JEE Mathematics Probability MCQs Set 18

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

Explanation:
1. Sample space S = {HH, HT, TH, TT}, so n(S) = 4.
2. Favourable outcomes are {HT, TH}, so n(A) = 2.
3. P(A) = 2/4 = 1/2.

Teacher's Note:
1. List the outcomes in order before counting, so that HT and TH are treated as different outcomes. Always check that the outcomes are equally likely.

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

Explanation:
1. A non-leap year has 52 weeks and 1 extra day.
2. The extra day can be any of the 7 days of the week, all equally likely.
3. P(53 Fridays) = 1/7.

Teacher's Note:
1. Count only the extra days beyond complete weeks. A non-leap year has one extra day and a leap year has two.

Answer: (c) \( \frac{5}{7} \)

Explanation:
1. A leap year has 52 weeks and 2 consecutive extra days.
2. There are 7 equally likely pairs of consecutive days: (Sun,Mon), (Mon,Tue), ..., (Sat,Sun).
3. A leap year has 53 Sundays if the pair contains Sunday: (Sat,Sun) or (Sun,Mon), that is 2 pairs.
4. P(only 52 Sundays) = 1 - 2/7 = 5/7.

Teacher's Note:
1. Use the complement when the required case contains most of the outcomes. Remember that two consecutive days are involved in a leap year.

Answer: (c) \( \frac{11}{36} \)

Explanation:
1. A = {(5,6),(6,5)}, so P(A) = 2/36.
2. B: each die shows 1, 3 or 5, so P(B) = (3/6)(3/6) = 9/36.
3. A and B are mutually exclusive because a sum of 11 needs one even number.
4. P(A or B) = 2/36 + 9/36 = 11/36.

Teacher's Note:
1. Check whether the events overlap before adding probabilities. If they are mutually exclusive, the addition is direct.

Answer: (c) \( \frac{1}{14} \)

Explanation:
1. Product zero: 00, 01, ..., 09 (10 tickets) and 10, 20, 30, 40 (4 tickets), so n = 14.
2. Among these, digit sum 8 occurs only for 08.
3. Required probability = 1/14.

Teacher's Note:
1. In conditional probability the given event becomes the new sample space. Count only inside it.

Answer: (d) \( \frac{1}{5} \)

Explanation:
1. Maximum 6: 6 is chosen and the other two come from {1,...,5}, so there are \( {}^5C_2 = 10 \) cases.
2. Minimum 3 as well: the other two numbers are 3 and one of {4, 5}, giving (3,4,6) and (3,5,6), that is 2 cases.
3. Required probability = 2/10 = 1/5.

Teacher's Note:
1. Fix the larger condition first and then count the cases of the second event inside it. Do not count the cases from the whole sample space.

Answer: (b) \( \frac{11}{28} \)

Explanation:
1. P = (12/16)(11/15)(10/14).
2. = 1320/3360 = 11/28.

Teacher's Note:
1. Without replacement the numerator and denominator both reduce by one at each step. This is the same as \( {}^{12}C_3/{}^{16}C_3 \).

Answer: (a) \( P(\bar E)-P(F) \)

Explanation:
1. \( P(\bar E\cap\bar F\cap G)=P(G)-P(G\cap E)-P(G\cap F)+P(E\cap F\cap G) \).
2. Using pairwise independence and \( P(E\cap F\cap G)=0 \): \( = P(G)[1-P(E)-P(F)] \).
3. Dividing by P(G): \( 1-P(E)-P(F)=P(\bar E)-P(F) \).

Teacher's Note:
1. Write the complement intersection by inclusion-exclusion inside G. Pairwise independence lets each pair factor out P(G).

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

Explanation:
1. Let A1: six occurs, A2: six does not occur; E: he reports six.
2. P(A1) = 1/6, P(E|A1) = 2/3; P(A2) = 5/6, P(E|A2) = 1/3.
3. Bayes: P(A1|E) = (1/6)(2/3) / [(1/6)(2/3) + (5/6)(1/3)] = 2/7.

Teacher's Note:
1. Bayes' theorem reverses the conditioning. Multiply along each branch ending in the report and divide by the total.

Answer: (b) \( \frac{17}{40} \)

Explanation:
1. If a white ball is transferred (probability 3/8), B has 7 white out of 15.
2. If a black ball is transferred (probability 5/8), B has 6 white out of 15.
3. P = (3/8)(7/15) + (5/8)(6/15) = 21/120 + 30/120 = 51/120 = 17/40.

Teacher's Note:
1. This is a total probability problem: split by what was transferred. Remember that urn B has 15 balls after the transfer.

Answer: (a) \( \frac{1}{4} \)

Explanation:
1. A point is nearer to the centre if its distance from the centre is less than a/2.
2. Favourable area = \( \pi (a/2)^2 = \frac{\pi a^2}{4} \).
3. Total area = \( \pi a^2 \), so P = 1/4.

Teacher's Note:
1. The distance to the circumference is a - r, so the dividing circle has radius a/2. In geometric probability use the ratio of areas.

Answer: (b) \( \frac{5}{32} \)

Explanation:
1. Count the sequences of 7 tosses containing HHHH.
2. Listing by position of the first run of four heads gives 20 favourable sequences out of 128.
3. P = 20/128 = 5/32.

Teacher's Note:
1. For consecutive-run problems, listing by the starting position of the first run avoids double counting. The formula (n+2)/2^(m+1) with n = 3, m = 4 gives the same value.

Answer: (b) \( \frac{7}{256} \)

Explanation:
1. The run of 6 heads must be bounded by tails (or the ends).
2. Run at the start: HHHHHH T followed by 3 free tosses: 8 sequences.
3. Run at the end: 3 free tosses then T HHHHHH: 8 sequences.
4. Run in the middle (starting at position 2, 3 or 4): T, 6 heads, T and 2 free tosses: 3 x 4 = 12 sequences.
5. Total 28 out of 1024 = 7/256.

Teacher's Note:
1. Treat the two ends separately from the middle positions, because a tail is needed only on the inner side. Check small cases by listing when in doubt.

Answer: (d) \( \frac{1}{6} \)

Explanation:
1. Sum 10: 3 ways, sum 11: 2 ways, sum 12: 1 way, total 6.
2. P = 6/36 = 1/6.

Teacher's Note:
1. For two dice the number of ways to get sum k is k - 1 for k up to 7 and 13 - k afterwards. Add the ways for each sum required.

Answer: (b) \( \frac{5}{8} \)

Explanation:
1. The number of outcomes with sum at least 10 is 27 + 27 + 25 + 21 + 15 + 10 + 6 + 3 + 1 = 135.
2. P = 135/216 = 5/8.

Teacher's Note:
1. The number of ways to get sum k with three dice is the coefficient of x^k in (x + x² + ... + x^6)³. Add the counts for sums 10 to 18.

Answer: (b) \( \frac{4}{7} \)

Explanation:
1. p = 1/2, q = 1/2.
2. P(A wins) = p/(1 - q³) = (1/2)/(1 - 1/8) = 4/7.

Teacher's Note:
1. For a game with k players taking turns, the first player wins with probability p/(1 - q^k). The three winning probabilities are in the ratio 1 : q : q².

Answer: (c) \( \frac{0}{1} \)

Explanation:
1. If two letters are in correct envelopes the third has only one envelope left, which is its own.
2. So exactly one wrong placement is impossible.
3. P = 0.

Teacher's Note:
1. If n - 1 letters are in the right envelopes, the last one is forced into the right envelope too. Hence exactly one wrong is impossible.

Answer: (d) \( \frac{1}{3} \)

Explanation:
1. Number of derangements of 3 objects = 3!(1 - 1 + 1/2 - 1/6) = 2.
2. P = 2/6 = 1/3.

Teacher's Note:
1. Derangements of 3, 4, 5 objects are 2, 9, 44. Divide by n! for the probability.

Answer: (b) \( \frac{1}{12} \)

Explanation:
1. The vowels are O, A, I, I (4 letters with I repeated).
2. Number of distinct orders of the vowels = 4!/2! = 12.
3. Only one of these 12 orders is the original, so P = 1/12.

Teacher's Note:
1. The vowels occupy fixed positions among themselves, so only their relative order matters. The repetition of I must be divided out.

Answer: (d) \( \frac{1}{330} \)

Explanation:
1. Vowels: O, A, I, I (4!/2! arrangements). Consonants: P, R, B, B, L, T, Y (7!/2! arrangements).
2. Total arrangements = 11!/(2! 2!).
3. P = (4!/2!)(7!/2!) / (11!/(2!2!)) = 4!7!/11! = 1/330.

Teacher's Note:
1. Keep vowel positions and consonant positions fixed and count the arrangements inside each group. Divide by the total number of distinct arrangements.

Answer: (c) \( \frac{{}^{10}C_6\times 2^6}{{}^{20}C_6} \)

Explanation:
1. Total ways = \( {}^{20}C_6 \).
2. Choose 6 different pairs: \( {}^{10}C_6 \), and one shoe from each: \( 2^6 \).
3. P = \( \frac{{}^{10}C_6\times 2^6}{{}^{20}C_6} \).

Teacher's Note:
1. To avoid a pair, pick the pairs first and then one shoe from each chosen pair. The complement gives the probability of at least one pair.

Answer: (a) \( \frac{42}{323} \)

Explanation:
1. Choose 2 complete pairs: \( {}^{10}C_2 \).
2. The other 2 shoes come from 2 different pairs among the remaining 8, one shoe from each: \( {}^8C_2\times 2^2 \).
3. P = 45 x 28 x 4 / 38760 = 5040/38760 = 42/323.

Teacher's Note:
1. Exactly two pairs means the other two shoes must not form a pair. Count them as one shoe from each of two different pairs.

Answer: (c) \( \left(\frac78\right)^5 \)

Explanation:
1. For each element there are \( 2^3 = 8 \) ways of belonging or not belonging to the three subsets.
2. The element is in all three in only 1 way, so 7 of 8 ways are favourable.
3. The 5 elements are independent: P = (7/8)^5.

Teacher's Note:
1. Treat each element separately. An element belongs to the common part only if it is in all three subsets.

Answer: (b) \( \left(\frac34\right)^n \)

Explanation:
1. Each element has 4 choices: in both, in P only, in Q only, in neither.
2. Only 'in both' is not allowed, so 3 choices per element are favourable.
3. P = 3^n/4^n.

Teacher's Note:
1. Count choices element by element, not subset by subset. The total number of ways to choose (P, Q) is 4^n.

Answer: (b) \( \frac{81}{10000} \)

Explanation:
1. Choose the 3 biscuits that go to the particular beggar: \( {}^5C_3 = 10 \).
2. The other 2 biscuits go to the other 9 beggars: \( 9^2 = 81 \).
3. Total ways = 10^5. P = 810/100000 = 81/10000.

Teacher's Note:
1. Each biscuit is a separate trial with 10 possible receivers. This is the binomial pattern with probability 1/10.