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JEE • MATHEMATICS • PROBABILITY

Probability practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 151-180 of 1130 questions

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151 Medium
151 of 1130
Application +4 | -1
Probability of throwing 16 in one throw with three dice is
View answer and explanation
Correct option: A
136\frac{1}{{36}}
Option A (136\frac{1}{{36}}) is correct because there are 6 favorable outcomes out of 216 total possible outcomes when throwing three dice.
152 Medium
152 of 1130
Application +4 | -1
A,B{A}, {B} and C{C} are three mutually exclusive and exhaustive events. If 13P(C)=12P(A)=P(B)\frac{1}{3} {P}({C})=\frac{1}{2} {P}({A})={P}({B}), then P(B)={P}({B})=
View answer and explanation
Correct option: C
16\frac{1}{6}
Option C (16\frac{1}{6}) is correct because it calculates the probability of event B given that A, B, and C are mutually exclusive and exhaustive.
153 Medium
153 of 1130
Application +4 | -1
10 persons sit around a circular table with 10 numbered chairs. The probability that the two particular persons AA and BB are always together is
View answer and explanation
Correct option: A
29\frac{2}{9}
Option A (29\frac{{2}}{{9}}) is correct because the probability that two particular persons A and B are always together is 2/9.
154 Easy
154 of 1130
Analyzing +4 | -1
One Indian and four American men and their wifes are to be seated randomly around a circular table. Then the conditional probability that the Indian man is seated adjacent to his wife given that each American man is seated adjacent to his wife, is
View answer and explanation
Correct option: C
25\frac{2}{5}
Option C (2/5) is correct because it represents the conditional probability that the Indian man is seated next to his wife, given that each American man is seated next to his wife.
155 Easy
155 of 1130
Application +4 | -1
An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colours is
View answer and explanation
Correct option: D
2/72 / 7
Option D (2/7) is correct because it calculates the probability of drawing three balls of different colors from the urn.
156 Easy
156 of 1130
Application +4 | -1
If the range of a random variable XX is {0,1,2,3,4,…} with P(X=k)=(k+1)a3kP\left( {X = k} \right) = \frac{{\left( {k + 1} \right)a}}{{{3^k}}} for k≥0,k \ge 0, then aa is equal to
View answer and explanation
Correct option: B
49\frac{4}{9}
Option B (49\frac{4}{9}) is correct because it is the value of 'a' that makes the sum of probabilities for all possible values of X equal to 1.
157 Easy
157 of 1130
Understanding +4 | -1
Probability of all 3 digit numbers having all the digits same is
View answer and explanation
Correct option: A
1100\frac{1}{{100}}
Option A (1100\frac{1}{{100}}) is correct because there are 9 favorable cases (111, 222, ..., 999) out of 900 total 3-digit numbers.
158 Easy
158 of 1130
Analyzing +4 | -1
For two events AA and BB, if P(A)=P(AB)=14P\left( A \right) = P\left( {\frac{A}{B}} \right) = \frac{1}{4} and P(BA)=12,P\left( {\frac{B}{A}} \right) = \frac{1}{2}, then
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because given the probabilities, events A and B are independent, and the conditional probabilities P(A′B)P(\frac{{A'}}{B}) and P(B′A)P(\frac{{B'}}{A}) can be derived.
159 Easy
159 of 1130
Application +4 | -1
A random variable X takes values −1,0,1,2-1,0,1,2 with probabilities 1+3p4,1−p4,1+2p4,1−4p4\frac{1+3 p}{4}, \frac{1-p}{4}, \frac{1+2 p}{4}, \frac{1-4 p}{4} respectively, where pp varies over RR. Then the minimum and maximum values of the mean of XX are respectively
View answer and explanation
Correct option: D
−116-\frac{1}{16} and 54\frac{5}{4}
Option D (−116-\frac{1}{16} and 54\frac{5}{4}) is correct because the minimum and maximum values of the mean are found by considering the range of possible values for p.
160 Easy
160 of 1130
Application +4 | -1
Two integers xx and yy are chosen with replacement out of the set {0,1,2,3,…,10}.\{0,1,2,3, \ldots, 10\}. Then the probability that ∣x−y∣>5,|x-y|>5, is
View answer and explanation
Correct option: B
30121\frac{30}{121}
Option B (30121\frac{30}{121}) is correct because it is the probability that |x-y|>5, calculated by finding the number of favorable pairs and dividing by the total number of pairs.
161 Medium
161 of 1130
Application +4 | -1
The odds in favour of a book reviewed by three independent critics are, respectively, 5:2,4:35: 2,4: 3 and 3:43: 4. The probability that majority of the critics give favourable remark is
View answer and explanation
Correct option: B
209343\frac{209}{343}
Option B (209343\frac{209}{343}) is correct because it represents the probability that a majority of the three critics give a favorable review.
162 Easy
162 of 1130
Application +4 | -1
A die is tossed thrice. A success is getting 1 or 6 on a toss. The mean and the variance of number of successes
View answer and explanation
Correct option: A
μ=1,σ2=2/3\mu=1, \sigma^2=2 / 3
Option A (μ=1,σ2=2/3\mu=1, \sigma^2=2 / 3) is correct because the mean and variance are calculated using the formulas for binomial distribution: mean = np and variance = npq.
163 Medium
163 of 1130
Application +4 | -1
x1,x2,x3,……x50\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \ldots \ldots \mathrm{x}_{50} are fifty real numbers such that xr<xr+1x_r < x_{r+1} for r=1,2,3,….49r=1,2,3, \ldots .49. Five numbers out of these are picked up at random and they are arranged in increasing order. The probability that the five numbers have x20x_{20} as the middle number is :
View answer and explanation
Correct option: B
30C2×19C250C5\frac{{ }^{30} \mathrm{C}_2 \times{ }^{19} \mathrm{C}_2}{{ }^{50} \mathrm{C}_5}
Option B (19C2×30C350C5\frac{{{{}^{19}C_2 \times {{}^{30}C_3}}}}{{{{}^{50}C_5}}}) is correct because it correctly calculates the probability of selecting five numbers with x20x_{{20}} as the middle number.
164 Medium
164 of 1130
Application +4 | -1
A contest consists of predicting the results (win, draw or defeat) of 10 football matches. The probability that one entry contains at least 5 correct answers is
View answer and explanation
Correct option: A
12585310\frac{12585}{3^{10}}
Option A (12585310\frac{12585}{3^{10}}) is correct because it represents the probability of having at least 5 correct predictions out of 10 matches, considering each match has 3 possible outcomes.
165 Medium
165 of 1130
Analyzing +4 | -1
Two persons each make a single throw with a pair of dice. The probability that the throws are equal, is
View answer and explanation
Correct option: A
73648\frac{73}{648}
Option A (73648\frac{{73}}{{648}}) is correct because it represents the probability calculated by dividing the number of ways the throws are equal by the total number of possible outcomes.
166 Easy
166 of 1130
Application +4 | -1
A coin is tossed three times. The probability of getting head and tail alternatively, is
View answer and explanation
Correct option: C
14\frac{1}{4}
Option C (213\frac{2}{{13}}) is correct because the mean of the number of aces is calculated using the probabilities of selecting and not selecting aces in two draws with replacement.
167 Easy
167 of 1130
Application +4 | -1
A man and his wife appear for an interview for two posts. The probability of the man’s selection is 1/5 and that of his wife’s selection is 1/7. The probability that at least one of them is selected, is
View answer and explanation
Correct option: D
1135\frac{{11}}{{35}}
Option D (1135\frac{{11}}{{35}}) is correct because it calculates the probability of at least one selection using the formula for the union of two independent events.
168 Easy
168 of 1130
Application +4 | -1
A die is tossed twice. The probability of having a number greater than 4 on each toss is
View answer and explanation
Correct option: B
19\frac{1}{9}
Option B (19\frac{1}{9}) is correct because it represents the probability of getting a number greater than 4 on each of the two tosses of a die.
169 Easy
169 of 1130
Application +4 | -1
A bag A contains 2 white and 3 red balls, and a bag B contains 4 white and 5 red balls. One ball is drawn at random from one of the bags and it is found to be red. Let pp be the probability that it was drawn from the bag B then find the value of [1p]\left[\frac{1}{p}\right] where [.] denotes G.I.F.
View answer and explanation
Correct option: A
2
Option A (2) is correct because the probability that the red ball was drawn from bag B is 25/52, and the greatest integer function of its reciprocal is 2.
170 Medium
170 of 1130
Application +4 | -1
If three students A, B, CA,\,B,\,C can solve a problem with probabilities 13, 14\frac{1}{3},\,\frac{1}{4} and 15\frac{1}{5} respectively, then the probability that the problem will be solved is
View answer and explanation
Correct option: A
35\frac{3}{5}
Option A (3/5) is correct because it calculates the probability that at least one student solves the problem by finding the complement (probability that none solve it).
171 Medium
171 of 1130
Application +4 | -1
Let NN denote the sum of the numbers obtained when two dice are rolled. If the probability that 2N<N!2^N < N ! is mn\frac{m}{n}, where mm and nn are coprime, then 4m−3n4 m-3 n is equal to
View answer and explanation
Correct option: D
8
Option D (8) is correct because it involves calculating the probability of the event 2N<N!2^N < N! and then finding the value of 4m−3n4m - 3n.
172 Easy
172 of 1130
Application +4 | -1
Three persons, A, B and C\mathrm{C}, fire at a target in turn, starting with A\mathrm{A}. Their probability of hitting the target are 0.4,0.30.4,0.3 and 0.2 respectively. The probability of two hits is
View answer and explanation
Correct option: B
0.188
Option B (0.188) is correct because it is the sum of probabilities of exactly two hits by A, B, and C in all possible combinations.
173 Medium
173 of 1130
Understanding +4 | -1
A fair coin is tossed repeatedly. If the tail appears on first four tosses, then the probability of the head appearing on the fifth toss, equals
View answer and explanation
Correct option: A
12\frac{1}{2}
Option A (12\frac{1}{2}) is correct because coin tosses are independent events, so the previous results don't affect the probability of the next toss.
174 Easy
174 of 1130
Analyzing +4 | -1
If nn integers taken at random are multiplied together, then the probability that the last digit of the product is 1, 3, 7 or 9, is
View answer and explanation
Correct option: A
2n5n\frac{{{2^n}}}{{{5^n}}}
Option A, 2n5n\frac{{{2^n}}}{{{5^n}}}, is correct because it represents the probability of the last digit being 1, 3, 7, or 9 after multiplying n random integers.
175 Easy
175 of 1130
Application +4 | -1
The probability of getting the sum more than 7 when a pair of dice is tossed is
View answer and explanation
Correct option: D
512\frac{5}{12}
Option D (512\frac{5}{12}) is correct because it represents the probability of getting a sum greater than 7 when tossing a pair of dice, calculated by counting favorable outcomes and dividing by the total possible outcomes.
176 Easy
176 of 1130
Application +4 | -1
′X′'X' speaks truth in 60% and ′Y′'Y' in 50% of the cases. The probability that they contradict each other narrating the same incident is
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (1/2) is correct because it correctly calculates the probability of the two people contradicting each other by considering the cases where one speaks the truth and the other lies.
177 Easy
177 of 1130
Application +4 | -1
Two coins and a die are tossed. The probability that both coins fall heads and the die shows a 3 or 6, is
View answer and explanation
Correct option: B
1/121/12
Option B (112\frac{1}{12}) is correct because it multiplies the probabilities of each independent event occurring.
178 Medium
178 of 1130
Application +4 | -1
Let a die be rolled nn times. Let the probability of getting odd numbers seven times be equal to the probability of getting odd numbers nine times. If the probability of getting even numbers twice is k215\frac{k}{2^{15}}, then kk is equal to
View answer and explanation
Correct option: A
60
Option A (60) is correct because after setting up the probabilities and solving for n, the probability of getting even numbers twice can be calculated, leading to k=60.
179 Easy
179 of 1130
Application +4 | -1
The probability that in a family of 5 members, exactly 2 members have birthday on sunday, is
View answer and explanation
Correct option: D
10×6375\frac{{10 \times {6^3}}}{{{7^5}}}
Option D is correct because it applies the binomial probability formula to calculate the probability of exactly 2 members having a birthday on Sunday.
180 Easy
180 of 1130
Application +4 | -1
Let a die is loaded in such a way that prime number faces are twice as likely to occur as a non-prime number faces. Then, the probability that an odd number will be show up when the die is tossed, is
View answer and explanation
Correct option: D
59\frac{5}{9}
Option D is correct because the probability is calculated by assigning weights to prime and non-prime numbers and then finding the probability of an odd number appearing.
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