JEE Mathematics Probability MCQs Set 26

Answer: (a) 1/5

Explanation:
1. Sums less than 6: sums 2, 3, 4, 5 occur in 1, 2, 3, 4 outcomes, so 10 outcomes.
2. Sum 3: 2 outcomes. \(P=\frac2{10}=\frac15\).

Teacher's Note:
1. Restrict to the conditioning event, then count.

Answer: (c) The outcomes must be dependent on each other

Explanation:
1. A binomial setting needs a fixed number of independent trials, each with two outcomes and a constant success probability.
2. Outcomes depending on each other violates independence, so that is not a requirement.

Teacher's Note:
1. Binomial trials must be independent.

Answer: (d) \( {}^5C_4(0.7)^4(0.3) \)

Explanation:
1. Swimmer probability is \(0.7\). Choose which 4 of the 5 are swimmers: \(\binom54\).
2. \(P=\binom54(0.7)^4(0.3)^1\).

Teacher's Note:
1. Binomial formula: nCr p^r q^(n-r) with r the number of swimmers.

Answer: (d) 10

Explanation:
1. \(E(X^2)=1\cdot\frac1{10}+4\cdot\frac2{10}+9\cdot\frac3{10}+16\cdot\frac4{10}=\frac{1+8+27+64}{10}=10\).

Teacher's Note:
1. Square each value of X before weighting.

Answer: (b) 10/13

Explanation:
1. Both correct: \(\frac13\cdot\frac14=\frac1{12}\). Both wrong: \(\frac23\cdot\frac34=\frac12\), and same wrong answer: \(\frac12\cdot\frac1{20}=\frac1{40}\).
2. P(correct | same answer) \(=\frac{1/12}{1/12+1/40}=\frac{10}{13}\).

Teacher's Note:
1. Bayes' theorem. The stem now says the 1/20 is the chance of a common error given both are wrong, which is the reading that matches the options.

Answer: (a) \( \left[0,\frac12\right] \)

Explanation:
1. P(at least one failure) \(=1-p^5\ge\frac{31}{32}\), so \(p^5\le\frac1{32}\).
2. Thus \(p\le\frac12\), and with \(p\ge0\) the interval is \([0,\frac12]\).

Teacher's Note:
1. Complement of 'no failure' is 1 minus p to the power 5.

Answer: (a) Independent but not equally likely

Explanation:
1. \(P(A\cup B)=\frac56\) and \(P(A)=\frac34\), so \(P(B)=\frac56+\frac14-\frac34=\frac13\).
2. \(P(A)P(B)=\frac34\cdot\frac13=\frac14=P(A\cap B)\): independent. \(P(A)\ne P(B)\): not equally likely.

Teacher's Note:
1. Recover P(B) from inclusion-exclusion, then test the product rule.

Answer: (c) 833

Explanation:
1. The numbers are \(2k\), \(k=1,\dots,50\), so variance \(=4\cdot\mathrm{Var}(1,\dots,50)=4\cdot\frac{50^2-1}{12}\).
2. \(=4\cdot\frac{2499}{12}=833\).

Teacher's Note:
1. Doubling the data multiplies the variance by 4; the variance of 1..n is (n^2-1)/12.

Answer: (b) \( \frac{11}{3^5} \)

Explanation:
1. \(P(4)=\binom54\left(\frac13\right)^4\frac23=\frac{10}{3^5}\), \(P(5)=\frac1{3^5}\).
2. Total \(=\frac{11}{3^5}\).

Teacher's Note:
1. Add the binomial probabilities of 4 and 5 correct answers.

Answer: (d) 5/17

Explanation:
1. \(A'\cup B'=(A\cap B)'\), with probability \(1-\frac3{20}=\frac{17}{20}\).
2. \(A\cap(A\cap B)'=A\cap B'\), with probability \(\frac25-\frac3{20}=\frac14\).
3. \(P=\frac{1/4}{17/20}=\frac5{17}\).

Teacher's Note:
1. Use De Morgan: A' or B' is the complement of A and B.

Answer: (b) \( \frac{256}{729} \)

Explanation:
1. \(p=\frac23\), \(q=\frac13\).
2. \(P(5)=6\cdot\frac{2^5}{3^6}=\frac{192}{729}\), \(P(6)=\frac{64}{729}\).
3. Total \(=\frac{256}{729}\).

Teacher's Note:
1. Twice as often as it fails means success probability 2/3.

Answer: (b) \( \frac{{}^{20}C_{10}}{2^{20}} \)

Explanation:
1. Total ordered pairs: \(2^{10}\cdot2^{10}=2^{20}\).
2. Equal size \(k\): \(\sum_k\binom{10}k^2=\binom{20}{10}\) (Vandermonde).
3. \(P=\frac{{}^{20}C_{10}}{2^{20}}\).

Teacher's Note:
1. Use the identity sum of C(n,k)^2 = C(2n,n).

Answer: (d) 64/127

Explanation:
1. Non-empty subsets: \(2^7-1=127\). A fixed element lies in \(2^6=64\) of them.
2. \(P=\frac{64}{127}\).

Teacher's Note:
1. Fix x first; the number of subsets containing it is 2^(n-1).

Answer: (b) \(3-p\)

Explanation:
1. \(\binom n2p^2q^{n-2}=\binom n3p^3q^{n-3}\) gives \(3q=(n-2)p\).
2. So \(np=3q+2p=3(1-p)+2p=3-p\).

Teacher's Note:
1. Equate the two probabilities and solve for np.

Answer: (a) Both the statements are true

Explanation:
1. \(P(A/E)P(E)=P(A\cap E)\), and \(P(E/A)=\frac{P(A\cap E)}{P(A)}\ge P(A\cap E)\) because \(P(A)\le1\).
2. So both statements hold.

Teacher's Note:
1. Statement I is the same as Statement II after using the multiplication rule.

Answer: (a) 4/7

Explanation:
1. \((1-p)(1-q)=\frac{15}{49}\) gives \(p+q=pq+\frac{34}{49}\); exactly one: \(p+q-2pq=\frac{26}{49}\).
2. Subtracting gives \(pq=\frac8{49}\) and \(p+q=\frac{6}{7}\), so \(p,q\) are the roots of \(t^2-\frac67t+\frac8{49}=0\): \(\frac27\) and \(\frac47\).
3. The more probable event has probability \(\frac47\).

Teacher's Note:
1. Form a quadratic from the sum and product of the two probabilities.

Answer: (c) 205

Explanation:
1. Total points \(=12\): \(\binom{12}3=220\) selections.
2. Subtract collinear triples: \(\binom33+\binom43+\binom53=1+4+10=15\).
3. Triangles \(=220-15=205\).

Teacher's Note:
1. Remove the triples of points lying on the same side.

Answer: (d) 3

Explanation:
1. P(sum 7) \(=\frac6{36}\), P(sum 5) \(=\frac4{36}\).
2. Only these two sums matter: \(p=\frac{6}{6+4}=\frac35\), so \(5p=3\).

Teacher's Note:
1. Condition on the first roll that is either 5 or 7.

Answer: (b) 5

Explanation:
1. P(white from A) \(=\frac58\), P(white from B) \(=\frac34\), bags equally likely.
2. P(B | white) \(=\frac{3/4}{5/8+3/4}=\frac{6}{11}\), so \(q-p=5\).

Teacher's Note:
1. Bayes' theorem with equal priors.

Answer: (d) 26

Explanation:
1. \(P(A\cup B)=0.42+0.48-0.16=0.74\).
2. \(P((A\cup B)')=0.26\), so the answer is 26.

Teacher's Note:
1. Complement of the union.

Answer: (a) 11

Explanation:
1. Favourable: (2,5), (4,3), (6,1): 3 outcomes out of 36.
2. \(P=\frac3{36}=\frac1{12}\), so \(q-p=11\).

Teacher's Note:
1. List the outcomes explicitly.

Answer: (c) 3

Explanation:
1. \(P(A\cup B)=\frac13+\frac14-\frac15=\frac{23}{60}\), so \(P(A'\cap B')=\frac{37}{60}\).
2. \(P(B'/A')=\frac{37/60}{2/3}=\frac{37}{40}\), so \(q-p=3\).

Teacher's Note:
1. De Morgan for the numerator.

Answer: (b) 5

Explanation:
1. P(first box) = P(second box) \(=\frac12\); white probabilities \(\frac49\) and \(\frac59\).
2. \(P=\frac{4/9}{4/9+5/9}=\frac49\), so \(q-p=5\).

Teacher's Note:
1. Bayes' theorem.

Answer: (c) 6

Explanation:
1. \(C\) wins in round \(k\) if all earlier draws are black and \(C\) draws red: \(\left(\frac12\right)^{2}\cdot\frac12\) in round 1, and so on.
2. \(P=\frac{(1/2)^3}{1-(1/2)^3}=\frac17\), so \(q-p=6\).

Teacher's Note:
1. Sum the geometric series over rounds.