JEE Mathematics Probability MCQs Set 24

Answer: (b) \( \frac{(p+1)m+pn}{(m+n)(p+q+1)} \)

Explanation:
1. If the moved ball is white (probability \( \frac{m}{m+n} \)), urn II has \( p+1 \) white balls among \( p+q+1 \).
2. If it is black (probability \( \frac{n}{m+n} \)), urn II has \( p \) white balls among \( p+q+1 \).
3. Total probability \( =\frac{m}{m+n}\cdot\frac{p+1}{p+q+1}+\frac{n}{m+n}\cdot\frac{p}{p+q+1}=\frac{(p+1)m+pn}{(m+n)(p+q+1)} \).

Teacher's Note:
1. Urn II always holds p + q + 1 balls after the transfer.
2. Split on the colour of the transferred ball.

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

Explanation:
1. If the fourth card is the eight of diamonds (probability \( \frac{1}{4} \)), the top card is a diamond with probability \( \frac{12}{51} \).
2. Otherwise (probability \( \frac{3}{4} \)) it is a diamond with probability \( \frac{13}{51} \).
3. Total \( =\frac14\cdot\frac{12}{51}+\frac34\cdot\frac{13}{51}=\frac{51}{204}=\frac14 \).

Teacher's Note:
1. Condition on the suit of the eight.
2. The answer equals the unconditional chance of a diamond.

Answer: (b) \( \frac{3}{32} \)

Explanation:
1. Count the 9-bit strings containing HHHHH by listing the position of the first run of five heads.
2. Run at positions 1-5: 16 strings; first run at positions 2-6 (preceded by a tail): 8; 3-7: 8; 4-8: 8; 5-9: 8. Total 48.
3. Probability \( =\frac{48}{512}=\frac{3}{32} \).

Teacher's Note:
1. A run starting after position 1 must be preceded by a tail to avoid double counting.
2. Total outcomes are 512.

Answer: (b) \( \frac{32}{41} \)

Explanation:
1. Two-headed coin chosen: probability \( \frac{1}{10} \), five heads certain.
2. Ordinary coin: probability \( \frac{9}{10} \), five heads with probability \( \frac1{32} \).
3. Bayes: \( \frac{\frac1{10}}{\frac1{10}+\frac9{10}\cdot\frac1{32}}=\frac{32}{41} \).

Teacher's Note:
1. Use Bayes' theorem with two hypotheses.
2. Compare 32 : 9 as odds.

Answer: (c) Rs. 240

Explanation:
1. Exactly one hit: A hits and B misses = 0.8 x 0.6 = 0.48; B hits and A misses = 0.2 x 0.4 = 0.08.
2. Given exactly one hit, P(A fired it) = 0.48 / 0.56 = 6/7.
3. A's share = (6/7) x 280 = Rs. 240.

Teacher's Note:
1. Condition on the event 'exactly one hole'.
2. Divide in the ratio 0.48 : 0.08 = 6 : 1.

Answer: (c) 0.127

Explanation:
1. Good numbers: 26, 62, 34, 43, so P(good) = 4/90 = 2/45.
2. P(none good in three draws) = \( \left(\frac{43}{45}\right)^3\approx0.8725 \).
3. Required probability = 1 - 0.8725 = 0.1275, which is 0.127 to three decimals.

Teacher's Note:
1. List all two-digit numbers whose digits multiply to 12.
2. Use the complement.

Answer: (d) 82

Explanation:
1. The digits sum to 45 minus the two left-out digits, so divisibility by 9 needs the omitted pair to have a sum divisible by 9, that is, sum 9.
2. Such pairs: (1,8), (2,7), (3,6), (4,5): 4 of the \( \binom92=36 \) choices.
3. So \( \frac pq=\frac{4}{36}=\frac19 \) and \( p^2+q^2=1+81=82 \).

Teacher's Note:
1. Total of 1 to 9 is 45, a multiple of 9.
2. Order of the digits does not affect divisibility.

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

Explanation:
1. Let z be Z's probability. Exactly two hits: \( \frac23\cdot\frac34(1-z)+\frac23\cdot\frac14z+\frac13\cdot\frac34z=\frac{11}{24} \).
2. This gives \( \frac12(1-z)+\frac16z+\frac14z=\frac{11}{24} \), that is \( \frac12-\frac{z}{12}=\frac{11}{24} \).
3. So \( z=\frac12 \).

Teacher's Note:
1. Add the three ways in which exactly two of the three can hit.
2. The equation is linear in z.

Answer: (d) 54

Explanation:
1. P(heart) = P(diamond) = 1/4 and P(black) = 1/2.
2. Probability \( =\frac{6!}{2!\,2!\,2!}\left(\frac14\right)^2\left(\frac14\right)^2\left(\frac12\right)^2=\frac{90}{1024}=\frac{45}{512} \).
3. Then \( \frac{pq}{2}=\frac{45\times512}{2}=11520=2^8\cdot3^2\cdot5 \), which has \( 9\times3\times2=54 \) divisors.

Teacher's Note:
1. Use the multinomial arrangement 6!/(2! 2! 2!).
2. Number of divisors = product of (exponent + 1).

Answer: (c) 6

Explanation:
1. Only the relative order of the three volumes matters.
2. Of their 3! = 6 equally likely orders exactly one is increasing, so the probability is 1/6.
3. The reciprocal is 6.

Teacher's Note:
1. The other 37 books do not affect the relative order.
2. All 6 orders are equally likely.

Answer: (d) \( \frac{569}{1496} \)

Explanation:
1. The selection probabilities are 2/34, 5/34, 10/34, 17/34, which add to 1.
2. P(white) \( =\frac{2}{34}\cdot\frac12+\frac5{34}\cdot\frac25+\frac{10}{34}\cdot\frac38+\frac{17}{34}\cdot\frac4{11} \).
3. This equals \( \frac{1}{34}\left(1+2+\frac{15}{4}+\frac{68}{11}\right)=\frac{569}{1496} \).

Teacher's Note:
1. Use the total probability theorem over the four urns.
2. Keep the selection weights unequal.

Answer: (c) \( \frac{14}{41} \)

Explanation:
1. P(says high) = 0.3 x 0.9 + 0.7 x 0.2 = 0.27 + 0.14 = 0.41.
2. P(low grade and says high) = 0.7 x 0.2 = 0.14.
3. Bayes: 0.14 / 0.41 = 14/41.

Teacher's Note:
1. The 'prior' proportions are 3 : 7.
2. Compute the total chance that he says 'high grade' first.

Answer: (c) \( \frac{11}{16} \)

Explanation:
1. Ruin on the first toss: T, probability 1/2.
2. Ruin on the third toss: HTT, probability 1/8.
3. Ruin on the fifth toss: HHTTT or HTHTT, probability 2/32 = 1/16.
4. Total \( \frac12+\frac18+\frac1{16}=\frac{11}{16} \).

Teacher's Note:
1. Ruin can only occur after an odd number of tosses.
2. List the paths that first reach 0 at toss 1, 3, 5.

Answer: (b) \( \frac{10}{19} \)

Explanation:
1. The rupee stays in the first purse during the first transfer with probability 1/10.
2. Otherwise (probability 9/10) it is in the second purse (19 coins), and it comes back with probability 9/19.
3. Total \( \frac1{10}+\frac9{10}\cdot\frac9{19}=\frac{19+81}{190}=\frac{10}{19} \).

Teacher's Note:
1. Two disjoint cases: the rupee never leaves, or it leaves and returns.
2. The second purse holds 19 coins at the moment of the return transfer.

Answer: (b) \( \frac{13}{19} \)

Explanation:
1. PARALLELOGRAM has 13 letters, so 12 consecutive pairs, and 'RA' occurs twice (PA-RA and G-RA-M).
2. PARALLELOPIPED has 14 letters, so 13 consecutive pairs, and 'RA' occurs once.
3. Bayes: \( \frac{\frac12\cdot\frac{2}{12}}{\frac12\cdot\frac2{12}+\frac12\cdot\frac1{13}}=\frac{1/6}{1/6+1/13}=\frac{13}{19} \).

Teacher's Note:
1. Count pairs of consecutive letters in each word.
2. Treat the visible pair as equally likely to be any pair.

Answer: (d) 28 : 3

Explanation:
1. P(good and says good) = 0.7 x 0.8 = 0.56.
2. P(bad and says good) = 0.3 x 0.2 = 0.06.
3. Odds = 0.56 : 0.06 = 28 : 3.

Teacher's Note:
1. Odds in favour = favourable : unfavourable, here good : bad given the report.
2. Do not use the prior odds 7 : 3.

Answer: (c) 5

Explanation:
1. P(head) = \( \frac{1}{n+1}\cdot1+\frac{n}{n+1}\cdot\frac12=\frac{n+2}{2(n+1)} \).
2. Set \( \frac{n+2}{2(n+1)}=\frac7{12} \): \( 12n+24=14n+14 \).
3. So n = 5.

Teacher's Note:
1. Split on whether the two-headed coin is chosen.
2. Cross-multiply and solve the linear equation.

Answer: (c) 24 : 19

Explanation:
1. P(defective) = 0.35 x 0.02 + 0.25 x 0.01 + 0.40 x 0.03 = 0.007 + 0.0025 + 0.012 = 0.0215.
2. P(C | defective) = 0.012 / 0.0215 = 24/43.
3. Odds in favour = 24 : (43 - 24) = 24 : 19.

Teacher's Note:
1. Odds in favour = p : (1 - p).
2. 24 : 43 is the probability, not the odds.

Answer: (d) 36

Explanation:
1. Let W mean the first-game winner wins a game; after a win the chance of W winning again is 2/3, after a loss it is 1/3.
2. Summing over the 16 outcome sequences of the next four games with at least three wins gives \( \frac49 \).
3. Hence p = 4, q = 9 and pq = 36.

Teacher's Note:
1. The chance of winning a game depends on who won the game before it.
2. Enumerate sequences with 3 or 4 wins.

Answer: (a) 11

Explanation:
1. Two children: probability 1/2, grandchildren total 4 when the pair has (1,3), (3,1) or (2,2): probability \( \frac1{4}\cdot\frac14\cdot2+\frac12\cdot\frac12=\frac38 \), contributing \( \frac12\cdot\frac38=\frac3{16} \).
2. One child (1/4): that child has 4 children, impossible. Three children (1/4): sum 4 from three children means (1,1,2) in 3 orders: \( 3\cdot\frac14\cdot\frac14\cdot\frac12=\frac{3}{32} \), contributing \( \frac14\cdot\frac3{32}=\frac{3}{128} \).
3. Total \( \frac{24}{128}+\frac{3}{128}=\frac{27}{128} \), so q = 128 and the digit sum is 11.

Teacher's Note:
1. Condition on the number of children of the couple.
2. Count the ordered ways the grandchildren total 4.

Answer: (a) (i) and (iii) only

Explanation:
1. A = {1,3,5} and B = {2,4,6} are disjoint with union S, so (i) is true.
2. A and C share 5 and so are not mutually exclusive: (ii) is false.
3. A, C, D: {1,3,5} U {4,5,6} U {1,2} = S, so (iii) is true; B, C, D miss the outcome 3, so (iv) is false.

Teacher's Note:
1. Write each event as a set.
2. An exhaustive system must cover all six outcomes.

Answer: (d) (i), (iii) and (iv) only

Explanation:
1. Exactly one occurs means M only or N only: \( P(M\cap\bar N)+P(\bar M\cap N) \), which is (iv).
2. Since \( P(M\cap\bar N)=P(M)-P(M\cap N) \), this equals (i).
3. (ii) is P(M U N), not the required event. For (iii): \( (1-P(M))+(1-P(N))-2(1-P(M\cup N))=2P(M\cup N)-P(M)-P(N)=P(M)+P(N)-2P(M\cap N) \), so (iii) is also correct.

Teacher's Note:
1. Exactly one = A only + B only.
2. Rewrite (iii) with P(M' and N') = 1 - P(M U N).

Answer: (c) (i), (ii) and (iii) only

Explanation:
1. Each \( P(E_i)=\frac12 \) and each pair meets in exactly one ticket, so \( P(E_i\cap E_j)=\frac14=P(E_i)P(E_j) \): the events are pairwise independent.
2. All three occur only for 222: \( P(E_1\cap E_2\cap E_3)=\frac14 \).
3. But \( P(E_1)P(E_2)P(E_3)=\frac18\neq\frac14 \), so the three are not mutually independent.

Teacher's Note:
1. Pairwise independence is weaker than mutual independence.
2. Compare the triple intersection with the product.

Answer: (d) \( \frac{7}{12} \)

Explanation:
1. P(success) = 0.1 x 0.8 + 0.2 x 0.6 + 0.7 x 0.4 = 0.08 + 0.12 + 0.28 = 0.48.
2. P(4 hours and success) = 0.28.
3. Bayes: 0.28 / 0.48 = 7/12.

Teacher's Note:
1. Compute P(success) with the total probability theorem first.
2. Divide the 4-hour branch by the total.