JEE Mathematics Probability MCQs Set 29

Multiple Choice Questions (MCQs) for JEE Mathematics: Probability

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

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Multiple Choice Questions

Question 1: The mean and variance of seven observations are 8 and 16, respectively. If 5 of the observations are 2, 4, 10, 12, 14, then the product of the remaining two observations is
(a) 45
(b) 49
(c) 48
(d) 40
Show Answer & Explanation

Answer: (c) 48

Explanation:
1. Let the remaining observations be \( a,b \). Mean: \( \frac{42+a+b}7=8\Rightarrow a+b=14 \).
2. Variance: \( \frac{a^2+b^2+460}7-64=16\Rightarrow a^2+b^2=100 \).
3. \( 2ab=(a+b)^2-(a^2+b^2)=196-100=96 \), so \( ab=48 \). Answer: (c)

Teacher's Note:
1. Use the mean to get \( a+b \).
2. Use the variance to get \( a^2+b^2 \).

Question 2: The minimum number of times one has to toss a fair coin so that the probability of observing at least one head is at least 90% is
(a) 2
(b) 3
(c) 5
(d) 4
Show Answer & Explanation

Answer: (d) 4

Explanation:
1. \( P(\text{at least one head})=1-\frac1{2^n}\ge\frac{90}{100}\Rightarrow2^n\ge10 \).
2. \( 2^3=8\lt10 \) and \( 2^4=16\ge10 \), so the minimum is \( n=4 \). Answer: (d)

Teacher's Note:
1. Use the complement: no heads in \( n \) tosses.
2. Find the smallest \( n \) with \( 2^n\ge10 \).

Question 3: If the standard deviation of the numbers \( -1,0,1,k \) is \( \sqrt5 \) where \( k\gt0 \), then \( k \) is equal to
(a) \( 2\sqrt{\frac{10}3} \)
(b) \( 2\sqrt6 \)
(c) \( 4\sqrt{\frac53} \)
(d) \( \sqrt6 \)
Show Answer & Explanation

Answer: (b) \( 2\sqrt6 \)

Explanation:
1. Mean \( =\frac k4 \). Variance \( =\frac{1+0+1+k^2}4-\frac{k^2}{16}=\frac{8+3k^2}{16} \).
2. \( \frac{8+3k^2}{16}=5\Rightarrow3k^2=72\Rightarrow k^2=24 \), so \( k=2\sqrt6 \) (as \( k\gt0 \)). Answer: (b)

Teacher's Note:
1. Write the variance as mean of squares minus square of mean.
2. Solve for \( k^2 \).

Question 4: The mean and the median of the following ten numbers in increasing order 10, 22, 26, 29, 34, \( x \), 42, 67, 70, \( y \) are 42 and 35 respectively, then \( \dfrac yx \) is equal to
(a) \( \frac73 \)
(b) \( \frac72 \)
(c) \( \frac83 \)
(d) \( \frac94 \)
Show Answer & Explanation

Answer: (a) \( \frac73 \)

Explanation:
1. Mean: \( \frac{300+x+y}{10}=42\Rightarrow x+y=120 \).
2. Median (average of 5th and 6th values): \( \frac{34+x}2=35\Rightarrow x=36 \). Then \( y=84 \).
3. \( \frac yx=\frac{84}{36}=\frac73 \). Answer: (a)

Teacher's Note:
1. Use the mean to get \( x+y \).
2. Use the two middle values for the median.

Question 5: If for some \( x\in\mathbb R \), the frequency distribution of the marks obtained by 20 students in a test is: marks 2, 3, 5, 7 with frequencies \( (x+1)^2,\ 2x-5,\ x^2-3x,\ x \) respectively. Then the mean of the marks is
(a) 3.0
(b) 2.8
(c) 2.5
(d) 3.2
Show Answer & Explanation

Answer: (b) 2.8

Explanation:
1. The frequencies add to 20: \( (x+1)^2+(2x-5)+(x^2-3x)+x=20\Rightarrow2x^2+x-24=0 \), i.e. \( (x+4)(x-3)=0 \), so \( x=3 \) or \( -4 \). A frequency cannot be negative, so \( x=3 \).
2. The frequencies are \( 16,1,0,3 \). Mean \( =\frac{2\cdot16+3\cdot1+5\cdot0+7\cdot3}{20}=\frac{56}{20}=2.8 \). Answer: (b)

Teacher's Note:
1. Use \( \sum f=20 \) to find \( x \).
2. Compute \( \sum fx/\sum f \).

Question 6: Assume that each born child is equally likely to be a boy or a girl. If two families have two children each, then the conditional probability that all children are girls given that at least two are girls is
(a) \( \frac1{17} \)
(b) \( \frac1{12} \)
(c) \( \frac1{10} \)
(d) \( \frac1{11} \)
Show Answer & Explanation

Answer: (d) \( \frac1{11} \)

Explanation:
1. There are 4 children. \( P(\text{4 girls})=\frac1{16} \).
2. \( P(\text{at least 2 girls})=\frac{{}^4C_2+{}^4C_3+{}^4C_4}{16}=\frac{6+4+1}{16}=\frac{11}{16} \).
3. Required probability \( =\frac{1/16}{11/16}=\frac1{11} \). Answer: (d)

Teacher's Note:
1. Count the favourable outcomes among the 16.
2. Use the conditional probability formula.

Question 7: The minimum number of times a fair coin must be tossed so that the probability of getting at least one head is more than 99% is
(a) 8
(b) 6
(c) 7
(d) 5
Show Answer & Explanation

Answer: (c) 7

Explanation:
1. \( 1-\left(\frac12\right)^n\gt\frac{99}{100}\Rightarrow\left(\frac12\right)^n\lt\frac1{100}\Rightarrow2^n\gt100 \).
2. \( 2^6=64\lt100 \) and \( 2^7=128\gt100 \), so the minimum is \( n=7 \). Answer: (c)

Teacher's Note:
1. Use the complement event.
2. Find the smallest \( n \) with \( 2^n\gt100 \).

Question 8: Let a random variable \( X \) have a binomial distribution with mean 8 and variance 4. If \( P(X\le2)=\dfrac k{2^{16}} \), then \( k \) is equal to
(a) 17
(b) 121
(c) 1
(d) 137
Show Answer & Explanation

Answer: (d) 137

Explanation:
1. \( np=8 \) and \( npq=4 \) give \( q=\frac12 \), \( p=\frac12 \) and \( n=16 \).
2. \( P(X\le2)=\frac{{}^{16}C_0+{}^{16}C_1+{}^{16}C_2}{2^{16}}=\frac{1+16+120}{2^{16}}=\frac{137}{2^{16}} \).
3. So \( k=137 \). Answer: (d)

Teacher's Note:
1. Find \( n,p \) from the mean and variance.
2. Add the three binomial probabilities.

Question 9: If the data \( x_1,x_2,\ldots,x_{10} \) is such that the mean of first four of these is 11, the mean of the remaining six is 16 and the sum of squares of all of these is 2000, then the standard deviation of this data is
(a) \( 2\sqrt2 \)
(b) 2
(c) 4
(d) \( \sqrt2 \)
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. The sum of the first four is 44 and of the remaining six is 96, so the total is 140 and the mean is 14.
2. Variance \( =\frac{2000}{10}-14^2=200-196=4 \).
3. The standard deviation is \( 2 \). Answer: (b)

Teacher's Note:
1. Find the overall mean.
2. Use variance \( =\overline{x^2}-\bar x^2 \).

Question 10: For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability that the candidate can solve any problem is \( \frac45 \), then the probability that he is unable to solve less than two problems is
(a) \( \frac{201}5\left(\frac15\right)^{49} \)
(b) \( \frac{316}{25}\left(\frac45\right)^{48} \)
(c) \( \frac{54}5\left(\frac45\right)^{49} \)
(d) \( \frac{164}{25}\left(\frac15\right)^{48} \)
Show Answer & Explanation

Answer: (c) \( \frac{54}5\left(\frac45\right)^{49} \)

Explanation:
1. Let \( X \) be the number of problems he cannot solve: \( X\sim B\left(50,\frac15\right) \). We need \( P(X\lt2)=P(X=0)+P(X=1) \).
2. \( P(X=0)=\left(\frac45\right)^{50} \) and \( P(X=1)=50\cdot\frac15\left(\frac45\right)^{49}=10\left(\frac45\right)^{49} \).
3. Sum \( =\left(\frac45\right)^{49}\left(\frac45+10\right)=\frac{54}5\left(\frac45\right)^{49} \). Answer: (c)

Teacher's Note:
1. Define \( X \) as the number of unsolved problems.
2. Add \( P(X=0) \) and \( P(X=1) \).

Question 11: A person throws two fair dice. He wins Rs. 15 for throwing a doublet (same numbers on the two dice), wins Rs. 12 when the throw results in the sum of 9, and loses Rs. 6 for any other outcome on the throw. Then the expected gain/loss (in Rs.) of the person is
(a) \( \frac12 \) gain
(b) \( \frac14 \) loss
(c) \( \frac12 \) loss
(d) 2 gain
Show Answer & Explanation

Answer: (c) \( \frac12 \) loss

Explanation:
1. Doublets: 6 outcomes. Sum 9: 4 outcomes, \( (3,6),(4,5),(5,4),(6,3) \), none of which is a doublet. Others: \( 36-10=26 \) outcomes.
2. Expected gain \( =15\cdot\frac6{36}+12\cdot\frac4{36}-6\cdot\frac{26}{36}=\frac{90+48-156}{36}=-\frac12 \).
3. So the person expects a loss of Rs. \( \frac12 \). Answer: (c)

Teacher's Note:
1. Count outcomes for each payoff.
2. Compute the expected value.

Question 12: Two cards are drawn successively with replacement from a well shuffled deck of 52 cards. Let \( X \) denote the random variable of number of aces obtained in the two drawn cards. Then \( P(X=1)+P(X=2) \) equals
(a) \( \frac{25}{169} \)
(b) \( \frac{52}{169} \)
(c) \( \frac{49}{169} \)
(d) \( \frac{24}{169} \)
Show Answer & Explanation

Answer: (a) \( \frac{25}{169} \)

Explanation:
1. \( X \) is binomial with \( n=2 \) and \( p=\frac4{52}=\frac1{13} \).
2. \( P(X=1)=2\cdot\frac1{13}\cdot\frac{12}{13}=\frac{24}{169} \) and \( P(X=2)=\frac1{169} \).
3. The sum is \( \frac{25}{169} \). Answer: (a)

Teacher's Note:
1. Identify the binomial parameters.
2. Add the two probabilities.

Question 13: In a group of data, there are \( n \) observations \( x_1,x_2,\ldots,x_n \). If \( \sum_{i=1}^n(x_i+1)^2=9n \) and \( \sum_{i=1}^n(x_i-1)^2=5n \), the standard deviation of the data is
(a) 2
(b) \( \sqrt7 \)
(c) 5
(d) \( \sqrt5 \)
Show Answer & Explanation

Answer: (d) \( \sqrt5 \)

Explanation:
1. Subtract: \( \sum[(x_i+1)^2-(x_i-1)^2]=4\sum x_i=4n \), so the mean is 1.
2. Add: \( 2\sum x_i^2+2n=14n\Rightarrow\sum x_i^2=6n \), so the variance is \( 6-1=5 \).
3. The standard deviation is \( \sqrt5 \). Answer: (d)

Teacher's Note:
1. Subtract and add the two given sums.
2. Use variance \( =\overline{x^2}-\bar x^2 \).

Question 14: An unbiased coin is tossed. If the outcome is a head, then a pair of unbiased dice is rolled and the sum of the numbers obtained on them is noted. If the toss of the coin results in tail, then a card from a well-shuffled pack of nine cards numbered 1, 2, 3, ..., 9 is randomly picked and the number on the card is noted. The probability that the noted number is either 7 or 8 is
(a) \( \frac{15}{72} \)
(b) \( \frac{13}{36} \)
(c) \( \frac{19}{72} \)
(d) \( \frac{19}{36} \)
Show Answer & Explanation

Answer: (c) \( \frac{19}{72} \)

Explanation:
1. After a head: sum 7 or 8 has \( 6+5=11 \) outcomes out of 36, probability \( \frac{11}{36} \).
2. After a tail: the card is 7 or 8 with probability \( \frac29 \).
3. \( P=\frac12\cdot\frac{11}{36}+\frac12\cdot\frac29=\frac{11}{72}+\frac8{72}=\frac{19}{72} \). Answer: (c)

Teacher's Note:
1. Find each conditional probability.
2. Combine using total probability.

Question 15: The mean of five observations is 5 and their variance is 9.20. If three of the given five observations are 1, 3 and 8, then a ratio of other two observations is
(a) 4 : 9
(b) 6 : 7
(c) 10 : 3
(d) 5 : 8
Show Answer & Explanation

Answer: (a) 4 : 9

Explanation:
1. Let the other two be \( x,y \). Mean: \( \frac{12+x+y}5=5\Rightarrow x+y=13 \).
2. Variance: \( \frac{16+4+9+(x-5)^2+(y-5)^2}5=9.2\Rightarrow x^2+y^2-10(x+y)+79=46 \), so \( x^2+y^2=97 \).
3. With \( x+y=13 \): \( xy=\frac{169-97}2=36 \), so \( x,y=4,9 \) and the ratio is \( 4:9 \). Answer: (a)

Teacher's Note:
1. Get \( x+y \) from the mean.
2. Get \( x^2+y^2 \) from the variance.

Question 16: If the probability of hitting a target by a shooter in any shot, is \( \frac13 \), then the minimum number of independent shots at the target required by him so that the probability of hitting the target at least once is greater than \( \frac56 \), is
(a) 6
(b) 3
(c) 5
(d) 4
Show Answer & Explanation

Answer: (c) 5

Explanation:
1. \( 1-\left(\frac23\right)^n\gt\frac56\Rightarrow\left(\frac23\right)^n\lt\frac16 \).
2. For \( n=4 \): \( \frac{16}{81}\approx0.198\gt0.167 \). For \( n=5 \): \( \frac{32}{243}\approx0.132\lt0.167 \). So the minimum is \( n=5 \). Answer: (c)

Teacher's Note:
1. Use the complement event.
2. Test successive values of \( n \).

Question 17: The outcome of each of 30 items was observed; 10 items gave an outcome \( \frac12-d \) each, 10 items gave outcome \( \frac12 \) each and the remaining 10 items gave outcome \( \frac12+d \) each. If the variance of this outcome data is \( \frac43 \), then \( |d| \) equals
(a) \( \frac23 \)
(b) \( \frac{\sqrt5}2 \)
(c) \( \sqrt2 \)
(d) 2
Show Answer & Explanation

Answer: (c) \( \sqrt2 \)

Explanation:
1. The mean is \( \frac12 \) by symmetry.
2. Variance \( =\frac{10d^2+0+10d^2}{30}=\frac{2d^2}3 \).
3. \( \frac{2d^2}3=\frac43\Rightarrow d^2=2 \), so \( |d|=\sqrt2 \). Answer: (c)

Teacher's Note:
1. Find the mean by symmetry.
2. Compute the variance of the deviations.

Question 18: A bag contains 30 white balls and 10 red balls. 16 balls are drawn one by one randomly from the bag with replacement. If \( X \) is the number of white balls drawn, then \( \dfrac{\text{mean of }X}{\text{standard deviation of }X} \) is equal to
(a) \( \frac{4\sqrt3}3 \)
(b) 4
(c) \( 3\sqrt2 \)
(d) \( 4\sqrt3 \)
Show Answer & Explanation

Answer: (d) \( 4\sqrt3 \)

Explanation:
1. \( X\sim B\left(16,\frac34\right) \). Mean \( =12 \) and variance \( =16\cdot\frac34\cdot\frac14=3 \).
2. The ratio is \( \frac{12}{\sqrt3}=4\sqrt3 \). Answer: (d)

Teacher's Note:
1. Identify the binomial parameters.
2. Standard deviation is the square root of variance.

Question 19: Let \( S=\{1,2,\ldots,20\} \). A subset \( B \) of \( S \) is said to be nice, if the sum of the elements of \( B \) is 203. Then the probability that a randomly chosen subset of \( S \) is nice, is
(a) \( \frac6{2^{20}} \)
(b) \( \frac4{2^{20}} \)
(c) \( \frac7{2^{20}} \)
(d) \( \frac5{2^{20}} \)
Show Answer & Explanation

Answer: (d) \( \frac5{2^{20}} \)

Explanation:
1. \( S \) has \( 2^{20} \) subsets and the sum of all elements is \( \frac{20\cdot21}2=210 \).
2. A subset has sum 203 exactly when its complement has sum 7. The subsets with sum 7 are \( \{7\},\{1,6\},\{2,5\},\{3,4\},\{1,2,4\} \), which is 5 subsets.
3. Probability \( =\frac5{2^{20}} \). Answer: (d)

Teacher's Note:
1. Use the complement of a subset.
2. List subsets with sum 7.

Question 20: If the sum of the deviations of 50 observations from 30 is 50, then the mean of these observations is
(a) 50
(b) 30
(c) 51
(d) 31
Show Answer & Explanation

Answer: (d) 31

Explanation:
1. \( \sum(x_i-30)=50\Rightarrow\sum x_i-1500=50\Rightarrow\sum x_i=1550 \).
2. Mean \( =\frac{1550}{50}=31 \). Answer: (d)

Teacher's Note:
1. Expand the sum of deviations.
2. Divide by 50.

Question 21: In a random experiment, a fair die is rolled until two fours are obtained in succession. The probability that the experiment will end in the fifth throw of the die is equal to
(a) \( \frac{175}{6^5} \)
(b) \( \frac{225}{6^5} \)
(c) \( \frac{200}{6^5} \)
(d) \( \frac{150}{6^5} \)
Show Answer & Explanation

Answer: (a) \( \frac{175}{6^5} \)

Explanation:
1. There are \( 6^5 \) equally likely outcomes. Throws 4 and 5 must both be fours.
2. Throw 3 must not be a four (5 ways), and throws 1 and 2 must not be both fours (\( 36-1=35 \) ways).
3. Probability \( =\frac{5\cdot35}{6^5}=\frac{175}{6^5} \). Answer: (a)

Teacher's Note:
1. Fix the last two throws.
2. Avoid an earlier pair of fours.

Question 22: In a game, a man wins Rs. 100 if he gets 5 or 6 on a throw of a fair die and loses Rs. 50 for getting any other number on the die. If he decides to throw the die either till he gets a five or a six or to a maximum of three throws, then his expected gain/loss (in rupees) is
(a) \( \frac{400}3 \) loss
(b) \( \frac{400}9 \) loss
(c) 0
(d) \( \frac{400}3 \) gain
Show Answer & Explanation

Answer: (c) 0

Explanation:
1. \( p=\frac13 \) and \( q=\frac23 \). Gain on first success: Rs. 100; second: \( -50+100=50 \); third: \( -100+100=0 \); all three fail: \( -150 \).
2. Expected \( =\frac13(100)+\frac23\cdot\frac13(50)+\left(\frac23\right)^2\frac13(0)+\left(\frac23\right)^3(-150) \).
3. \( =\frac{100}3+\frac{100}9+0-\frac{1200}{27}=\frac{900+300-1200}{27}=0 \). Answer: (c)

Teacher's Note:
1. List the outcomes of the three throws.
2. Weight each by its probability.

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