JEE Mathematics Probability MCQs Set 25

Answer: (a) \( \frac{1}{9} \), \( \frac{1}{3} \)

Explanation:
1. Let the second event have probability \(p\); the first has \(p^2\). Odds against are \(\frac{1-p^2}{p^2}\) and \(\frac{1-p}{p}\).
2. The condition \(\frac{1-p^2}{p^2}=\left(\frac{1-p}{p}\right)^3\) gives \((1+p)p=(1-p)^2\), so \(3p=1\) after cancelling \(1-p\) (the root \(p=1\) is rejected).
3. Hence the chances are \(\frac19\) and \(\frac13\); check: odds against are 8 and 2, and \(2^3=8\).

Teacher's Note:
1. Odds against an event of probability p are (1-p)/p. Remember to discard the trivial root p = 1.

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

Explanation:
1. Odds against 5 : 2 give \(P=\frac27\); odds in favour 6 : 5 give \(P=\frac6{11}\).
2. P(none) \(=\frac57\cdot\frac5{11}=\frac{25}{77}\).
3. P(at least one) \(=1-\frac{25}{77}=\frac{52}{77}\).

Teacher's Note:
1. For independent events, P(at least one) is 1 minus the product of the probabilities that each fails.

Answer: (c) \( \frac{1}{12} \) \(\le P(B)\le\) \( \frac{3}{4} \)

Explanation:
1. \(P(A)=\frac23\) and \(P(A\cup B)=\frac34\).
2. Smallest: \(P(B)\ge P(A\cup B)-P(A)=\frac34-\frac23=\frac1{12}\) (when \(A\) and \(B\) are disjoint).
3. Largest: \(P(B)\le P(A\cup B)=\frac34\) (when \(B\supseteq A\) and \(B\) covers the whole union).

Teacher's Note:
1. Use P(B) >= P(A u B) - P(A) for the lower bound and P(B) <= P(A u B) for the upper bound.

Answer: (b) 1/4

Explanation:
1. Multiples of 6: \(\lfloor200/6\rfloor=33\); multiples of 8: 25; multiples of 24 (the lcm): 8.
2. Favourable \(=33+25-8=50\), so the probability is \(\frac{50}{200}=\frac14\).

Teacher's Note:
1. Use inclusion-exclusion with the lcm for numbers divisible by either of two divisors.

Answer: (a) 0.25

Explanation:
1. \(P(A)=0.7\) and \(P(\bar B)=0.6\), so \(P(A\cup\bar B)=0.7+0.6-0.5=0.8\).
2. \(B\cap(A\cup\bar B)=A\cap B\), and \(P(A\cap B)=P(A)-P(A\cap\bar B)=0.2\).
3. So the required probability is \(\frac{0.2}{0.8}=0.25\).

Teacher's Note:
1. Simplify the intersection B and (A or not-B) to A and B before dividing.

Answer: (a) 1/9

Explanation:
1. Given that both numbers are at most 10, the pair is any of \(\binom{10}{2}=45\) pairs from 1 to 10.
2. Pairs with smaller number 5: the larger is 6, 7, 8, 9 or 10, that is 5 pairs.
3. \(P=\frac5{45}=\frac19\).

Teacher's Note:
1. Restrict the sample space to the condition, then count.

Answer: (b) 5/12

Explanation:
1. Unequal outcomes: \(36-6=30\), split equally between 'first smaller' and 'first larger'.
2. \(P=\frac{15}{36}=\frac5{12}\).

Teacher's Note:
1. By symmetry, half of the unequal outcomes favour each ordering.

Answer: (d) 1/2

Explanation:
1. \(P(A)P(B)=\frac18\) and \((1-P(A))(1-P(B))=\frac38\) give \(1-(P(A)+P(B))+\frac18=\frac38\), so \(P(A)+P(B)=\frac34\).
2. \(P(A),P(B)\) are the roots of \(t^2-\frac34t+\frac18=0\), that is \(\frac12\) and \(\frac14\).
3. Since \(P(A)>P(B)\), \(P(A)=\frac12\).

Teacher's Note:
1. Use sum and product of the two probabilities to form a quadratic. The source did not say which of the two values belongs to A, so the condition P(A) > P(B) is stated.

Answer: (d) 5/38

Explanation:
1. \(P=\frac{10}{20}\cdot\frac{9}{19}\cdot\frac{10}{18}\).
2. \(=\frac{900}{6840}=\frac5{38}\).

Teacher's Note:
1. Multiply the conditional probabilities of each successive draw.

Answer: (c) 1 : 1

Explanation:
1. Let \(p=\frac m{m+n}\), \(q=1-p\). A wins on draws 1, 3, 5, ...: \(P(A)=p+q^2p+\dots=\frac{p}{1-q^2}\); B wins on draws 2, 4, ...: \(P(B)=\frac{qp}{1-q^2}\).
2. Condition \(\frac p{1-q^2}=2\cdot\frac{qp}{1-q^2}\) gives \(q=\frac12\).
3. So \(p=q\), that is \(m:n=1:1\).

Teacher's Note:
1. In alternating-draw games P(A wins) = p/(1-q^2) and P(B wins) = qp/(1-q^2), so their ratio is just 1 : q.

Answer: (c) 5/24

Explanation:
1. Both bags contain a ticket numbered 4, and each bag is chosen with probability \(\frac12\).
2. \(P=\frac12\cdot\frac14+\frac12\cdot\frac16=\frac18+\frac1{12}=\frac5{24}\).

Teacher's Note:
1. Total probability: weight each bag's chance of showing the number by the probability of choosing that bag.

Answer: (a) 2/3

Explanation:
1. Missing card red (prior \(\frac12\)): 25 red cards remain, probability of 13 red \(\frac{\binom{25}{13}}{\binom{51}{13}}\). Missing card black: 26 red remain, probability \(\frac{\binom{26}{13}}{\binom{51}{13}}\).
2. \(\binom{26}{13}=\frac{26}{13}\binom{25}{12}\) and \(\binom{25}{13}=\binom{25}{12}\), so the ratio of likelihoods is 2 : 1.
3. P(black missing) \(=\frac{2}{2+1}=\frac23\).

Teacher's Note:
1. Compare likelihoods of the observation under each possible missing card.

Answer: (c) \( \frac{n(n-1)}{(m+n)(m+n-1)} \)

Explanation:
1. Total arrangements of identical coins: \(\binom{m+n}{n}\). With ten paise coins at both ends, the remaining \(n-2\) ten paise coins go in \(m+n-2\) places: \(\binom{m+n-2}{n-2}\).
2. Ratio \(=\frac{(m+n-2)!\,n!\,m!}{(n-2)!\,m!\,(m+n)!}=\frac{n(n-1)}{(m+n)(m+n-1)}\).
3. Check: \(m=3,n=4\) gives \(\frac{\binom52}{\binom74}=\frac{10}{35}=\frac27=\frac{12}{42}\).

Teacher's Note:
1. The same answer follows by choosing the two end coins one after another: n/(m+n) times (n-1)/(m+n-1).

Answer: (d) \( \frac{110}{9} \) \(\left(\frac23\right)^{10}\)

Explanation:
1. Each ball goes to any of the 3 boxes: \(3^{12}\) outcomes. Choose the 3 balls for box 1 in \(\binom{12}{3}=220\) ways; the other 9 go to the 2 other boxes in \(2^9\) ways.
2. \(P=\frac{220\cdot2^9}{3^{12}}=\frac{110\cdot2^{10}}{9\cdot3^{10}}=\frac{110}9\left(\frac23\right)^{10}\).

Teacher's Note:
1. This is a binomial probability: 3 successes out of 12 with success chance 1/3.

Answer: (b) 4/9

Explanation:
1. Both serve: \(\binom73=35\) committees. Neither serves: \(\binom75=21\). Total \(\binom95=126\).
2. \(P=\frac{35+21}{126}=\frac{56}{126}=\frac49\).

Teacher's Note:
1. Add the two mutually exclusive cases: both included and both excluded.

Answer: (d) 10

Explanation:
1. \(P(H)=\frac n{2n+1}\cdot1+\frac{n+1}{2n+1}\cdot\frac12=\frac{3n+1}{2(2n+1)}\).
2. \(\frac{3n+1}{2(2n+1)}=\frac{31}{42}\) gives \(21(3n+1)=31(2n+1)\), so \(n=10\).

Teacher's Note:
1. Total probability over the two kinds of coin.

Answer: (a) \(<\frac14\)

Explanation:
1. \(\bar E_1\cap\bar E_2\) is the complement of \(E_1\cup E_2\), so the intersection is the empty event.
2. Its probability is \(0\), which is less than \(\frac14\).

Teacher's Note:
1. An event and its complement can never occur together.

Answer: (c) 561/15925

Explanation:
1. Exactly one ace in the first 17 cards and the 18th card is an ace.
2. \(P=\frac{\binom41\binom{48}{16}}{\binom{52}{17}}\cdot\frac{3}{35}=\frac{561}{15925}\).

Teacher's Note:
1. 'First time on the k-th draw' means exactly one success in k-1 draws followed by a success.

Answer: (b) 1/6

Explanation:
1. Among the relative orders of \(A,B,C\), all \(3!=6\) are equally likely.
2. Only one is \(A\) then \(B\) then \(C\): \(P=\frac16\).

Teacher's Note:
1. The other five speakers do not affect the relative order of A, B and C.

Answer: (c) \( \frac{84\times5^6}{6^{10}} \)

Explanation:
1. Three sixes in the first nine throws: \(\binom93=84\) ways, with six non-sixes: \(5^6\).
2. The tenth throw is a six, so \(P=\frac{84\cdot5^6}{6^{10}}\).

Teacher's Note:
1. Only the first ten throws matter; the last ten are irrelevant.

Answer: (c) 9/25

Explanation:
1. P(no test on either day) \(=\left(\frac45\right)^2=\frac{16}{25}\).
2. P(at least one test missed) \(=1-\frac{16}{25}=\frac9{25}\).

Teacher's Note:
1. Use the complement for 'at least one'.

Answer: (d) 24/91

Explanation:
1. \(P=\frac{\binom{10}3}{\binom{15}3}=\frac{120}{455}=\frac{24}{91}\).

Teacher's Note:
1. Choose all items from the original ones only.

Answer: (d) 4

Explanation:
1. \(E(X)=30\cdot\frac15+10\cdot\frac3{10}-10\cdot\frac12=6+3-5=4\).

Teacher's Note:
1. E(X) is the sum of value times probability.

Answer: (d) 7/12

Explanation:
1. \(P(A\cap B)=\frac3{10}+\frac25-\frac35=\frac1{10}\).
2. \(P(B|A)=\frac{1/10}{3/10}=\frac13\), \(P(A|B)=\frac{1/10}{2/5}=\frac14\).
3. Sum \(=\frac13+\frac14=\frac7{12}\).

Teacher's Note:
1. Find the intersection by inclusion-exclusion first.

Answer: (d) None of the above is correct

Explanation:
1. Independent events with positive probabilities cannot be mutually exclusive, since \(P(A\cap B)=P(A)P(B)>0\).
2. Nothing forces \(P(A)+P(B)=1\): a head on a coin and a six on a die are independent with probabilities \(\frac12\) and \(\frac16\), whose sum is not 1.
3. Hence none of the statements is correct.

Teacher's Note:
1. Independent and mutually exclusive are different ideas.

Answer: (b) 0.188

Explanation:
1. Each fires once: \(P(AB\bar C)=0.4\cdot0.3\cdot0.8=0.096\); \(P(A\bar BC)=0.4\cdot0.7\cdot0.2=0.056\); \(P(\bar ABC)=0.6\cdot0.3\cdot0.2=0.036\).
2. Sum \(=0.188\).

Teacher's Note:
1. Add the three arrangements in which exactly two of the independent shooters hit.