JEE Mathematics Probability MCQs Set 21

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Multiple Choice Questions

Question 1: Two symmetrical dice are thrown. The probability of throwing a doublet such that their sum is greater than 9 is
(a) \( \frac{1}{36} \)
(b) \( \frac{1}{72} \)
(c) \( \frac{1}{18} \)
(d) \( \frac{1}{9} \)
Show Answer & Explanation

Answer: (c) \( \frac{1}{18} \)

Explanation:
1. Doublets with sum greater than 9: (5,5) and (6,6).
2. P = 2/36 = 1/18.

Teacher's Note:
1. List doublets first, then apply the sum condition.

Question 2: Two symmetrical dice are rolled. The probability that the sum of the points on them is divisible by 5 is
(a) \( \frac{7}{36} \)
(b) \( \frac{2}{9} \)
(c) \( \frac{4}{9} \)
(d) \( \frac{5}{9} \)
Show Answer & Explanation

Answer: (a) \( \frac{7}{36} \)

Explanation:
1. Sum 5: 4 ways; sum 10: 3 ways.
2. P = 7/36.

Teacher's Note:
1. The possible sums are 2 to 12, so only 5 and 10 are divisible by 5.

Question 3: Two unbiased dice with faces numbered 1 to 6 are thrown. Let \( E_k=\{(a,b): ab=k\} \) and \( p_k=P(E_k) \). Which of the following is correct?
(a) \( p_1<p_{11}<p_4<p_6 \)
(b) \( p_{36}<p_{11}<p_6<p_4 \)
(c) \( p_{36}<p_6<p_2<p_4 \)
(d) \( p_1<p_{30}<p_4<p_6 \)
Show Answer & Explanation

Answer: (d) \( p_1<p_{30}<p_4<p_6 \)

Explanation:
1. Number of outcomes: ab = 1: (1,1): 1; ab = 30: (5,6),(6,5): 2; ab = 4: (1,4),(4,1),(2,2): 3; ab = 6: (1,6),(6,1),(2,3),(3,2): 4.
2. Also p_36 has 1 outcome, p_2 has 2 and p_11 has 0.
3. So p_1 < p_30 < p_4 < p_6 is correct; options 2, 3, 4 each contain a false inequality.

Teacher's Note:
1. Count the factorisations of k with both factors from 1 to 6. A prime above 6 such as 11 has probability 0.

Question 4: A coin and a six-faced die, both unbiased, are thrown simultaneously. The probability of getting a head on the coin and an odd number on the die is
(a) 3/4
(b) 1/4
(c) 1/2
(d) 2/3
Show Answer & Explanation

Answer: (b) 1/4

Explanation:
1. P = (1/2)(3/6) = 1/4.

Teacher's Note:
1. The coin and die are independent, so multiply the two probabilities.

Question 5: The six faces of an unbiased die are numbered 2, 3, 5, 7, 11 and 13. If two such dice are thrown, the probability that the sum on the uppermost faces is an odd number is
(a) 25/36
(b) 13/18
(c) 5/36
(d) 5/18
Show Answer & Explanation

Answer: (d) 5/18

Explanation:
1. Five faces are odd and only 2 is even.
2. The sum is odd when one die shows the even face and the other an odd face.
3. P = 2 x (1/6)(5/6) = 10/36 = 5/18.

Teacher's Note:
1. Odd plus even gives odd, and odd plus odd or even plus even gives even. Count the mixed outcomes in both orders.

Question 6: The probability that a leap year will have 53 Sundays and 53 Mondays is
(a) \( \frac{4}{7} \)
(b) \( \frac{2}{7} \)
(c) \( \frac{3}{7} \)
(d) \( \frac{1}{7} \)
Show Answer & Explanation

Answer: (d) \( \frac{1}{7} \)

Explanation:
1. The two extra days must be Sunday and Monday, in that order.
2. Only 1 of the 7 equally likely pairs of consecutive days qualifies.
3. P = 1/7.

Teacher's Note:
1. In a leap year the two extra days are consecutive. Each of the 7 possible pairs is equally likely.

Question 7: The probability that the February of a leap year will have 5 Saturdays is
(a) \( \frac{2}{7} \)
(b) \( \frac{4}{7} \)
(c) \( \frac{3}{7} \)
(d) \( \frac{1}{7} \)
Show Answer & Explanation

Answer: (d) \( \frac{1}{7} \)

Explanation:
1. February of a leap year has 29 days = 4 weeks + 1 day.
2. The extra day must be a Saturday: 1 of 7.
3. P = 1/7.

Teacher's Note:
1. Only the 29th day of February can create a fifth occurrence of a weekday.

Question 8: The probability that the 13th day of a randomly chosen month is a Friday is, by the usual simplification (treating the weekday as equally likely among the 7 days),
(a) \( \frac{1}{7} \)
(b) \( \frac{1}{84} \)
(c) \( \frac{1}{42} \)
(d) \( \frac{1}{12} \)
Show Answer & Explanation

Answer: (a) \( \frac{1}{7} \)

Explanation:
1. Each of the 7 weekdays is equally likely for any given date.
2. P = 1/7.

Teacher's Note:
1. In an exam such questions assume the 13th falls on each weekday with equal chance. The month is irrelevant.

Question 9: In a non-leap year, the probability of getting 53 Sundays or 53 Tuesdays or 53 Thursdays is
(a) \( \frac{3}{7} \)
(b) \( \frac{2}{7} \)
(c) \( \frac{1}{7} \)
(d) \( \frac{4}{7} \)
Show Answer & Explanation

Answer: (a) \( \frac{3}{7} \)

Explanation:
1. A non-leap year has one extra day.
2. It must be Sunday, Tuesday or Thursday: 3 of the 7 days.
3. P = 3/7.

Teacher's Note:
1. Only one extra day is available, so the three events are mutually exclusive. Add their probabilities 1/7 each.

Question 10: It is given that a leap year has 53 Sundays. The probability that it has 53 Mondays is
(a) \( \frac{1}{2} \)
(b) \( \frac{1}{7} \)
(c) \( \frac{6}{7} \)
(d) \( \frac{2}{7} \)
Show Answer & Explanation

Answer: (a) \( \frac{1}{2} \)

Explanation:
1. The extra pair contains Sunday: (Sat,Sun) or (Sun,Mon).
2. 53 Mondays needs (Sun,Mon): 1 of these 2 pairs.
3. P = 1/2.

Teacher's Note:
1. Reduce the sample space to the pairs containing Sunday. Then count the pairs that also include Monday.

Question 11: A pack of 52 cards is distributed equally among four players. The probability that a particular hand has 5 spades, 3 clubs, 3 hearts and the remaining diamonds is
(a) \( \frac{{}^4C_1\cdot{}^{13}C_5\cdot{}^{13}C_3\cdot{}^{13}C_3\cdot{}^{13}C_2}{{}^{52}C_{13}} \)
(b) \( \frac{{}^4C_1\cdot{}^{13}C_3\cdot{}^{13}C_3\cdot{}^{13}C_3\cdot{}^{13}C_2}{{}^{52}C_{13}} \)
(c) \( \frac{{}^{13}C_5\cdot{}^{13}C_3\cdot{}^{13}C_3\cdot{}^{13}C_2}{{}^{52}C_{13}} \)
(d) \( \frac{{}^{13}C_5\cdot{}^{26}C_6\cdot{}^{13}C_2}{{}^{52}C_{13}} \)
Show Answer & Explanation

Answer: (c) \( \frac{{}^{13}C_5\cdot{}^{13}C_3\cdot{}^{13}C_3\cdot{}^{13}C_2}{{}^{52}C_{13}} \)

Explanation:
1. The hand has 13 cards: 5 + 3 + 3 + 2 = 13, so the remaining are 2 diamonds.
2. Choose from each suit: \( {}^{13}C_5\cdot{}^{13}C_3\cdot{}^{13}C_3\cdot{}^{13}C_2 \).
3. Total ways \( {}^{52}C_{13} \).

Teacher's Note:
1. The suits are specified in order, so no extra permutation factor is needed. Check that the counts add to 13.

Question 12: A card is drawn from an ordinary pack of 52 playing cards and a gambler bets that it is a spade or an ace. The probability that he wins the bet is
(a) \( \frac{4}{13} \)
(b) \( \frac{1}{13} \)
(c) \( \frac{2}{13} \)
(d) \( \frac{3}{13} \)
Show Answer & Explanation

Answer: (a) \( \frac{4}{13} \)

Explanation:
1. Spades 13 plus aces 4 minus the ace of spades counted twice = 16.
2. P = 16/52 = 4/13.

Teacher's Note:
1. Use inclusion-exclusion because the ace of spades belongs to both groups.

Question 13: In shuffling a pack of cards, four cards are accidentally dropped. The probability that the dropped cards are one from each suit is
(a) \( \frac{{}^{13}C_4}{{}^{52}C_4} \)
(b) \( \frac{13!}{{}^{52}C_4} \)
(c) \( \frac{{}^{13}C_4}{13^4} \)
(d) \( \frac{13^4}{{}^{52}C_4} \)
Show Answer & Explanation

Answer: (d) \( \frac{13^4}{{}^{52}C_4} \)

Explanation:
1. Choose one card from each of the 4 suits: 13^4 ways.
2. Total \( {}^{52}C_4 \) ways.

Teacher's Note:
1. The four suits give four independent choices of 13 cards each. Do not multiply by 4! because the cards are not arranged.

Question 14: The face cards are removed from a well-shuffled pack of 52 cards. Out of the remaining cards, 4 are drawn at random. The probability that they belong to different suits is
(a) \( \frac{13^4}{{}^{40}C_4} \)
(b) \( \frac{10^4}{{}^{40}C_4} \)
(c) \( \frac{13^4}{{}^{52}C_4} \)
(d) \( \frac{{}^{13}C_4}{{}^{40}C_4} \)
Show Answer & Explanation

Answer: (b) \( \frac{10^4}{{}^{40}C_4} \)

Explanation:
1. Removing the 12 face cards leaves 40 cards, 10 in each suit.
2. One card from each suit: 10^4 ways.
3. P = 10^4/\( {}^{40}C_4 \).

Teacher's Note:
1. After removing the face cards each suit has 10 cards, not 13.

Question 15: In a game of bridge, the probability that a particular player has all the 13 cards of red colour is
(a) \( \frac{13\times4}{{}^{52}C_{13}} \)
(b) \( \frac{13^4}{{}^{52}C_{13}} \)
(c) \( \frac{{}^{26}C_{13}}{{}^{52}C_{13}} \)
(d) \( \frac{13^2}{{}^{52}C_{13}} \)
Show Answer & Explanation

Answer: (c) \( \frac{{}^{26}C_{13}}{{}^{52}C_{13}} \)

Explanation:
1. There are 26 red cards.
2. The favourable hands choose 13 of them: \( {}^{26}C_{13} \).
3. P = \( {}^{26}C_{13}/{}^{52}C_{13} \).

Teacher's Note:
1. Treat the hand as a selection of 13 cards from 52. Red cards number 26.

Question 16: From a pack of cards, 2 cards are chosen at random. The probability that one card is a 10 which is not a heart and the other is a heart is
(a) \( \frac{8}{663} \)
(b) \( \frac{33}{34} \)
(c) \( \frac{1}{102} \)
(d) \( \frac{1}{34} \)
Show Answer & Explanation

Answer: (d) \( \frac{1}{34} \)

Explanation:
1. The 10 can be any of 3 tens other than hearts and the other card any of 13 hearts.
2. Favourable = 3 x 13 = 39.
3. P = 39/1326 = 1/34.

Teacher's Note:
1. Count the two cards as an unordered pair: one from each group. The groups do not overlap.

Question 17: A person draws 2 cards from a well-shuffled pack of cards. The cards are then replaced and another person draws 2 cards after shuffling. The probability that there is exactly 1 common card is
(a) \( \frac{75}{663} \)
(b) \( \frac{100}{663} \)
(c) \( \frac{25}{663} \)
(d) \( \frac{50}{663} \)
Show Answer & Explanation

Answer: (d) \( \frac{50}{663} \)

Explanation:
1. Let the first draw be a fixed pair. The second pair must contain exactly one of those 2 cards and one of the other 50.
2. Favourable = 2 x 50 = 100.
3. P = 100/\( {}^{52}C_2 \) = 100/1326 = 50/663.

Teacher's Note:
1. Fix the first pair and count the second pairs by overlap. Exactly one common card means one card from the pair and one from outside.

Question 18: Seven white balls and three black balls are randomly arranged in a row. The probability that no two black balls are placed adjacently is
(a) \( \frac{1}{3} \)
(b) \( \frac{7}{15} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{2}{15} \)
Show Answer & Explanation

Answer: (b) \( \frac{7}{15} \)

Explanation:
1. Arrange the 7 white balls, creating 8 gaps.
2. Choose 3 gaps for the black balls: \( {}^8C_3 = 56 \).
3. Total arrangements \( {}^{10}C_3 = 120 \).
4. P = 56/120 = 7/15.

Teacher's Note:
1. The gap method places the restricted items in the spaces between the others. Treat balls of the same colour as identical.

Question 19: A box contains 40 balls of the same shape and weight. Among them 10 are white, 16 are red and the rest are black. If two balls are drawn, the probability that one is red and one is black is
(a) 3/4
(b) 56/195
(c) 1
(d) 1/4
Show Answer & Explanation

Answer: (b) 56/195

Explanation:
1. Black balls = 40 - 26 = 14.
2. Favourable = 16 x 14 = 224.
3. Total \( {}^{40}C_2 = 780 \).
4. P = 224/780 = 56/195.

Teacher's Note:
1. Count the black balls from the remaining ones. Choose one from each colour group.

Question 20: An urn contains nine balls, of which three are red, four are blue and two are green. Three balls are drawn at random without replacement. The probability that the three balls have different colours is
(a) \( \frac{1}{3} \)
(b) \( \frac{2}{7} \)
(c) \( \frac{2}{23} \)
(d) \( \frac{1}{21} \)
Show Answer & Explanation

Answer: (b) \( \frac{2}{7} \)

Explanation:
1. One of each colour: 3 x 4 x 2 = 24 ways.
2. Total \( {}^9C_3 = 84 \).
3. P = 24/84 = 2/7.

Teacher's Note:
1. One ball from each colour class can be chosen independently. Multiply the class sizes.

Question 21: A bag contains 4 red, 3 black and 5 white balls. Three balls are drawn at random. The probability of drawing 2 balls of the same colour and one of a different colour is
(a) \( \frac{15}{44} \)
(b) \( \frac{131}{195} \)
(c) \( \frac{5}{44} \)
(d) \( \frac{29}{44} \)
Show Answer & Explanation

Answer: (d) \( \frac{29}{44} \)

Explanation:
1. Red pair: \( {}^4C_2\times8 = 48 \); black pair: \( {}^3C_2\times9=27 \); white pair: \( {}^5C_2\times7=70 \).
2. Favourable = 145. Total \( {}^{12}C_3 = 220 \).
3. P = 145/220 = 29/44.

Teacher's Note:
1. The third ball must be of another colour, so it comes from the balls of the other two colours. Count each colour pair separately.

Question 22: From a bag containing 4 white and 5 black balls, 3 balls are drawn at random. The odds against these being all black balls are
(a) 37 to 5
(b) 5 to 42
(c) 42 to 5
(d) 5 to 37
Show Answer & Explanation

Answer: (a) 37 to 5

Explanation:
1. P(all black) = \( {}^5C_3/{}^9C_3 = 10/84 = 5/42 \).
2. Odds against = (1 - 5/42) : 5/42 = 37 : 5.

Teacher's Note:
1. Odds against equal the probability of failure to the probability of success. Convert to a ratio of whole numbers.

Question 23: Box X contains 2 white and 3 black balls and bag Y contains 4 white and 2 black balls. One bag is selected at random and a ball is drawn from it. The probability that the ball is white is
(a) 2/15
(b) 7/15
(c) 14/15
(d) 8/15
Show Answer & Explanation

Answer: (d) 8/15

Explanation:
1. P = (1/2)(2/5) + (1/2)(4/6) = 1/5 + 1/3 = 8/15.

Teacher's Note:
1. Use the total probability rule over the two containers. Each has chance 1/2 of being chosen.

Question 24: Box A contains 3 red and 2 black balls and box B contains 2 red and 3 black balls. One ball is drawn at random from A and placed in B. Then one ball is drawn at random from B and placed in A. The probability that the compositions of balls in the two boxes remain unaltered is
(a) \( \frac{4}{15} \)
(b) \( \frac{9}{30} \)
(c) \( \frac{16}{30} \)
(d) \( \frac{17}{30} \)
Show Answer & Explanation

Answer: (d) \( \frac{17}{30} \)

Explanation:
1. Same colour is returned.
2. Red then red: (3/5)(3/6) = 9/30.
3. Black then black: (2/5)(4/6) = 8/30.
4. P = 17/30.

Teacher's Note:
1. After a transfer box B has 6 balls. The colours of the two balls must match for the boxes to be unchanged.

Question 25: Three 1 x 1 squares of a chess board (8 x 8) are chosen at random. The chance that all three are white is
(a) \( \frac{{}^{32}C_3}{{}^{64}C_3} \)
(b) \( \frac{{}^{16}C_3}{{}^{64}C_3} \)
(c) \( \frac{{}^{4}C_3}{{}^{64}C_3} \)
(d) \( \frac{{}^{8}C_3}{{}^{64}C_3} \)
Show Answer & Explanation

Answer: (a) \( \frac{{}^{32}C_3}{{}^{64}C_3} \)

Explanation:
1. A chess board has 32 white squares.
2. P = \( {}^{32}C_3/{}^{64}C_3 \).

Teacher's Note:
1. Half of the 64 squares are white.

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