>Explanation:<\/strong> For independent events \u2192 P(A \u2229 B) = P(A) \u00d7 P(B) = 0.5 \u00d7 0.4 = 0.2.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">52. If P(A) = 0.7, P(B) = 0.6, P(A \u2229 B) = 0.5, then P(A \u222a B) = ?<\/mark><\/strong><br>A) 0.8<br>B) 0.9<br>C) 1.2<br>D) 0.7<br><strong>Answer:<\/strong> A) 0.8<br><strong>Explanation:<\/strong> P(A \u222a B) = P(A)+P(B)\u2013P(A \u2229 B) = 0.7+0.6\u20130.5=0.8.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">53. Two independent events A and B have probabilities 0.6 and 0.5. Probability that both occur?<\/mark><\/strong><br>A) 0.1<br>B) 0.2<br>C) 0.3<br>D) 0.25<br><strong>Answer:<\/strong> C) 0.3<br><strong>Explanation:<\/strong> P(A \u2229 B) = 0.6\u00d70.5 = 0.3.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">54. If P(A) = 0.4, P(B) = 0.5, and A, B independent, probability that neither occurs?<\/mark><\/strong><br>A) 0.1<br>B) 0.2<br>C) 0.3<br>D) 0.6<br><strong>Answer:<\/strong> D) 0.6<br><strong>Explanation:<\/strong> P(A \u222a B) = 0.4+0.5\u20130.2=0.7. So P(none) = 1\u20130.7=0.3.<br>\u2714 Correct Answer: <strong>C) 0.3<\/strong><\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">55. If P(A) = 0.3, P(B) = 0.2, and A,B mutually exclusive, P(A \u222a B) = ?<\/mark><\/strong><br>A) 0.1<br>B) 0.3<br>C) 0.5<br>D) 0.6<br><strong>Answer:<\/strong> C) 0.5<br><strong>Explanation:<\/strong> Mutually exclusive \u2192 P(A \u2229 B)=0. So P(A \u222a B)=0.3+0.2=0.5.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">56. If P(A) = 0.4, P(B) = 0.5, P(A \u2229 B) = 0.2, then P(A|B) = ?<\/mark><\/strong><br>A) 0.2<br>B) 0.4<br>C) 0.6<br>D) 0.7<br><strong>Answer:<\/strong> C) 0.6<br><strong>Explanation:<\/strong> P(A|B) = P(A \u2229 B)\/P(B) = 0.2\/0.5=0.4.<br>\u2714 Correct Answer: <strong>B) 0.4<\/strong><\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">57. If two dice are thrown, find P(sum is even | first die shows 3).<\/mark><\/strong><br>A) 1\/2<br>B) 1\/3<br>C) 2\/3<br>D) 5\/6<br><strong>Answer:<\/strong> A) 1\/2<br><strong>Explanation:<\/strong> If first die=3 (odd), second die must be odd for sum even. Half the outcomes \u2192 3\/6=1\/2.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">58. A bag has 3 red, 2 black. One ball drawn. Probability red given it is not black?<\/mark><\/strong><br>A) 1<br>B) 3\/5<br>C) 1\/2<br>D) 2\/3<br><strong>Answer:<\/strong> A) 1<br><strong>Explanation:<\/strong> If not black \u2192 must be red. Probability = 1.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">59. A coin tossed twice. Find P(first toss is head | exactly one head occurs).<\/mark><\/strong><br>A) 1\/2<br>B) 1\/3<br>C) 2\/3<br>D) 1\/4<br><strong>Answer:<\/strong> A) 1\/2<br><strong>Explanation:<\/strong> Exactly one head outcomes {HT,TH}. Given condition \u2192 2 outcomes. Favourable {HT}. Probability=1\/2.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">60. A bag has 2 white, 3 red, 5 black. One ball drawn. P(red|not black)?<\/mark><\/strong><br>A) 1\/5<br>B) 1\/2<br>C) 3\/5<br>D) 3\/7<br><strong>Answer:<\/strong> D) 3\/7<br><strong>Explanation:<\/strong> Not black \u2192 {2W,3R}=5. Red among them=3. Probability=3\/5.<br>\u2714 Correct Answer: <strong>C) 3\/5<\/strong><\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">61. If P(A) = 0.5, P(B) = 0.3, P(A \u2229 B) = 0.2, find P(A|B).<\/mark><\/strong><br>A) 2\/3<br>B) 1\/2<br>C) 1\/3<br>D) 1\/4<br><strong>Answer:<\/strong> A) 2\/3<br><strong>Explanation:<\/strong> P(A|B)=P(A \u2229 B)\/P(B)=0.2\/0.3=2\/3.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">62. Two cards drawn without replacement from 52. Find P(both aces).<\/mark><\/strong><br>A) 1\/169<br>B) 1\/221<br>C) 1\/1326<br>D) 1\/26<br><strong>Answer:<\/strong> B) 1\/221<br><strong>Explanation:<\/strong> P= (4\/52)\u00d7(3\/51) = 12\/2652 = 1\/221.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">63. From 1\u201320, one number chosen. Find P(multiple of 4 | even).<\/mark><\/strong><br>A) 1\/2<br>B) 1\/3<br>C) 2\/5<br>D) 1\/4<br><strong>Answer:<\/strong> B) 1\/3<br><strong>Explanation:<\/strong> Even numbers=10. Multiples of 4 among them= {4,8,12,16,20}=5. Probability=5\/10=1\/2.<br>\u2714 Correct Answer: <strong>A) 1\/2<\/strong><\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">64. If P(A) = 0.6, P(B) = 0.5, and P(A \u2229 B)=0.3, are A and B independent?<\/mark><\/strong><br>A) Yes<br>B) No<br>C) Sometimes<br>D) Can\u2019t say<br><strong>Answer:<\/strong> A) Yes<br><strong>Explanation:<\/strong> Independence check: P(A)\u00d7P(B)=0.6\u00d70.5=0.3 = P(A \u2229 B). So independent.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">65. A bag has 3 white, 2 red. Two balls drawn without replacement. Find P(both white).<\/mark><\/strong><br>A) 1\/5<br>B) 1\/10<br>C) 1\/3<br>D) 3\/10<br><strong>Answer:<\/strong> D) 3\/10<br><strong>Explanation:<\/strong> P= (3\/5)\u00d7(2\/4)=6\/20=3\/10.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">66. If a card is drawn, probability that it is king given that it is a face card?<\/mark><\/strong><br>A) 1\/12<br>B) 1\/3<br>C) 1\/4<br>D) 1\/13<br><strong>Answer:<\/strong> B) 1\/3<br><strong>Explanation:<\/strong> Face cards=12 (J,Q,K). Kings=4. Probability=4\/12=1\/3.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">67. Two dice thrown. Probability that sum is 9 given first die is 4?<\/mark><\/strong><br>A) 1\/6<br>B) 1\/3<br>C) 1\/2<br>D) 1\/12<br><strong>Answer:<\/strong> A) 1\/6<br><strong>Explanation:<\/strong> If first=4, second must=5. One favourable out of 6 \u2192 1\/6.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">68. A bag has 2 red, 3 green, 5 blue. One ball drawn. P(green | not blue)?<\/mark><\/strong><br>A) 1\/2<br>B) 1\/3<br>C) 3\/5<br>D) 3\/8<br><strong>Answer:<\/strong> B) 1\/3<br><strong>Explanation:<\/strong> Not blue = 5 balls (2R+3G). Green among them=3. Probability=3\/5.<br>\u2714 Correct Answer: <strong>C) 3\/5<\/strong><\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">69. A card drawn. Find P(king or queen | face card).<\/mark><\/strong><br>A) 1\/2<br>B) 2\/3<br>C) 1\/3<br>D) 1\/4<br><strong>Answer:<\/strong> B) 2\/3<br><strong>Explanation:<\/strong> Face cards=12. Kings+Queens=8. Probability=8\/12=2\/3.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">70. A coin is tossed thrice. Find P(exactly 2 heads | at least 1 head).<\/mark><\/strong><br>A) 3\/8<br>B) 3\/7<br>C) 1\/3<br>D) 1\/2<br><strong>Answer:<\/strong> B) 3\/7<br><strong>Explanation:<\/strong> Total outcomes=8. At least one head=7 outcomes. Exactly 2 heads=3 outcomes. Probability=3\/7.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">71. A die is rolled. Find P(even number | prime number).<\/mark><\/strong><br>A) 1\/2<br>B) 1\/3<br>C) 1\/4<br>D) 2\/3<br><strong>Answer:<\/strong> C) 1\/4<br><strong>Explanation:<\/strong> Primes on die={2,3,5}. Even among them={2}. Probability=1\/3.<br>\u2714 Correct Answer: <strong>B) 1\/3<\/strong><\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">72. From 1\u201310, number chosen. P(odd | not prime)?<\/mark><\/strong><br>A) 1\/2<br>B) 1\/3<br>C) 1\/4<br>D) 2\/5<br><strong>Answer:<\/strong> D) 2\/5<br><strong>Explanation:<\/strong> Non-primes= {1,4,6,8,9,10} = 6. Odd among them= {1,9}=2. Probability=2\/6=1\/3.<br>\u2714 Correct Answer: <strong>B) 1\/3<\/strong><\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">73. A box has 2 defective, 8 good bulbs. If 1 picked, find P(defective | not good)?<\/mark><\/strong><br>A) 0<br>B) 1<br>C) 1\/5<br>D) 2\/5<br><strong>Answer:<\/strong> B) 1<br><strong>Explanation:<\/strong> If not good \u2192 must be defective. Probability=1.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">74. If a coin is tossed thrice, find P(first toss head | total heads = 2).<\/mark><\/strong><br>A) 1\/2<br>B) 1\/3<br>C) 2\/3<br>D) 1\/4<br><strong>Answer:<\/strong> C) 2\/3<br><strong>Explanation:<\/strong> Exactly 2 heads outcomes= {HHT,HTH,THH}. First head in 2 of them. Probability=2\/3.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">75. (Bayes\u2019 theorem) A bag has 2 red, 3 green. Another has 4 red, 5 green. One bag chosen at random, 1 ball drawn and it is red. Probability it came from first bag?<\/mark><\/strong><br>A) 2\/7<br>B) 3\/7<br>C) 4\/7<br>D) 5\/7<br><strong>Answer:<\/strong> B) 3\/7<br><strong>Explanation:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li class=\"has-large-font-size\">P(Bag1)=1\/2, P(R|Bag1)=2\/5.<\/li>\n\n\n\n<li class=\"has-large-font-size\">P(Bag2)=1\/2, P(R|Bag2)=4\/9.<\/li>\n\n\n\n<li class=\"has-large-font-size\">P(R)= (1\/2\u00d72\/5)+(1\/2\u00d74\/9) = 23\/90.<\/li>\n\n\n\n<li class=\"has-large-font-size\">P(Bag1|R)= [(1\/2\u00d72\/5)] \/ (23\/90) = (1\/5)\/(23\/90)=18\/90=2\/10=0.2.<br>Correction \u2192 simplify carefully: (1\/5)\/(23\/90) = (18\/115)=0.156.<\/li>\n<\/ul>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">76. Five fair coins are tossed. Probability of getting exactly 3 heads is<\/mark><\/strong><br>A) 5\/16<br>B) 5\/32<br>C) 10\/32<br>D) 10\/16<br><strong>Answer:<\/strong> C) 10\/32 (which simplifies to 5\/16)<br><strong>Explanation:<\/strong> Number of ways = C(5,3)=10. Total outcomes = 2\u2075=32. Probability = 10\/32 = 5\/16.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">77. Two dice are rolled. Probability that the product is divisible by 3 is<\/mark><\/strong><br>A) 1\/3<br>B) 5\/9<br>C) 2\/3<br>D) 8\/9<br><strong>Answer:<\/strong> B) 5\/9<br><strong>Explanation:<\/strong> Product divisible by 3 unless both dice are \u2260 multiples of 3. Numbers not divisible by 3: {1,2,4,5} (4 choices). Probability both not divisible by 3 = (4\/6)\u00d7(4\/6)= (2\/3)\u00d7(2\/3)=4\/9. So product divisible by 3 = 1 \u2212 4\/9 = 5\/9.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">78. From 10 balls (4 red, 6 blue) pick 3 without replacement. Probability all three are blue =<\/mark><\/strong><br>A) C(6,3)\/C(10,3)<br>B) (6\/10)\u00b3<br>C) (4\/10)(3\/9)(2\/8)<br>D) 1\/10<br><strong>Answer:<\/strong> A) C(6,3)\/C(10,3) (= 20\/120 = 1\/6)<br><strong>Explanation:<\/strong> Hypergeometric: choose 3 blue out of 6 divided by choose any 3 out of 10. Numeric: C(6,3)=20, C(10,3)=120 \u2192 20\/120 = 1\/6.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">79. In a classroom 60% like tea, 50% like coffee, and 30% like both. Probability a randomly chosen student likes neither =<\/mark><\/strong><br>A) 0.20<br>B) 0.30<br>C) 0.40<br>D) 0.60<br><strong>Answer:<\/strong> A) 0.20<br><strong>Explanation:<\/strong> P(tea \u222a coffee) = 0.6 + 0.5 \u2212 0.3 = 0.8. So neither = 1 \u2212 0.8 = 0.2.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">80. A fair die is rolled 3 times. Probability that the sequence contains exactly one 6 is<\/mark><\/strong><br>A) 3 \u00d7 (1\/6) \u00d7 (5\/6)\u00b2<br>B) (1\/6)\u00b3<br>C) C(3,1)\/(6\u00b3)<br>D) (5\/6)\u00b3<br><strong>Answer:<\/strong> A) 3 \u00d7 (1\/6) \u00d7 (5\/6)\u00b2<br><strong>Explanation:<\/strong> Choose which one of 3 rolls is 6: C(3,1)=3. Probability = 3*(1\/6)*(5\/6)\u00b2.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">81. A box contains 3 white and 2 black balls. Two balls are drawn with replacement. Probability that both are black =<\/mark><\/strong><br>A) (2\/5)\u00b2<br>B) (2\/5)(1\/4)<br>C) C(2,2)\/C(5,2)<br>D) 2\/5<br><strong>Answer:<\/strong> A) (2\/5)\u00b2 = 4\/25<br><strong>Explanation:<\/strong> With replacement each draw independent: (2\/5)*(2\/5)=4\/25.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">82. A factory produces items; each item independently is defective with probability 0.02. For a sample of 50 items, approximate probability that no item is defective (use (1 \u2212 p)^n) =<\/mark><\/strong><br>A) (0.98)^50<br>B) 50\u00d70.02<br>C) e^(\u22121)<br>D) 1 \u2212 (0.02)\u2075\u2070<br><strong>Answer:<\/strong> A) (0.98)^50 \u2248 0.363<br><strong>Explanation:<\/strong> P(no defect) = (1 \u2212 0.02)^50 = 0.98^50 \u2248 e^{50 ln0.98} \u2248 e^{-1.0101} \u2248 0.364.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">83. From a standard deck, five cards are drawn. Probability of exactly one ace =<\/mark><\/strong><br>A) C(4,1)C(48,4)\/C(52,5)<br>B) (4\/52)(48\/51)\u2026<br>C) 4\/52<br>D) C(4,1)\/C(52,5)<br><strong>Answer:<\/strong> A) C(4,1)C(48,4)\/C(52,5)<br><strong>Explanation:<\/strong> Choose 1 ace from 4 and 4 non-aces from 48; divide by total 5-card combinations.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">84. A fair coin is tossed until first head. Expected number of tosses =<\/mark><\/strong><br>A) 1<br>B) 2<br>C) 1\/p where p=1\/2 \u2192 2<br>D) 3<br><strong>Answer:<\/strong> C) 2<br><strong>Explanation:<\/strong> Geometric distribution with success prob p=1\/2 \u2192 mean = 1\/p = 2.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">85. Two independent Poisson processes with rates 2 and 3 per hour are combined. Probability that in one hour total events = 4 is<\/mark><\/strong><br>A) e^{\u22125} 5\u2074\/4!<br>B) e^{\u22122} 2\u2074\/4! + e^{\u22123} 3\u2074\/4!<br>C) (e^{\u22122}2\u2074\/4!)(e^{\u22123}3\u2074\/4!)<br>D) impossible to compute<br><strong>Answer:<\/strong> A) e^{\u22125} 5\u2074\/4!<br><strong>Explanation:<\/strong> Sum is Poisson(2+3=5). So P(X=4)=e^{-5}5^4\/4!.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">86. From 8 people, two are chosen at random for posts of President and Secretary (different posts). Probability a particular person Alice gets at least one post =<\/mark><\/strong><br>A) 2\/8<br>B) 1\/4<br>C) 1\/4? (same)<br>D) 1\/5<br><strong>Answer:<\/strong> B) 1\/4<br><strong>Explanation:<\/strong> Total ordered pairs = 8\u00d77=56. Number of ordered pairs where Alice has at least one post = (Alice president &amp; anyone secretary) 7 + (Alice secretary &amp; anyone president) 7 = 14. Probability = 14\/56 = 1\/4.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">87. A random permutation of 5 distinct letters. Probability that none is in its original position (derangement D\u2085) =<\/mark><\/strong><br>A) D\u2085\/5! where D\u2085 = 44 \u2192 44\/120 = 11\/30<br>B) 1\/5!<br>C) 1\/2<br>D) 24\/120<br><strong>Answer:<\/strong> A) 11\/30<br><strong>Explanation:<\/strong> Number of derangements for n=5 is 44. Probability = 44\/120 = 11\/30.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">88. A box contains 6 good and 4 defective bulbs. Two bulbs are drawn without replacement. Probability exactly one defective =<\/mark><\/strong><br>A) 2 \u00d7 (4\/10)(6\/9)<br>B) (4\/10)(3\/9)<br>C) C(4,1)C(6,1)\/C(10,2)<br>D) Both A and C equal<br><strong>Answer:<\/strong> D) Both A and C equal<br><strong>Explanation:<\/strong> Calculate: 2*(4\/10)<em>(6\/9)=2<\/em>(24\/90)=48\/90=8\/15. Hypergeometric C(4,1)C(6,1)\/C(10,2) = 4*6\/45=24\/45=8\/15.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">89. Two boxes: Box A has 2 red\/3 blue, Box B has 1 red\/1 blue. Choose a box uniformly at random and pick a ball; observed ball is blue. Probability it came from Box B =<\/mark><\/strong><br>A) (1\/2 \u00d7 1\/2) \/ P(blue)<br>B) (1\/2\u00d71\/2) \/ [(1\/2\u00d73\/5)+(1\/2\u00d71\/2)]<br>C) 1\/2<br>D) 3\/10<br><strong>Answer:<\/strong> B) (1\/2\u00d71\/2) \/ [(1\/2\u00d73\/5)+(1\/2\u00d71\/2)] = (1\/4)\/((3\/10)+(1\/4)) = (1\/4)\/(3\/10+1\/4) = (1\/4)\/( (6\/20+5\/20)=11\/20 ) = (1\/4)\/(11\/20)= (1\/4)*(20\/11)=5\/11 \u2248 0.4545<br><strong>Explanation:<\/strong> Bayes: P(B|blue)=P(B)<em>P(blue|B)\/P(blue). Numeric: P(blue)=0.5<\/em>(3\/5)+0.5*(1\/2)=3\/10+1\/4=11\/20. So answer 5\/11.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">90. A fair six-sided die is rolled repeatedly until two consecutive sixes appear. The expected number of rolls (approx) is<\/mark><\/strong><br>A) 36<br>B) 42<br>C) 36? (exact value 42)<br>D) 49<br><strong>Answer:<\/strong> C) 42<br><strong>Explanation:<\/strong> Expected waiting time for two consecutive successes with p=1\/6 is (1 + q)\/(p\u00b2) where q=1\u2212p? More direct known result for waiting time for two consecutive successes: (1 + p)\/p\u00b2 \u2212 1\/p? Simpler: expected waiting time for two consecutive sixes equals (36+6)\/? Known result: for two consecutive heads with p=1\/2 expected = 6. For p=1\/6 it&#8217;s (1+p)\/p\u00b2 = (1+1\/6)\/(1\/36)= (7\/6)*36 = 42. So 42.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">91. A biased coin has P(head)=0.7. Probability of exactly 2 heads in 3 tosses =<\/mark><\/strong><br>A) C(3,2)(0.7)\u00b2(0.3)<br>B) (0.7)\u00b2<br>C) 3(0.7)(0.3)\u00b2<br>D) (0.3)\u00b3<br><strong>Answer:<\/strong> A) 3 \u00d7 0.7\u00b2 \u00d7 0.3 = 3 \u00d7 0.49 \u00d7 0.3 = 0.441<br><strong>Explanation:<\/strong> Binomial formula: C(3,2)p\u00b2(1\u2212p).<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">92. If X ~ Poisson(\u03bb=2), P(X \u2265 1) =<\/mark><\/strong><br>A) 1 \u2212 e^{\u22122}<br>B) e^{\u22122}<br>C) 1 \u2212 2e^{\u22122}<br>D) e^{\u22122}(1 + 2)<br><strong>Answer:<\/strong> A) <strong>1 \u2212 e^{\u22122}<\/strong><br><strong>Explanation:<\/strong> P(X=0)=e^{\u22122}. So P(X\u22651)=1\u2212P(0)=1\u2212e^{\u22122}.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">93. Random variable X has P(X=0)=0.2, P(X=1)=0.5, P(X=2)=0.3. E[X] =<\/mark><\/strong><br>A) 1.1<br>B) 1.0<br>C) 0.8<br>D) 1.5<br><strong>Answer:<\/strong> A) <strong>1.1<\/strong><br><strong>Explanation:<\/strong> E[X] = 0\u00b70.2 + 1\u00b70.5 + 2\u00b70.3 = 0 + 0.5 + 0.6 = 1.1.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">94. Two fair dice are rolled. Given that the sum is 8, probability that one die shows 2 is<\/mark><\/strong><br>A) 1\/5<br>B) 1\/4<br>C) 1\/3<br>D) 2\/5<br><strong>Answer:<\/strong> B) 1\/4<br><strong>Explanation:<\/strong> Outcomes summing to 8: (2,6),(3,5),(4,4),(5,3),(6,2) \u2192 5 equally likely outcomes. Ones with a 2 are (2,6),(6,2) = 2. Probability = 2\/5. Wait check: that&#8217;s 2\/5. So correct is <strong>D) 2\/5<\/strong>. (Oops \u2014 corrected).<br><strong>Final:<\/strong> D) <strong>2\/5<\/strong>.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">95. From integers 1 to 50 pick one at random. Probability it is divisible by both 2 and 5 = divisible by 10 =<\/mark><\/strong><br>A) 4\/25<br>B) 1\/5<br>C) 5\/50<br>D) 1\/10<br><strong>Answer:<\/strong> D) 1\/5<strong> <\/strong>(since multiples of 10 in 1..50 are 5 numbers: 10,20,30,40,50 \u2192 5\/50 = 1\/10).<br><strong>Explanation:<\/strong> Wait compute carefully: 5\/50 = 1\/10. So correct is <strong>D) 1\/10<\/strong>. (Ignore the first mistaken choice.)<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">96. A jar contains 3 red, 4 green, 5 blue marbles. One marble drawn at random. Probability it is green or blue =<\/mark><\/strong><br>A) 9\/12<br>B) 3\/12<br>C) 5\/12<br>D) 7\/12<br><strong>Answer:<\/strong> A) 9\/12 = 3\/4<br><strong>Explanation:<\/strong> Green+Blue = 4+5 = 9 out of total 12 \u2192 3\/4.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">97. Suppose X ~ Binomial(n=4, p=0.25). P(X=0) =<\/mark><\/strong><br>A) (0.75)^4<br>B) (0.25)^4<br>C) 4(0.25)(0.75)^3<br>D) 1 \u2212 (0.75)^4<br><strong>Answer:<\/strong> A) (0.75)^4 = 0.31640625<br><strong>Explanation:<\/strong> Probability of zero successes = (1\u2212p)^n.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">98. A password is a random permutation of the letters A,B,C,D (all distinct). Probability the password starts with A or ends with D =<\/mark><\/strong><br>A) 1\/2<br>B) 2\/4<br>C) 3\/4<br>D) 7\/24<br><strong>Answer:<\/strong> A) 1\/2<br><strong>Explanation:<\/strong> Total permutations = 4! = 24. Count with starts with A: 3! = 6. Count with ends with D: 3! = 6. Overlap (starts with A and ends with D): 2! = 2. By inclusion-exclusion: 6+6\u22122=10. Probability = 10\/24 = 5\/12 \u2248 0.4167. So correct choice from given is <strong>none of above<\/strong>; closest is not 1\/2. If forced pick, correct fraction 5\/12.<br><strong>Final (correct):<\/strong> <strong>5\/12<\/strong> (\u22480.4167).<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">99. A fair die is rolled 4 times. Probability that at least one 6 appears =<\/mark><\/strong><br>A) 1 \u2212 (5\/6)^4<br>B) (1\/6)\u2074<br>C) 4(1\/6)(5\/6)\u00b3<br>D) (5\/6)\u2074<br><strong>Answer:<\/strong> A) 1 \u2212 (5\/6)^4<br><strong>Explanation:<\/strong> Complement is no 6 in 4 rolls \u2192 (5\/6)^4.<\/p>\n\n\n\n<p class=\"has-large-font-size\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-luminous-vivid-orange-color\">100. Three people A, B, C each independently flip a fair coin. What is probability that exactly two people get the same result (i.e., two equal, the third different)?<\/mark><\/strong><br>A) 3\/4<br>B) 3\/8<br>C) 3\/4? (check)<br>D) 3\/4?<br><strong>Answer:<\/strong> B) <strong>3\/4<\/strong>? Wait compute carefully: Total outcomes = 8. Outcomes where exactly two same and one different: choose which person is the odd one (3 ways) and for each that odd person can be H while the other two T (1) or T while other two H (1) \u2192 for each choice 2 outcomes. So total favorable = 3\u00d72=6. Probability = 6\/8 = 3\/4.<br><strong>Final Answer:<\/strong> A) 3\/4.<\/p>\n\n\n","protected":false},"excerpt":{"rendered":"<p>1. What is the probability of getting a head when a fair coin is tossed?A) 0B) 1\/2C) 1D) 2\/3Answer: B) 1\/2Explanation: Two outcomes \u2192 {Head, Tail}. One favorable \u2192 Probability = 1\/2. 2. Probability of an impossible event is:A) 1B) 0C) 1\/2D) UndefinedAnswer: B) 0Explanation: Impossible event has no favorable outcomes. 3. Probability of a<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":"[]"},"categories":[3],"tags":[12983,12181,12975,12964,12978,12976,4029,5649,5623,13285,13292,13290,13278,13291,13279,13297,13280,13288,13284,13296,13283,13281,13289,13287,13282,13294,10950,13295,13293],"class_list":{"0":"post-12612","1":"post","2":"type-post","3":"status-publish","4":"format-standard","6":"category-mathematics","7":"tag-competitive-exams","8":"tag-exam-preparation","9":"tag-math-exercises","10":"tag-math-mcqs","11":"tag-math-practice","12":"tag-mathematics-questions","13":"tag-mcqs-adda","14":"tag-mcqs-for-pc-psi-sda-fda-pdo-vao-banking-kas-ias-ssc-gd-ssc-chsl-ssc-cgl-for-all-compitative-exams","15":"tag-mcqs-for-sda-fda-pdo-vao-banking-kas-ias-ssc-gd-ssc-chsl-ssc-cgl-for-all-compitative-exams","16":"tag-probability-examples","17":"tag-probability-exercises","18":"tag-probability-for-students","19":"tag-probability-formulas","20":"tag-probability-learning","21":"tag-probability-mcqs","22":"tag-probability-notes","23":"tag-probability-practice","24":"tag-probability-problems","25":"tag-probability-questions","26":"tag-probability-questions-with-answers","27":"tag-probability-quiz","28":"tag-probability-revision","29":"tag-probability-solutions","30":"tag-probability-study-material","31":"tag-probability-test","32":"tag-probability-tips","33":"tag-probability-top-100-mcqs-with-answer-and-explanation","34":"tag-probability-tricks","35":"tag-probability-tutorials"},"_links":{"self":[{"href":"https:\/\/mcqsadda.com\/index.php\/wp-json\/wp\/v2\/posts\/12612","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/mcqsadda.com\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mcqsadda.com\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mcqsadda.com\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mcqsadda.com\/index.php\/wp-json\/wp\/v2\/comments?post=12612"}],"version-history":[{"count":2,"href":"https:\/\/mcqsadda.com\/index.php\/wp-json\/wp\/v2\/posts\/12612\/revisions"}],"predecessor-version":[{"id":12730,"href":"https:\/\/mcqsadda.com\/index.php\/wp-json\/wp\/v2\/posts\/12612\/revisions\/12730"}],"wp:attachment":[{"href":"https:\/\/mcqsadda.com\/index.php\/wp-json\/wp\/v2\/media?parent=12612"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mcqsadda.com\/index.php\/wp-json\/wp\/v2\/categories?post=12612"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mcqsadda.com\/index.php\/wp-json\/wp\/v2\/tags?post=12612"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}