JEE Mathematics Probability MCQs Set 32

Practice MCQs for JEE Mathematics Probability

Access targeted multiple-choice questions for Probability designed to align with the latest JEE academic syllabus for JEE Mathematics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

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Multiple Choice Questions

Question 1: A fair die is thrown twenty times. The probability that on the tenth throw the fourth six appears is
(a) \(\dfrac{{}^{20}C_{10}\times5^6}{6^{20}}\)
(b) \(\dfrac{120\times5^7}{6^{10}}\)
(c) \(\dfrac{84\times5^6}{6^{10}}\)
(d) None of these
Show Answer & Explanation

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

Explanation:
1. Three sixes in the first 9 throws, then a six: \({}^9C_3\left(\frac16\right)^3\left(\frac56\right)^6\cdot\frac16\).
2. \(=\dfrac{84\times5^6}{6^{10}}\).

Teacher's Note:
1. Write down the sample space or use the multiplication rule for independent events.
2. For conditional probability use P(A|B) = P(A and B)/P(B); for causes use Bayes theorem.

Question 2: Each child born is equally likely to be a boy or a girl. If a family has two children, the conditional probabilities that both are girls given that (i) the youngest is a girl, (ii) at least one is a girl, are
(a) \(\frac12\) and \(\frac14\)
(b) \(\frac13\) and \(\frac12\)
(c) \(\frac13\) and \(\frac14\)
(d) \(\frac12\) and \(\frac13\)
Show Answer & Explanation

Answer: (d) \(\frac12\) and \(\frac13\)

Explanation:
1. (i) Given the youngest is a girl, the other is a girl with probability \(\frac12\).
2. (ii) Sample space \(\{GG,GB,BG\}\) gives \(\frac13\).

Teacher's Note:
1. Write down the sample space or use the multiplication rule for independent events.
2. For conditional probability use P(A|B) = P(A and B)/P(B); for causes use Bayes theorem.

Question 3: In a test, an examinee either guesses, copies, or knows the answer to a four-choice question. The probability that he guesses is \(\frac13\), that he copies is \(\frac16\), and the probability his answer is correct given that he copied is \(\frac18\). The probability that he knew the answer given that he answered correctly is
(a) \(\frac1{4}\)
(b) \(\frac{3}{4}\)
(c) \(\frac{24}{29}\)
(d) \(\frac{5}{29}\)
Show Answer & Explanation

Answer: (c) \(\frac{24}{29}\)

Explanation:
1. P(knows) \(=\frac12\). P(correct) \(=\frac13\cdot\frac14+\frac16\cdot\frac18+\frac12\cdot1=\frac{29}{48}\).
2. P(knew | correct) \(=\dfrac{1/2}{29/48}=\dfrac{24}{29}\).

Teacher's Note:
1. Write down the sample space or use the multiplication rule for independent events.
2. For conditional probability use P(A|B) = P(A and B)/P(B); for causes use Bayes theorem.

Question 4: One ticket is selected at random from 100 tickets numbered 00, 01, ..., 99. If x1 and x2 denote the sum and product of the digits on the ticket, then P(x1 = 9 given x2 = 0) is
(a) \(\frac{19}{100}\)
(b) \(\frac2{19}\)
(c) \(\frac1{50}\)
(d) None of these
Show Answer & Explanation

Answer: (b) \(\frac2{19}\)

Explanation:
1. Tickets with product 0 contain a digit 0: 00 to 09 and 10, 20, ..., 90, giving 19 tickets.
2. Of these only 09 and 90 have digit sum 9, so the probability is \(\frac2{19}\).

Teacher's Note:
1. Write down the sample space or use the multiplication rule for independent events.
2. For conditional probability use P(A|B) = P(A and B)/P(B); for causes use Bayes theorem.

Question 5: A die is rolled three times. \(E_1\): each number is larger than the previous one. \(E_2\): the numbers (in order) form an increasing AP. Then \(P(E_2\mid E_1)\) is
(a) \(\frac3{10}\)
(b) \(\frac1{36}\)
(c) \(\frac{10}3\)
(d) \(\frac12\)
Show Answer & Explanation

Answer: (a) \(\frac3{10}\)

Explanation:
1. \(P(E_1)=\frac{{}^6C_3}{216}=\frac{20}{216}\). Increasing APs: common difference 1 gives 4 triples, difference 2 gives 2, so \(P(E_2)=\frac6{216}\).
2. \(E_2\subset E_1\), so \(P(E_2\mid E_1)=\frac6{20}=\frac3{10}\).

Teacher's Note:
1. Write down the sample space or use the multiplication rule for independent events.
2. For conditional probability use P(A|B) = P(A and B)/P(B); for causes use Bayes theorem.

Question 6: A student passes if he succeeds in tests I and II, or in tests I and III. The probabilities of passing tests I, II, III are \(p\), \(q\) and \(\frac12\). If the probability that the student is successful is \(\frac12\), then
(a) \(p^2+q=1\)
(b) \(pq+p=1\)
(c) \(pq-1=p\)
(d) \(pq+q=1\)
Show Answer & Explanation

Answer: (b) \(pq+p=1\)

Explanation:
1. \(P=p\left[q+\frac12-\frac q2\right]=\frac{p(q+1)}2=\frac12\).
2. So \(pq+p=1\).

Teacher's Note:
1. Write down the sample space or use the multiplication rule for independent events.
2. For conditional probability use P(A|B) = P(A and B)/P(B); for causes use Bayes theorem.

Question 7: In a class 30% of students like tea, 20% like coffee and 10% like both. A student is selected at random. The probability that he does not like tea given that he likes coffee is
(a) \(\frac34\)
(b) \(\frac13\)
(c) \(\frac12\)
(d) None of these
Show Answer & Explanation

Answer: (c) \(\frac12\)

Explanation:
1. P(coffee, not tea) \(=0.2-0.1=0.1\).
2. Divide by P(coffee) \(=0.2\): \(\frac12\).

Teacher's Note:
1. Write down the sample space or use the multiplication rule for independent events.
2. For conditional probability use P(A|B) = P(A and B)/P(B); for causes use Bayes theorem.

Question 8: Abhay speaks the truth 60% of the time. Hasan rolls a die blindfolded and asks Abhay to tell him if the outcome is a prime. Abhay says NO. The probability that the outcome is really prime is
(a) 0.4
(b) 0.5
(c) 0.6
(d) 0.75
Show Answer & Explanation

Answer: (a) 0.4

Explanation:
1. P(prime) \(=\frac12\). P(prime and he lies) \(=0.5\times0.4=0.2\); P(not prime and he tells the truth) \(=0.5\times0.6=0.3\).
2. P(prime | NO) \(=\frac{0.2}{0.5}=0.4\).

Teacher's Note:
1. Write down the sample space or use the multiplication rule for independent events.
2. For conditional probability use P(A|B) = P(A and B)/P(B); for causes use Bayes theorem.

Question 9: n letters are placed at random in n addressed envelopes. The probability that no letter is placed in the right envelope is
(a) \(\frac1{1!}-\frac1{2!}+\frac1{3!}-\cdots+(-1)^n\frac1{n!}\)
(b) \(\frac1{2!}+\frac1{3!}+\frac1{4!}-\cdots+\frac1{n!}\)
(c) \(\frac1{2!}-\frac1{3!}+\frac1{4!}-\cdots+(-1)^n\frac1{n!}\)
(d) None of these
Show Answer & Explanation

Answer: (c) \(\frac1{2!}-\frac1{3!}+\frac1{4!}-\cdots+(-1)^n\frac1{n!}\)

Explanation:
1. By inclusion-exclusion the number of derangements is \(n!\sum_{k=0}^n\frac{(-1)^k}{k!}\).
2. Divide by \(n!\): \(1-1+\frac1{2!}-\frac1{3!}+\cdots=\frac1{2!}-\frac1{3!}+\cdots+(-1)^n\frac1{n!}\).

Teacher's Note:
1. Write down the sample space or use the multiplication rule for independent events.
2. For conditional probability use P(A|B) = P(A and B)/P(B); for causes use Bayes theorem.

Question 10: In independent trials the probability of success on each trial is \(\frac14\). If the trials continue up to the second success, the probability that the second success occurs on the fourth or later trial is
(a) \(\frac{27}{32}\)
(b) \(\frac5{32}\)
(c) \(\frac{23}{32}\)
(d) \(\frac9{32}\)
Show Answer & Explanation

Answer: (a) \(\frac{27}{32}\)

Explanation:
1. P(2nd success on trial 2) \(=\frac1{16}\); on trial 3 \(=2\cdot\frac14\cdot\frac34\cdot\frac14=\frac3{32}\).
2. Required \(=1-\frac2{32}-\frac3{32}=\frac{27}{32}\).

Teacher's Note:
1. Write down the sample space or use the multiplication rule for independent events.
2. For conditional probability use P(A|B) = P(A and B)/P(B); for causes use Bayes theorem.

Question 11: Rajesh does not like to study. The probability that he will study is 1/3. If he studied, the probability that he will fail is 1/2, and if he did not study, the probability that he will fail is 3/4. Given that he failed, the probability that he did not study is
(a) \(\frac23\)
(b) \(\frac34\)
(c) \(\frac13\)
(d) None of these
Show Answer & Explanation

Answer: (b) \(\frac34\)

Explanation:
1. P(fail) \(=\frac13\cdot\frac12+\frac23\cdot\frac34=\frac16+\frac12=\frac23\).
2. P(not studied | fail) \(=\dfrac{1/2}{2/3}=\frac34\).

Teacher's Note:
1. Write down the sample space or use the multiplication rule for independent events.
2. For conditional probability use P(A|B) = P(A and B)/P(B); for causes use Bayes theorem.

Question 12: Two dice are thrown n times in succession. The probability of obtaining a double six at least once is
(a) \(\left(\frac1{36}\right)^n\)
(b) \(\left(\frac1{12}\right)^n\)
(c) \(1-\left(\frac{35}{36}\right)^n\)
(d) None of these
Show Answer & Explanation

Answer: (c) \(1-\left(\frac{35}{36}\right)^n\)

Explanation:
1. P(no double six in one throw) \(=\frac{35}{36}\), so at least once is \(1-\left(\frac{35}{36}\right)^n\).

Teacher's Note:
1. Write down the sample space or use the multiplication rule for independent events.
2. For conditional probability use P(A|B) = P(A and B)/P(B); for causes use Bayes theorem.

Question 13: A random variable X takes the values \(x_1,\dots,x_{10}\) with probabilities \(P(X=x_i)=ki\). Then \(k\) is equal to
(a) \(\frac1{10}\)
(b) \(\frac1{15}\)
(c) \(10\)
(d) \(\frac1{55}\)
Show Answer & Explanation

Answer: (d) \(\frac1{55}\)

Explanation:
1. \(k(1+2+\cdots+10)=55k=1\), so \(k=\frac1{55}\).

Teacher's Note:
1. Write down the sample space or use the multiplication rule for independent events.
2. For conditional probability use P(A|B) = P(A and B)/P(B); for causes use Bayes theorem.

Question 14: A bag contains p white and q black balls. Two players A and B alternately draw a ball each, replacing it, until one draws a white ball and wins. A begins. If the probability of A winning is three times that of B, then \(p:q\) is
(a) \(3:4\)
(b) \(4:3\)
(c) \(2:1\)
(d) \(1:2\)
Show Answer & Explanation

Answer: (c) \(2:1\)

Explanation:
1. Let \(a=\frac p{p+q}\). \(P(A)=\frac{a}{1-(1-a)^2}=\frac1{2-a}\).
2. \(P(A)=\frac34\) gives \(a=\frac23\), so \(p:q=2:1\).

Teacher's Note:
1. Write down the sample space or use the multiplication rule for independent events.
2. For conditional probability use P(A|B) = P(A and B)/P(B); for causes use Bayes theorem.

Question 15: A box contains 20 identical balls of which 10 are blue and 10 are green. Balls are drawn at random one at a time with replacement. The probability that a blue ball is drawn the 4th time on the 7th draw is
(a) \(\frac{27}{32}\)
(b) \(\frac5{32}\)
(c) \(\frac5{64}\)
(d) \(\frac12\)
Show Answer & Explanation

Answer: (b) \(\frac5{32}\)

Explanation:
1. Exactly 3 blue in the first 6 draws, then blue: \({}^6C_3\left(\frac12\right)^7=\frac{20}{128}=\frac5{32}\).

Teacher's Note:
1. Write down the sample space or use the multiplication rule for independent events.
2. For conditional probability use P(A|B) = P(A and B)/P(B); for causes use Bayes theorem.

Question 16: The mean and the variance of a binomial distribution are 4 and 2. The probability of 2 successes is
(a) \(\frac{28}{256}\)
(b) \(\frac{219}{256}\)
(c) \(\frac{128}{256}\)
(d) \(\frac{37}{256}\)
Show Answer & Explanation

Answer: (a) \(\frac{28}{256}\)

Explanation:
1. \(np=4\) and \(npq=2\) give \(q=\frac12\), \(n=8\).
2. \(P(2)={}^8C_2\left(\frac12\right)^8=\frac{28}{256}\).

Teacher's Note:
1. Write down the sample space or use the multiplication rule for independent events.
2. For conditional probability use P(A|B) = P(A and B)/P(B); for causes use Bayes theorem.

Question 17: The probability of the simultaneous occurrence of two events A and B is p. If the probability that exactly one of the events occurs is q, then P(A') + P(B') equals
(a) \(2+2q-p\)
(b) \(2-2p-q\)
(c) \(2-p-q\)
(d) \(2-2q-p\)
Show Answer & Explanation

Answer: (b) \(2-2p-q\)

Explanation:
1. \(P(A)+P(B)=2p+q\) (the intersection is counted twice, exactly-one once).
2. \(P(A')+P(B')=2-(2p+q)=2-2p-q\).

Teacher's Note:
1. Write down the sample space or use the multiplication rule for independent events.
2. For conditional probability use P(A|B) = P(A and B)/P(B); for causes use Bayes theorem.

Question 18: A father has 3 children with at least one boy. The probability that he has 2 boys and 1 girl is
(a) \(\frac14\)
(b) \(\frac13\)
(c) \(\frac23\)
(d) \(\frac37\)
Show Answer & Explanation

Answer: (d) \(\frac37\)

Explanation:
1. Outcomes with at least one boy: \(8-1=7\); with exactly 2 boys and 1 girl: 3.
2. Probability \(=\frac37\).

Teacher's Note:
1. Write down the sample space or use the multiplication rule for independent events.
2. For conditional probability use P(A|B) = P(A and B)/P(B); for causes use Bayes theorem.

Question 19: A man is known to speak the truth 3 out of 4 times. He throws a die and reports that it is a six. The probability that it is actually a six is
(a) \(\frac38\)
(b) \(\frac15\)
(c) \(\frac34\)
(d) \(\frac12\)
Show Answer & Explanation

Answer: (a) \(\frac38\)

Explanation:
1. \(\dfrac{\frac16\cdot\frac34}{\frac16\cdot\frac34+\frac56\cdot\frac14}=\dfrac3{3+5}=\dfrac38\).

Teacher's Note:
1. Write down the sample space or use the multiplication rule for independent events.
2. For conditional probability use P(A|B) = P(A and B)/P(B); for causes use Bayes theorem.

Question 20: The probability of India winning a test match against the Westindies is 1/2, assuming independence from match to match. The probability that in a 5 match series India's second win occurs at the third test is
(a) \(\frac23\)
(b) \(\frac12\)
(c) \(\frac14\)
(d) \(\frac18\)
Show Answer & Explanation

Answer: (c) \(\frac14\)

Explanation:
1. Exactly one win in the first two matches and a win in the third: \(2\cdot\left(\frac12\right)^3=\frac14\).

Teacher's Note:
1. Write down the sample space or use the multiplication rule for independent events.
2. For conditional probability use P(A|B) = P(A and B)/P(B); for causes use Bayes theorem.

Question 21: If \(E_1,E_2\) are events with \(P(E_1)=\frac14\), \(P(E_2\mid E_1)=\frac12\) and \(P(E_1\mid E_2)=\frac14\), which statement is incorrect?
(a) \(E_1\) and \(E_2\) are exhaustive
(b) \(E_1\) and \(E_2\) are independent
(c) \(E_2\) is twice as likely to occur as \(E_1\)
(d) Probabilities of \(E_1\cap E_2\), \(E_1\), \(E_2\) are in G.P.
Show Answer & Explanation

Answer: (a) \(E_1\) and \(E_2\) are exhaustive

Explanation:
1. \(P(E_1\cap E_2)=\frac18\), so \(P(E_2)=\frac{1/8}{1/4}=\frac12\).
2. \(P(E_1\cup E_2)=\frac14+\frac12-\frac18=\frac58\ne1\), so they are not exhaustive; the others hold.

Teacher's Note:
1. Write down the sample space or use the multiplication rule for independent events.
2. For conditional probability use P(A|B) = P(A and B)/P(B); for causes use Bayes theorem.

Question 22: A fair die is tossed eight times. The probability that a third six is observed on the eighth throw is
(a) \(\dfrac{{}^7C_3\,5^3}{6^8}\)
(b) \(\dfrac{{}^7C_2\,5^5}{6^8}\)
(c) \(\dfrac{{}^7C_6\,5^6}{6^8}\)
(d) None of these
Show Answer & Explanation

Answer: (b) \(\dfrac{{}^7C_2\,5^5}{6^8}\)

Explanation:
1. Two sixes in the first seven throws and a six on the eighth: \({}^7C_2\left(\frac16\right)^2\left(\frac56\right)^5\cdot\frac16=\dfrac{{}^7C_2\,5^5}{6^8}\).

Teacher's Note:
1. Write down the sample space or use the multiplication rule for independent events.
2. For conditional probability use P(A|B) = P(A and B)/P(B); for causes use Bayes theorem.

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