Free Topic-Wise General Studies MCQs
Master Probability concepts for the UPSC CSAT. Solve high-standard practice MCQs covering dice, cards, and conditional probability with thorough mathematical explanations.
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Explanation: Calculate P(maximum ≤ 4) − P(maximum ≤ 3). P(max ≤ 4) = (4/6)² = 16/36, since both dice must show 4 or less. P(max ≤ 3) = (3/6)² = 9/36. The difference is 16/36 − 9/36 = 7/36.
Explanation: Use conditional probability: P(sum > 9 | sum > 6) = P(sum > 9) / P(sum > 6). There are 21 outcomes with sum > 6 and 6 outcomes with sum > 9 (sums of 10, 11, or 12). Hence, the probability is 6/21 = 2/7.
Explanation: Choose 2 red balls from 8 and 2 blue balls from 4: C(8,2) × C(4,2) = 28 × 6 = 168. The total number of 4-ball combinations is C(12,4) = 495. The probability is 168/495 = 56/165.
Explanation: Apply the complement rule: P(at least one six) = 1 − P(no sixes). The probability of not getting a six on one die is 5/6. For four independent dice, P(no sixes) = (5/6)⁴ = 625/1296. Therefore, the required probability is 1 − 625/1296 = 671/1296.
Explanation: Apply the complement rule: subtract the two extreme cases (all heads and all tails) from the total. There are 2⁵ = 32 outcomes. P(at least one of each) = 1 − 1/32 − 1/32 = 30/32 = 15/16.
Explanation: For the product to be odd, every die must show an odd number (1, 3, or 5). The probability for one die is 3/6 = 1/2. Since the dice are independent, multiply: (1/2) × (1/2) × (1/2) = 1/8.
Explanation: Count the favorable cases: exactly two white and one black is C(4,2) × C(6,1) = 36; exactly three white is C(4,3) = 4. Total favorable = 40. Divide by C(10,3) = 120. The probability is 40/120 = 1/3.
Explanation: There are 4 aces and 16 ten-value cards (10, Jack, Queen, King in each suit). The number of favorable 2-card combinations is 4 × 16 = 64. Divide by the total 2-card hands, C(52,2) = 1326, to obtain 64/1326 = 32/663.
Explanation: After drawing the first card (any suit), 39 of the remaining 52 cards belong to a different suit. Because the first card is replaced, the second draw is independent and the probability is 39/52 = 3/4.
Explanation: There are 2⁴ = 16 equally likely outcomes. The number of ways to arrange exactly two heads among four tosses is C(4,2) = 6. The probability is 6/16 = 3/8.
Explanation: The sum is odd when there is an odd number of heads: exactly 1 head or exactly 3 heads. There are C(3,1) = 3 ways for one head and C(3,3) = 1 way for three heads, giving 4 favorable outcomes out of 8. The probability is 4/8 = 1/2.
Explanation: Use the binomial formula with p = 3/5. P(exactly 3 heads) = C(4,3) × (3/5)³ × (2/5) = 216/625. P(exactly 4 heads) = (3/5)⁴ = 81/625. Sum: 216/625 + 81/625 = 297/625.
Explanation: Count the favorable outcomes as permutations of 4 distinct values from 6: 6 × 5 × 4 × 3 = 360. The total number of outcomes is 6⁴ = 1296. The probability is 360/1296 = 5/18.
Explanation: Given the first is a tail, we need exactly two heads among the remaining four tosses. The number of ways to choose which two of the four are heads is C(4,2) = 6. With 2⁴ = 16 outcomes for the remaining tosses, the probability is 6/16 = 3/8.
Explanation: List ordered pairs where the second value exceeds the first by at least 3: (1,4), (1,5), (1,6), (2,5), (2,6), (3,6). There are 6 favorable outcomes out of 36, so the probability is 6/36 = 1/6.
Explanation: Enumerate all ordered triples (a,b,c) with a+b+c ≥ 15. The favorable cases are those with sums 15, 16, 17, and 18. Direct enumeration yields 20 such triples out of 216 total, giving 20/216 = 5/54.
Explanation: The probability the first card is an ace is 4/52. Given that, the probability the second is also an ace is 3/51. Multiply these dependent probabilities: (4/52) × (3/51) = 12/2652 = 1/221.
Explanation: Use the complement: 1 − P(no red). The number of ways to choose 2 non-red balls from the 11 blue and green balls is C(11,2) = 55. The total is C(15,2) = 105. Thus, 1 − 55/105 = 50/105 = 10/21.
Explanation: Count outcomes where exactly two dice match. Choose the repeated value in 6 ways, the distinct value in 5 ways, and the position of the odd die in 3 ways: 6 × 5 × 3 = 90. With 216 total outcomes, the probability is 90/216 = 5/12.
Explanation: Identify prime sums between 2 and 12: 2, 3, 5, 7, 11. Count the ways to obtain each: 1 way for 2, 2 ways for 3, 4 ways for 5, 6 ways for 7, and 2 ways for 11. Total favorable outcomes = 15 out of 36, giving 15/36 = 5/12.
Explanation: Condition on the color of the transferred ball. If a red ball is moved (probability 3/5), Urn B then has 2 red and 4 blue balls, so P(red|moved red) = 2/6. If a blue ball is moved (probability 2/5), Urn B has 1 red and 5 blue, so P(red|moved blue) = 1/6. By total probability: (3/5)×(2/6) + (2/5)×(1/6) = 4/15.
Explanation: Use the complement rule: P(at least one king) = 1 − P(no kings). The number of ways to choose 3 non-kings is C(48,3) = 17,296. The total number of 3-card hands is C(52,3) = 22,100. Thus, 1 − 17,296/22,100 = 4,804/22,100 = 1201/5525.
Explanation: Use the binomial probability formula with n = 3, k = 2, p = 6/10 = 3/5: C(3,2) × (3/5)² × (2/5) = 3 × (9/25) × (2/5) = 54/125.
Explanation: Choose one card from each of the four suits: C(13,1)⁴ = 13⁴ = 28,561. The total number of 4-card hands is C(52,4) = 270,725. The probability is 28,561/270,725 = 2197/20,825.
Explanation: Choose one ball from each colour: C(2,1) × C(3,1) × C(4,1) = 24. The total number of 3-ball combinations is C(9,3) = 84. The probability is 24/84 = 2/7.
Explanation: Choose 2 red balls from 5 and 2 blue balls from 5: C(5,2) × C(5,2) = 10 × 10 = 100. The total number of 4-ball combinations is C(10,4) = 210. The probability is 100/210 = 10/21.
Explanation: The number of ways to choose one red and one green is C(6,1) × C(4,1) = 24. The total number of ways to choose any 2 balls is C(10,2) = 45. The probability is 24/45 = 8/15.
Explanation: Coin tosses are independent events. The outcome of the first toss does not influence the second. Therefore, the conditional probability equals the unconditional probability of a head: 1/2.
Explanation: Fix the first and fourth coins to match (either both heads or both tails). The middle two coins can be anything, giving 2 × 2² = 8 favorable outcomes out of 16 total. The probability is 8/16 = 1/2.
Explanation: Examine all 16 outcomes. Those containing 'HH' but not 'HHH' are HHTT, HHTH, HTHH, THHT, and TTHH. There are 5 such outcomes, giving a probability of 5/16.
Explanation: After removing one red ball, the urn contains 4 red and 3 blue balls. The probability of drawing a blue ball next is 3/7.
Explanation: Choose the rank for the triple in 13 ways and the suits in C(4,3) = 4 ways. Choose a different rank for the pair in 12 ways and the suits in C(4,2) = 6 ways. Favorable hands = 13 × 4 × 12 × 6 = 3,744. Divide by C(52,5) = 2,598,960 to obtain 3,744/2,598,960 = 6/4,165.
Explanation: The sample space has 36 equally likely outcomes. Identify all pairs (a,b) where a+b ∈ {4,8,12}: (1,3), (2,2), (3,1), (2,6), (3,5), (4,4), (5,3), (6,2), (6,6) — nine outcomes. Probability = 9/36 = 1/4.
Explanation: For the first ace to appear on the third draw, the first two cards must be non-aces and the third must be an ace. The probability is (48/52) × (47/51) × (4/50) = 9,024/132,600 = 376/5,525.
Explanation: Given the card is red, the reduced sample space consists of 26 cards (13 hearts and 13 diamonds). Exactly 13 of these are diamonds, so the conditional probability is 13/26 = 1/2.
Explanation: Use the binomial probability formula: C(3,2) × (2/3)² × (1/3)¹ = 3 × (4/9) × (1/3) = 12/27 = 4/9.
Explanation: The total exceeds 2 only for the pairs RR (score 4) and RB (score 3). There are C(3,2) = 3 ways for RR and 3 × 2 = 6 ways for RB, giving 9 favorable outcomes. The total number of pairs is C(6,2) = 15. The probability is 9/15 = 3/5.
Explanation: Choose the suit in 4 ways and then choose 3 cards from that suit's 13 cards: 4 × C(13,3) = 4 × 286 = 1,144. The total number of 3-card hands is C(52,3) = 22,100. The probability is 1,144/22,100 = 286/5,525.
Explanation: Apply the inclusion-exclusion principle. There are 13 spades, 12 face cards, and 3 cards that are both (Jack, Queen, King of spades). The probability is (13 + 12 − 3)/52 = 22/52 = 11/26.
Explanation: For the first red to appear on the third draw, the first two must be blue and the third red. Because of replacement, the draws are independent: (7/10) × (7/10) × (3/10) = 147/1000.
Explanation: Count the number of ways to choose 5 cards from one suit: C(13,5) = 1287. With 4 suits, there are 4 × 1287 = 5148 flush hands. Divide by the total number of 5-card hands, C(52,5) = 2,598,960. The probability is 5148/2,598,960 = 33/16,660.
Explanation: Given exactly two heads, the reduced sample space is {HHT, HTH, THH}. In two of these three equally likely outcomes, the first coin is a head. The conditional probability is therefore 2/3.
Explanation: The sample space has 2³ = 8 equally likely outcomes. The favorable outcomes with at least two heads are HHT, HTH, THH, and HHH — four outcomes. The probability is 4/8 = 1/2.
Explanation: Use the complement: P(at least two same suit) = 1 − P(all three different suits). The probability the second card differs from the first is 39/51, and the third differs from both is 26/50. Thus, 1 − (39/51) × (26/50) = 1 − 169/425 = 256/425.
Explanation: Apply the law of total probability. P(Red) = P(choose A) × P(Red|A) + P(choose B) × P(Red|B) = (1/2) × (2/5) + (1/2) × (4/5) = 1/5 + 2/5 = 3/5.
Explanation: List all ordered pairs with absolute difference 2: (1,3), (2,4), (3,5), (4,6), (3,1), (4,2), (5,3), (6,4). There are 8 favorable outcomes out of 36 equally likely outcomes, yielding 8/36 = 2/9.
Explanation: Because the draws are with replacement, they are independent. The information about the second draw does not affect the first. The probability that the first ball is red is therefore 3/8.
Explanation: Use the gap method: for k heads among 5 tosses with no two adjacent, the count is C(5−k+1, k). Sum over k = 0, 1, 2, 3: C(6,0) + C(5,1) + C(4,2) + C(3,3) = 1 + 5 + 6 + 1 = 13. With 32 total outcomes, the probability is 13/32.
Explanation: List all 8 outcomes. Those containing at least one 'HH' substring are HHT, THH, and HHH. There are 3 favorable outcomes, so the probability is 3/8.
Explanation: Given both are face cards, the sample space reduces to C(12,2) = 66 equally likely pairs. There are 6 red face cards (Jack, Queen, King of hearts and diamonds), so C(6,2) = 15 favorable pairs. The probability is 15/66 = 5/22.
Explanation: Transform the problem by setting yi = xi − 1, so y1 + y2 + y3 = 3 with yi ≥ 0. The number of non-negative integer solutions is C(3+3−1, 3−1) = C(5,2) = 10. With 6³ = 216 total outcomes, the probability is 10/216 = 5/108.
Explanation: Count the favorable cases: all white is C(5,3) = 10, and all black is C(5,3) = 10. Total favorable = 20. Divide by C(10,3) = 120. The probability is 20/120 = 1/6.
Explanation: There are 40 non-face cards. The number of 5-card hands with no face cards is C(40,5) = 658,008. Divide by the total number of 5-card hands, C(52,5) = 2,598,960. The simplified probability is 658,008/2,598,960 = 2109/8330.
Explanation: Given both are even, the reduced sample space has 3 × 3 = 9 equally likely outcomes: {2,4,6} × {2,4,6}. The pairs summing to 8 are (2,6), (4,4), and (6,2). Thus, the probability is 3/9 = 1/3.
Explanation: Use the inclusion-exclusion principle: P(multiple of 3 or 4) = P(multiple of 3) + P(multiple of 4) − P(multiple of 12). There are 12 outcomes for multiples of 3, 9 for multiples of 4, and 1 for 12. Hence, (12 + 9 − 1)/36 = 20/36 = 5/9.
Explanation: By symmetry, P(first > second) equals P(second > first). The probability of a tie is 6/36 = 1/6. The remaining 30/36 = 5/6 is split equally between the two strict inequalities, giving (5/6) ÷ 2 = 5/12.
Explanation: Sum the probabilities of the three mutually exclusive cases: (1) only the first fair coin is heads: (1/2)×(1/2)×(1/3) = 1/12; (2) only the second fair coin is heads: 1/12; (3) only the biased coin is heads: (1/2)×(1/2)×(2/3) = 2/12. Total = 4/12 = 1/3.
Explanation: There are 4 kings, 26 red cards, and 2 cards that are both (King of hearts and King of diamonds). By inclusion-exclusion, the probability is (4 + 26 − 2)/52 = 28/52 = 7/13.
Explanation: By symmetry, P(more heads) = P(more tails). The probability of a tie (3 heads and 3 tails) is C(6,3)/64 = 20/64 = 5/16. The remaining probability 1 − 5/16 = 11/16 is split equally, giving 11/32.
Explanation: The total number of ways to choose 2 balls from 10 is C(10,2) = 45. The number of ways to choose 2 red balls from 5 is C(5,2) = 10. The probability is 10/45 = 2/9.