JEE Mathematics Probability MCQs Set 33

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Review structured MCQ sets for JEE Mathematics Probability. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Chapter-wise Objective Questions: Probability

Navigate directly to the 50 objective questions for Probability using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: For events A, B: \(P(A\cap B')=0.20\), \(P(A'\cap B)=0.15\), \(P(A'\cap B')=0.1\). Then \(P(A\mid B)\) equals
(a) \(\frac2{11}\)
(b) \(\frac27\)
(c) \(\frac{11}{14}\)
(d) \(\frac17\)
Show Answer & Explanation

Answer: (c) \(\frac{11}{14}\)

Explanation:
1. \(P(A\cap B)=1-0.20-0.15-0.10=0.55\) and \(P(B)=0.55+0.15=0.70\).
2. \(P(A\mid B)=\frac{0.55}{0.70}=\frac{11}{14}\).

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: Three letters are written to three different persons and addresses on the three envelopes are also written. Without looking at the addresses, the letters are put in the envelopes. The probability that all the letters are not placed in their right envelopes is
(a) \(\frac12\)
(b) \(\frac13\)
(c) \(\frac16\)
(d) \(\frac56\)
Show Answer & Explanation

Answer: (d) \(\frac56\)

Explanation:
1. P(all in right envelopes) \(=\frac1{3!}=\frac16\), so the required probability is \(\frac56\).

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: If X follows a binomial distribution with \(n=8\) and \(p=\frac12\), then \(P(|X-4|\le2)\) is
(a) \(\frac{119}{228}\)
(b) \(\frac{19}{128}\)
(c) \(\frac{119}{128}\)
(d) \(\frac{18}{128}\)
Show Answer & Explanation

Answer: (c) \(\frac{119}{128}\)

Explanation:
1. Complement: \(X\in\{0,1,7,8\}\), probability \(\frac{1+8+8+1}{256}=\frac{18}{256}\).
2. So \(P=1-\frac{9}{128}=\frac{119}{128}\).

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: A random variable has the distribution \( P(x) = 0, 2p, 2p, 3p, p^2, 2p^2, 7p^2, 2p \) for x = 0, 1, ..., 7. The value of p is
(a) \(-1\)
(b) \(\frac1{10}\)
(c) \(\frac3{10}\)
(d) None of these
Show Answer & Explanation

Answer: (b) \(\frac1{10}\)

Explanation:
1. \(10p^2+9p=1\), so \((10p-1)(p+1)=0\).
2. \(p=\frac1{10}\) (since \(p>0\)).

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 boy throws stones at a target; the probability of hitting at any trial is \(\frac12\). The probability of hitting the target the 5th time on the 10th throw is
(a) \(\frac{63}{2^9}\)
(b) \(\frac5{2^{10}}\)
(c) \(\frac{{}^{10}C_5}{2^{10}}\)
(d) None of these
Show Answer & Explanation

Answer: (a) \(\frac{63}{2^9}\)

Explanation:
1. Four hits in the first nine throws, then a hit: \({}^9C_4\left(\frac12\right)^{10}=\frac{126}{1024}=\frac{63}{2^9}\).

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: There are n letters and n addressed envelopes. The probability that all the letters are not kept in the right envelopes is
(a) \(\frac1{n!}\)
(b) \(1-\frac1{n!}\)
(c) \(1-\frac1n\)
(d) None of these
Show Answer & Explanation

Answer: (b) \(1-\frac1{n!}\)

Explanation:
1. P(all right) \(=\frac1{n!}\), so the complement is \(1-\frac1{n!}\).

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 7: A man has a bunch of 10 keys of which only one can open the door. He chooses a key at random; each wrong key is discarded. The probability that the door is opened on the fifth trial is
(a) \(\frac12\)
(b) \(\frac{{}^{10}C_5}{10^5}\)
(c) \(\frac{5!}{10!}\)
(d) \(\frac1{10}\)
Show Answer & Explanation

Answer: (d) \(\frac1{10}\)

Explanation:
1. The correct key is equally likely to be in any position: \(\frac9{10}\cdot\frac89\cdot\frac78\cdot\frac67\cdot\frac16=\frac1{10}\).

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 8: The mode of the binomial distribution for which the mean and standard deviation are 10 and sqrt5 respectively is
(a) 10
(b) 7
(c) 8
(d) 9
Show Answer & Explanation

Answer: (a) 10

Explanation:
1. \(np=10\), \(npq=5\) give \(q=\frac12\), \(n=20\).
2. \((n+1)p=10.5\), so the mode is \(10\).

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 9: Three friends A, B and C play a game throwing a die one after the other; the one who first gets a composite number wins. A starts, then B, then C. The ratio of their winning probabilities is
(a) 8 : 6 : 5
(b) 10 : 5 : 4
(c) 9 : 6 : 4
(d) None of these
Show Answer & Explanation

Answer: (c) 9 : 6 : 4

Explanation:
1. P(composite) \(=\frac26=\frac13\) (4 or 6). \(P(A)=\dfrac{1/3}{1-(2/3)^3}=\frac9{19}\).
2. \(P(B)=\frac23P(A)=\frac6{19}\), \(P(C)=\frac49\cdot\frac9{19}=\frac4{19}\).

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 10: A is one of 6 horses in a race and is to be ridden by one of two jockeys B and C. It is 2 to 1 that B rides A, in which case all horses are equally likely to win. If C rides A, his chance of winning is trebled. The odds against A winning are
(a) 13 : 5
(b) 5 : 13
(c) 5 : 18
(d) None of these
Show Answer & Explanation

Answer: (a) 13 : 5

Explanation:
1. P(win) \(=\frac23\cdot\frac16+\frac13\cdot\frac12=\frac19+\frac16=\frac5{18}\).
2. Odds against \(=\frac{13}{18}:\frac5{18}=13:5\).

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: An anti-aircraft gun can take a maximum of four shots at a plane. The probabilities of hitting at the 1st, 2nd, 3rd and 4th shots are 0.4, 0.3, 0.2 and 0.1. The probability that at least one shot hits the plane is
(a) 0.3024
(b) 0.6976
(c) 0.72
(d) 0.6431
Show Answer & Explanation

Answer: (b) 0.6976

Explanation:
1. P(no hit) \(=0.6\times0.7\times0.8\times0.9=0.3024\).
2. P(at least one hit) \(=0.6976\).

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: If \(P(A)=P(A\mid B)=\frac14\) and \(P(B\mid A)=\frac12\), then \(P(\bar A\mid B)\) equals
(a) \(\frac14\)
(b) \(\frac12\)
(c) \(\frac34\)
(d) \(\frac{3}{8}\)
Show Answer & Explanation

Answer: (c) \(\frac34\)

Explanation:
1. \(P(A\cap B)=\frac14\cdot\frac12=\frac18\), so \(P(B)=\frac{1/8}{1/4}=\frac12\).
2. \(P(A\mid B)=P(A)\), i.e. independent, so \(P(\bar A\mid B)=1-\frac14=\frac34\).

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 13: One hundred identical coins, each with probability \(p\) of showing heads (\(0<p<1\)), are tossed. If the probability of heads on 50 coins equals that on 51 coins, the value of \(p\) is
(a) \(\frac12\)
(b) \(\frac{49}{101}\)
(c) \(\frac{50}{101}\)
(d) \(\frac{51}{101}\)
Show Answer & Explanation

Answer: (d) \(\frac{51}{101}\)

Explanation:
1. \({}^{100}C_{50}p^{50}q^{50}={}^{100}C_{51}p^{51}q^{49}\) gives \(q=\frac{50}{51}p\).
2. \(51(1-p)=50p\), so \(p=\frac{51}{101}\).

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: In a school 50% of students like chocolate, 30% like cake and 10% like both. A student is selected at random. The probability that he likes chocolate given that he likes cake is
(a) \(\frac25\)
(b) \(\frac35\)
(c) \(\frac13\)
(d) None of these
Show Answer & Explanation

Answer: (c) \(\frac13\)

Explanation:
1. \(P(C\mid K)=\frac{0.1}{0.3}=\frac13\).

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 15: The probability of guessing correctly at least 8 out of 10 answers on a true-false examination is
(a) \(\frac5{128}\)
(b) \(\frac7{128}\)
(c) \(\frac{19}{128}\)
(d) \(\frac{11}{128}\)
Show Answer & Explanation

Answer: (b) \(\frac7{128}\)

Explanation:
1. \(\frac{{}^{10}C_8+{}^{10}C_9+{}^{10}C_{10}}{2^{10}}=\frac{45+10+1}{1024}=\frac7{128}\).

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: A fair coin is tossed 99 times. If X is the number of heads, P(X = r) is maximum when r is
(a) 49 or 50
(b) 50 or 51
(c) 51
(d) None of these
Show Answer & Explanation

Answer: (a) 49 or 50

Explanation:
1. For \(n=99\), \((n+1)p=50\) is an integer, so the maximum occurs at \(r=49\) and \(r=50\).

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 that TB is detected when a person actually suffers is 0.99. The probability of a healthy person being diagnosed to have TB is 0.001. In a city, 1 in 1000 people suffers from TB. A person is selected at random and diagnosed to have TB. The probability that he actually has TB is
(a) \(\frac2{223}\)
(b) \(\frac{110}{221}\)
(c) \(\frac{110}{223}\)
(d) \(\frac1{221}\)
Show Answer & Explanation

Answer: (b) \(\frac{110}{221}\)

Explanation:
1. \(\dfrac{0.001\times0.99}{0.001\times0.99+0.999\times0.001}=\dfrac{990}{1989}=\dfrac{110}{221}\).

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 problem is given to three students A, B, C whose probabilities of solving it are \(\frac12,\frac13,\frac14\). The probability that the problem is solved is
(a) \(\frac12\)
(b) \(\frac23\)
(c) \(\frac13\)
(d) \(\frac34\)
Show Answer & Explanation

Answer: (d) \(\frac34\)

Explanation:
1. \(1-\frac12\cdot\frac23\cdot\frac34=1-\frac14=\frac34\).

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 19: A and B are independent witnesses. The probability that A speaks the truth is x and that B speaks the truth is y. A and B agree in a certain statement. The probability that the statement is true is
(a) \(\dfrac{xy}{1-x-y+2xy}\)
(b) \(\dfrac{x-y}{x+y}\)
(c) \(\dfrac{xy}{1+x+y+xy}\)
(d) \(\dfrac{x-y}{1-x-y+2xy}\)
Show Answer & Explanation

Answer: (a) \(\dfrac{xy}{1-x-y+2xy}\)

Explanation:
1. Both true: \(xy\); both false: \((1-x)(1-y)\).
2. Probability \(=\dfrac{xy}{xy+(1-x)(1-y)}=\dfrac{xy}{1-x-y+2xy}\).

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: A coin is tossed thrice. If E is the event of showing at least two heads and F is the event of showing head on the first throw, then P(E | F) is
(a) \(\frac43\)
(b) \(\frac14\)
(c) \(\frac34\)
(d) \(\frac12\)
Show Answer & Explanation

Answer: (c) \(\frac34\)

Explanation:
1. Given the first is a head, at least one head is needed in the other two throws: \(1-\frac14=\frac34\).

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 21: A man takes a step forward with probability 0.4 and backward with probability 0.6. The probability that at the end of eleven steps he is one step away from the starting point is
(a) \(462\times\left(\frac6{25}\right)^5\)
(b) \(\dfrac{2^5\cdot3^5}{5^{10}}\)
(c) \(231\times\dfrac{3^5}{5^{10}}\)
(d) None of these
Show Answer & Explanation

Answer: (a) \(462\times\left(\frac6{25}\right)^5\)

Explanation:
1. 6 forward and 5 backward or 5 forward and 6 backward: \({}^{11}C_5(0.4)^5(0.6)^6+{}^{11}C_5(0.4)^6(0.6)^5\).
2. \(=462(0.24)^5(0.6+0.4)=462\left(\frac6{25}\right)^5\).

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: For a biased die the probabilities of faces 1 to 6 are 0.10, 0.32, 0.21, 0.15, 0.05, 0.17. The die is tossed and it is told that either face 1 or face 2 has shown up. The probability that it is face 1 is
(a) \(\frac{16}{21}\)
(b) \(\frac5{21}\)
(c) \(\frac1{10}\)
(d) \(\frac5{16}\)
Show Answer & Explanation

Answer: (b) \(\frac5{21}\)

Explanation:
1. \(\dfrac{0.10}{0.10+0.32}=\dfrac{10}{42}=\dfrac5{21}\).

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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FAQs

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