JEE Mathematics Probability MCQs Set 23

Answer: (a) 43 : 34

Explanation:
1. Odds 8 : 3 against A give \( P(A)=\frac{3}{11} \); odds 5 : 2 against B give \( P(B)=\frac27 \).
2. Exactly one event must happen, so \( P(C)=1-\frac3{11}-\frac27=\frac{34}{77} \).
3. Odds against C = \( (1-P(C)) : P(C)=43:34 \).

Teacher's Note:
1. Odds against an event are (not happening) : (happening).
2. The three probabilities add up to 1.

Answer: (b) \( \frac{4}{5} \)

Explanation:
1. Total arrangements: \( \frac{6!}{3!\,2!}=60 \).
2. Treat BAN as one block; with the remaining letters A, A, N there are \( \frac{4!}{2!}=12 \) arrangements containing the pattern.
3. Required probability \( =1-\frac{12}{60}=\frac45 \).

Teacher's Note:
1. Count the unwanted case by gluing the three letters together.
2. B occurs only once, so the pattern cannot be counted twice.

Answer: (a) \( \frac{5}{9} \)

Explanation:
1. The number is even exactly when the second card shows an even digit: 0, 2, 4, 6 or 8 (5 choices).
2. The first card can then be any of the other 8 cards, so favourable ordered pairs = 5 x 8 = 40.
3. Total ordered pairs = 9 x 8 = 72, so the probability is \( \frac{40}{72}=\frac59 \).

Teacher's Note:
1. Only the last digit decides parity; a leading 0 does not matter.
2. Count ordered pairs, not combinations.

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

Explanation:
1. Only the relative order of A, B, C matters.
2. Their 3! = 6 possible orders are equally likely and exactly one is A, B, C.
3. So the probability is \( \frac{1}{6} \).

Teacher's Note:
1. Ignore the other 5 speakers; they do not change the relative order of A, B, C.
2. Each of the six orders of three persons is equally likely.

Answer: (c) 10

Explanation:
1. Number of triangles: \( \binom63=20 \).
2. A regular hexagon contains exactly 2 equilateral triangles (alternate vertices).
3. So \( p=\frac{2}{20}=\frac1{10} \) and 100p = 10.

Teacher's Note:
1. Equilateral triangles need vertices that are two steps apart.
2. There are only two ways to choose alternate vertices.

Answer: (d) \( \frac{2}{5} \) and \( \frac{4}{15} \)

Explanation:
1. Original chances: A = 1/2, B = 3/10, C = 1/5.
2. A now has \( \frac{1}{3} \), so B and C share \( \frac{2}{3} \) in the ratio 3 : 2.
3. B = \( \frac23\cdot\frac35=\frac25 \) and C = \( \frac23\cdot\frac25=\frac4{15} \).

Teacher's Note:
1. Total probability is still 1.
2. Share the remainder in the old ratio 3 : 2.

Answer: (b) \( \frac{3}{10} \)

Explanation:
1. Multiples of 5: 20; multiples of 8: 12; multiples of 40: 2.
2. Using inclusion-exclusion, 20 + 12 - 2 = 30 numbers qualify.
3. Probability \( =\frac{30}{100}=\frac3{10} \).

Teacher's Note:
1. Numbers divisible by both 5 and 8 are counted twice.
2. Both means divisible by LCM = 40.

Answer: (d) \( \frac{22}{425} \)

Explanation:
1. Choose the suit in 4 ways and 3 cards of that suit in \( \binom{13}{3}=286 \) ways: 1144 favourable cases.
2. Total ways \( \binom{52}3=22100 \).
3. Probability \( =\frac{1144}{22100}=\frac{22}{425} \).

Teacher's Note:
1. Do not forget the factor 4 for the choice of suit.
2. Reduce the fraction by dividing by 52.

Answer: (d) \( \frac{1}{462} \)

Explanation:
1. There are two alternating patterns (boy first or girl first).
2. Each pattern has 6! x 6! arrangements, so favourable = 2 x 6! x 6!.
3. Total = 12!, so the probability is \( \frac{2\,(6!)^2}{12!}=\frac{2}{924}=\frac1{462} \).

Teacher's Note:
1. Use 12!/(6! 6!) = C(12, 6) = 924.
2. Two patterns: BGBG... and GBGB....

Answer: (b) \( \frac{\binom{13}{4}\,4^4}{\binom{52}{4}} \)

Explanation:
1. Choose 4 different ranks from 13 in \( \binom{13}{4} \) ways.
2. For each chosen rank pick any of its 4 suits: \( 4^4 \) ways.
3. Probability \( =\frac{\binom{13}{4}4^4}{\binom{52}{4}}=\frac{2816}{4165} \).

Teacher's Note:
1. Choose ranks first, then a suit for each rank.
2. The cards are drawn as a set, so order is not counted.

Answer: (a) \( \frac{5}{21} \)

Explanation:
1. Sum of memberships = 40 + 50 + 60 = 150 = 1(70) + 2(x) + 3(10), where x is the number in exactly two clubs.
2. So 2x = 150 - 70 - 30 = 50 and x = 25.
3. Total persons = 70 + 25 + 10 = 105, so probability \( =\frac{25}{105}=\frac5{21} \).

Teacher's Note:
1. Each person is counted once for every club he belongs to.
2. Total persons = exactly one + exactly two + all three.

Answer: (b) 0.56

Explanation:
1. P(A and not B) = 0.4 - 0.16 = 0.24.
2. P(B and not A) = 0.48 - 0.16 = 0.32.
3. Exactly one occurs: 0.24 + 0.32 = 0.56.

Teacher's Note:
1. Exactly one = P(A) + P(B) - 2P(A and B).
2. 0.4 + 0.48 - 0.32 = 0.56.

Answer: (b) \( \frac{2}{3} \)

Explanation:
1. Sample space (older, younger): BB, BG, GB, GG.
2. At least one boy leaves BB, BG, GB, which are equally likely.
3. The other child is a girl in BG and GB, so the probability is \( \frac{2}{3} \).

Teacher's Note:
1. Condition on 'at least one boy', not on a named child.
2. If the older child is specified as a boy, the answer would be 1/2.

Answer: (d) \( \frac{52}{77} \)

Explanation:
1. First event: \( P=\frac27 \); second event: \( P=\frac6{11} \).
2. Neither happens with probability \( \frac57\cdot\frac5{11}=\frac{25}{77} \).
3. At least one: \( 1-\frac{25}{77}=\frac{52}{77} \).

Teacher's Note:
1. Use independence: P(neither) = product of the complements.
2. Convert odds to probabilities first.

Answer: (b) \( \frac{3}{5} \)

Explanation:
1. Treat AB as one block: arrangements with A, B together = 2 x 5! = 240.
2. Now count those with C, D adjacent: blocks AB and CD plus E, F give 4! x 2 x 2 = 96.
3. So C, D separated in 240 - 96 = 144 cases and the conditional probability is \( \frac{144}{240}=\frac35 \).

Teacher's Note:
1. Conditional probability = favourable inside the reduced sample space.
2. Count the complement (C and D together) and subtract.

Answer: (b) \( \frac{17}{125} \)

Explanation:
1. P(class XI) = 2/5 and P(class XII) = 3/5.
2. P(brilliant | XI) = 5/50 = 1/10 and P(brilliant | XII) = 8/50 = 4/25.
3. Total probability \( =\frac25\cdot\frac1{10}+\frac35\cdot\frac4{25}=\frac{17}{125} \).

Teacher's Note:
1. Use the theorem of total probability.
2. Weight each class by its chance of being chosen.

Answer: (c) \( \frac{8}{29} \)

Explanation:
1. P(blue) = \( \frac12\cdot\frac27+\frac12\cdot\frac68=\frac{29}{56} \).
2. P(box I and blue) = \( \frac12\cdot\frac27=\frac17=\frac8{56} \).
3. Bayes' theorem: \( \frac{8/56}{29/56}=\frac8{29} \).

Teacher's Note:
1. Required probability = P(box I and blue) / P(blue).
2. This is a Bayes' theorem problem.

Answer: (d) \( \frac{1}{17} \)

Explanation:
1. P(both kings) = \( \frac4{52}\cdot\frac3{51} \).
2. P(second is a king) = 4/52 = 1/13.
3. Ratio: \( \frac{(4\cdot3)/(52\cdot51)}{1/13}=\frac3{51}=\frac1{17} \).

Teacher's Note:
1. By symmetry, fixing the second card as a king leaves 51 cards, 3 of them kings.
2. Conditional probability = P(both) / P(second).

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

Explanation:
1. \( P(F\cap M')=P(F)-P(M\cap F)=0.7-0.65=0.05 \).
2. P(M') = 1 - 0.8 = 0.2.
3. Required \( =\frac{0.05}{0.2}=\frac14 \).

Teacher's Note:
1. Split F into the parts with M and without M.
2. Divide by P(M'), not P(F).

Answer: (d) \( \frac{7}{8192} \)

Explanation:
1. Number of ways: \( \binom{14}{13}=14 \).
2. Each sequence has probability \( (1/2)^{14}=\frac1{16384} \).
3. Probability \( =\frac{14}{16384}=\frac7{8192} \).

Teacher's Note:
1. Use the binomial formula with p = 1/2.
2. C(14, 13) = 14.

Answer: (c) \( \binom{10}{2}\left(\frac1{15}\right)^2\left(\frac{14}{15}\right)^8 \)

Explanation:
1. Non-empty subsets of 4 options: \( 2^4-1=15 \), only one of which is the correct set, so \( p=\frac1{15} \) per question.
2. Exactly two correct out of 10 is binomial with n = 10, r = 2.
3. Probability \( =\binom{10}{2}\left(\frac1{15}\right)^2\left(\frac{14}{15}\right)^8 \).

Teacher's Note:
1. Count the possible answers per question before using the binomial formula.
2. Empty selection is not an attempt.

Answer: (a) \( \frac{7}{128} \)

Explanation:
1. With p = 1/2: \( \binom n3=\binom n4 \) gives n = 7.
2. P(exactly once) = \( \binom71\left(\frac12\right)^7=\frac7{128} \).

Teacher's Note:
1. C(n, 3) = C(n, 4) means 3 + 4 = n.
2. All terms carry (1/2)^n.

Answer: (c) 54 and 45

Explanation:
1. P(A wins) = \( \frac16\left[1+\left(\frac56\right)^2+\dots\right]=\frac{1/6}{1-25/36}=\frac6{11} \).
2. P(B wins) = 5/11.
3. Expectations: 99 x 6/11 = 54 and 99 x 5/11 = 45.

Teacher's Note:
1. Sum the infinite geometric series of A's winning turns.
2. Expectation = stake x probability of winning.

Answer: (b) 99

Explanation:
1. For n = 99, the binomial coefficients \( \binom{99}{r} \) are largest at r = 49 and r = 50 (equal).
2. Sum of possible r = 49 + 50 = 99.

Teacher's Note:
1. For odd n the two middle terms are equal and largest.
2. The middle terms are r = (n - 1)/2 and (n + 1)/2.

Answer: (b) \( \frac{17}{21} \)

Explanation:
1. Real roots need \( q^2-3q-4\ge0 \), i.e. \( (q-4)(q+1)\ge0 \), so \( q\le-1 \) or \( q\ge4 \).
2. Integers: -10 to -1 (10 values) and 4 to 10 (7 values): 17 favourable.
3. Total integers: 21, so the probability is \( \frac{17}{21} \).

Teacher's Note:
1. Use the discriminant condition D >= 0.
2. Count the integers in each valid range.