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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 361-390 of 1130 questions

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361 Medium
361 of 1130
Application +4 | -1
Let S={M=[aij],aij∈}{0,1,2},{1≤i,j≤2}S=\left\{M=\left[a_{i j}\right], a_{i j} \in\right\}\{0,1,2\},\{1 \leq i, j \leq 2\} be a sample space and A{M∈S:MA\{M \in S: M is invertible }\} be an even. Then P(A)P(A) is equal to
View answer and explanation
Correct option: D
5081\frac{50}{81}
Option D (50/81) is correct because it represents the probability of the matrix being invertible, calculated by subtracting the probability of the matrix being non-invertible from 1.
362 Medium
362 of 1130
Analyzing +4 | -1
If P(B)=34,  P(A∩B∩Cˉ)=13  P\left( B \right) = \frac{3}{4},\;P\left( {A \cap B \cap \bar C} \right) = \frac{1}{3}\; and P(Aˉ∩B∩Cˉ)=13,  P\left( {\bar A \cap B \cap \bar C} \right) = \frac{1}{3},\; then P(B∩C)P\left( {B \cap C} \right) is
View answer and explanation
Correct option: A
112\frac{1}{{12}}
Option A (112\frac{1}{{12}}) is correct because it is the calculated probability of the intersection of events B and C, given the provided probabilities.
363 Medium
363 of 1130
Analyzing +4 | -1
If nn positive integers are taken at random and multiplied together, the probability that the last digit of the product is 2, 4, 6 or 8, is
View answer and explanation
Correct option: C
4n−2n5n\frac{{{4^n} - {2^n}}}{{{5^n}}}
Option C (4n−2n5n\frac{{{4^n} - {2^n}}}{{{5^n}}}) is correct because it calculates the probability that the product of n random integers ends in 2, 4, 6, or 8.
364 Easy
364 of 1130
Application +4 | -1
If AA and BB are two independent events, then the probability that only one of AA and BB occur is
View answer and explanation
Correct option: A
P(A)+P(B)−2  P  (A∩B)P\left( A \right) + P\left( B \right) - 2\;P\;\left( {A \cap B} \right)
Option A (P(A)+P(B)−2P(A∩B)P(A) + P(B) - 2P(A \cap B)) is correct because it represents the probability that only one of A and B occurs, calculated using the probabilities of A and B and their intersection.
365 Medium
365 of 1130
Analyzing +4 | -1
In a bag there are 10 balls of either Black or White. Let AkA_k be the event which contains exactly kk white balls and its probability is proportional to k2k^2 Now an another ball is drawn randomly. If P(E)P(E) be the probability that the ball drawn is white and P(E)=abP(E)=\frac{a}{b} Where HCF⁡(a,b)=1\operatorname{HCF}(a, b)=1 then the value of ∣a−b∣|\mathrm{a}-\mathrm{b}| equals
View answer and explanation
Correct option: D
3
Option D (3) is correct because it calculates the probability of drawing a white ball and then finds the absolute difference between the numerator and denominator of the simplified probability.
366 Easy
366 of 1130
Application +4 | -1
A person draws out two balls successively from a bag containing 6 red and 4 white balls. The probability that at least one of them will be red, is
View answer and explanation
Correct option: A
7890\frac{{78}}{{90}}
Option A (7890\frac{{78}}{{90}}) is correct because it calculates the probability of at least one red ball being drawn by considering all possible favorable outcomes (WR, RW, RR).
367 Medium
367 of 1130
Application +4 | -1
The number of times a die must be tossed to obtain a 6 at least once with probability exceeding 0.9 is at least
View answer and explanation
Correct option: A
13
Option A (13) is correct because it is the smallest integer value of n for which the probability of obtaining a 6 at least once exceeds 0.9.
368 Easy
368 of 1130
Understanding +4 | -1
Let AA and BB be two non-null events such that A⊂BA \subset B. Then, which of the following statements is always correct.
View answer and explanation
Correct option: B
P(A/B)≥P(A)P(A/B) \ge P(A)
Option B (P(A/B)≥P(A)P(A/B) \ge P(A)) is correct because since A is a subset of B, P(A/B) = P(A)/P(B), and since P(B) is less than or equal to 1, P(A/B) is greater than or equal to P(A).
369 Easy
369 of 1130
Application +4 | -1
Let XX be a binomially distributed random variable with mean 4 and variance 43\frac{4}{3}. Then, 54P(X≤2)54 P(X \leq 2) is equal to
View answer and explanation
Correct option: B
14627\frac{146}{27}
Option B is correct because we can find n and p from the mean and variance, then calculate P(X <= 2) and multiply by 54.
370 Easy
370 of 1130
Understanding +4 | -1
Which of the following is true?

I. If the empty set ϕ\phi and the sample space describe events, then ϕ\phi is an impossible event.

II. In the above statement, the whole sample space S{S} is called the sure event.
View answer and explanation
Correct option: C
Both I and II are true
Option C (Both I and II are true) is correct because the empty set is indeed an impossible event, and the entire sample space is a sure event.
371 Easy
371 of 1130
Application +4 | -1
An integer is chosen at random from first two hundred numbers. Then, the probability that the integer chosen is divisible by 6 or 8 is
View answer and explanation
Correct option: A
14\frac{1}{4}
Option A (14\frac{1}{4}) is correct because the probability of choosing a number divisible by 6 or 8 is calculated using the principle of inclusion-exclusion.
372 Easy
372 of 1130
Application +4 | -1
From a pack of 52 cards, two cards are drawn one by one without replacement. The probability that first drawn card is a king and second is a queen, is
View answer and explanation
Correct option: C
4663\frac{4}{663}
Option C (4663\frac{{4}}{{663}}) is correct because the probability of drawing a king first and then a queen without replacement is calculated as 452×451\frac{{4}}{{52}} \times \frac{{4}}{{51}}.
373 Easy
373 of 1130
Analyzing +4 | -1
The probabilities of three events A, B and C are given by P(A) = 0.6, P(B) = 0.4 and P(C) = 0.5. If P(A∪B)=0.8,   P(A∩C)=0.3,    P(A∩B∩C)=0.2,    P(B∩C)=βP\left( {A \cup B} \right) = 0.8,\,\,\,P\left( {A \cap C} \right) = 0.3,\,\,\,\,P\left( {A \cap B \cap C} \right) = 0.2,\,\,\,\,P\left( {B \cap C} \right) = \beta and P(A∪B∪C)=α,P\left( {A \cup B \cup C} \right) = \alpha , where 0.85 ≤α≤0.95,0.85\, \le \alpha \le 0.95, then β\beta lies in the interval
View answer and explanation
Correct option: B
[0.25, 0.35]
Option B ([0.25, 0.35]) is the correct interval for β\beta because it satisfies the probability constraints given the range of α\alpha and the inclusion-exclusion principle.
374 Medium
374 of 1130
Analyzing +4 | -1
Let E′E^{\prime} denote the complement of an event EE. Let E,FE, F, GG be pairwise independent events such that P(G)>0P(G)>0 and P(E∩F∩G)=0P(E \cap F \cap G)=0. Then P(E′∩F′/G)P\left(E^{\prime} \cap F^{\prime} / G\right) equals
View answer and explanation
Correct option: C
P(E′)−P(F)P\left(E^{\prime}\right)-P(F)
The correct answer is C (P(E′)−P(F)P(E')-P(F)) because it correctly applies the properties of pairwise independent events and conditional probability to simplify the expression.
375 Easy
375 of 1130
Understanding +4 | -1
The probability that a number selected at random from the set of numbers 1, 2, 3, 4, …, 100 is a cube, is
View answer and explanation
Correct option: A
125\frac{1}{{25}}
Option A (125\frac{1}{{25}}) is correct because there are 4 perfect cubes (1, 8, 27, 64) between 1 and 100, so the probability is 4/100 = 1/25.
376 Easy
376 of 1130
Analyzing +4 | -1
If A\mathrm{A} and B\mathrm{B} are two events such that P(A)=47,P(A∩B)=328P(A)=\frac{4}{7}, P(A \cap B)=\frac{3}{28} and the conditional probability P(AAc∪Bc)P\left(\frac{A}{A^c \cup B^c}\right) (where AcA^c denotes the compliment of the event A\mathrm{A} ) is equal to λ\lambda, then the value of 13λ\frac{13}{\lambda} is equal to
View answer and explanation
Correct option: A
25
The correct answer is 25 because the conditional probability P(AAc∪Bc)P(\frac{A}{A^c \cup B^c}) can be simplified using De Morgan's law and the given probabilities to find λ\lambda, then calculate 13λ\frac{13}{\lambda}.
377 Medium
377 of 1130
Application +4 | -1
A bag contains 5 white, 6 green, 7 blue, 8 red and 9 black balls. If 3 balls are drawn one by one with replacement after each drawing, then the probability that 3rd3^{\text {rd}} ball drawn is green is, if it is known that second ball is green:
View answer and explanation
Correct option: C
635\frac{6}{35}
Option C (6/35) is correct because the probability of drawing a green ball on the third draw is independent of the second draw due to replacement, and the total number of green balls is 6 out of 35 total balls.
378 Easy
378 of 1130
Application +4 | -1
Bag A contains 3 white, 7 red balls and Bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from the bag A, if the ball drawn is white, is
View answer and explanation
Correct option: A
13\frac{1}{3}
Option A (13\frac{1}{3}) is correct because it is calculated using Bayes' theorem to find the probability of selecting bag A given that a white ball was drawn.
379 Easy
379 of 1130
Application +4 | -1
A basket contains 5 apples and 7 oranges and another basket contains 4 apples and 8 oranges. One fruit is picked out from each basket. The probability that the fruits are both apples or both oranges, is
View answer and explanation
Correct option: D
76/144
Option D (76/144) is correct because it calculates the probability of both fruits being apples OR both fruits being oranges, summing the individual probabilities.
380 Hard
380 of 1130
Application +4 | -1
Let AA be a set containing nn elements. A subset PP of the set AA is chosen at random. The set AA is reconstructed by replacing the elements of PP, and another subset QQ of AA is chosen at random. The probability that P∩QP \cap Q contains exactly m(m<n)m(m < n) elements, is
View answer and explanation
Correct option: C
nCm×3n−m4n\frac{{ }^n C_m \times 3^{n-m}}{4^n}
Option C (nCm×3n−m4n\frac{{^n C_m \times 3^{{n-m}}}}{{4^n}}) is correct because it calculates the probability of P and Q having exactly m elements in common by considering the combinations and dividing by the total possible outcomes.
381 Easy
381 of 1130
Application +4 | -1
If a bag contains 50 tickets, numbered 1,2,3,…,501,2,3, \ldots, 50 of which five are drawn at random and arranged in ascending order of magnitude (x1<x2<x3<x4<x5)\left( {{x_1} < {x_2} < {x_3} < {x_4} < {x_5}} \right). The probability that x3=30x_3=30, is
View answer and explanation
Correct option: A
20C2×29C250C5\frac{{ }^{20} C_2 \times{ }^{29} C_2}{{ }^{50} C_5}
Option A is correct because the probability is calculated by choosing two numbers less than 30 and two numbers greater than 30, then dividing by the total number of ways to choose five numbers.
382 Medium
382 of 1130
Application +4 | -1
A letter is known to have come from either TATANAGAR or CALCUTTA. On the envelope, just two consecutive letters, TA, are visible. The probability that the letter has come from CALCUTTA is
View answer and explanation
Correct option: A
411\frac{4}{11}
Option A (411\frac{4}{11}) is correct because it is calculated using Bayes' theorem, considering the probabilities of the letter coming from each city given the observed letters.
383 Easy
383 of 1130
Application +4 | -1
The probability that a leap year selected at random will contain either 53 Thursdays or 53 Fridays, is
View answer and explanation
Correct option: A
37\frac{3}{7}
Option A (37\frac{3}{7}) is correct because the probability is calculated by considering the possible pairs of days in a leap year and finding the probability of having either 53 Thursdays or 53 Fridays.
384 Medium
384 of 1130
Application +4 | -1
A box contains 15 transistors, 5 of which are defective. An inspector takes out one transistor at random, examines it for defects and replaces it. After it has replaced another inspector does the same thing and then so does a third inspector. The probability that atleast one of the inspectors finds a defective transistor, is equal to
View answer and explanation
Correct option: C
19/27
Option C (19/27) is correct because it represents the probability of at least one defective transistor being found, calculated as 1 minus the probability of all transistors being non-defective.
385 Easy
385 of 1130
Understanding +4 | -1
If AA and BB are two events, the probability that exactly one of them occurs is not given by
View answer and explanation
Correct option: D
P(Aˉ)+P(Bˉ)−2P(Aˉ∩Bˉ)P(\bar{A})+P(\bar{B})-2 P(\bar{A} \cap \bar{B})
Option D is correct because P(Aˉ)+P(Bˉ)−2P(Aˉ∩Bˉ)P(\bar{A})+P(\bar{B})-2 P(\bar{A} \cap \bar{B}) represents the probability of exactly one of the complements occurring, not exactly one of the original events.
386 Hard
386 of 1130
Application +4 | -1
Three dice are thrown. The probability of getting a sum which is a perfect square, is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the calculated probability of getting a sum that is a perfect square is 17108\frac{17}{108}, which is not among the given options.
387 Easy
387 of 1130
Application +4 | -1
A police-man fires six bullets on a dacoit. The Probability that the dacoit will be killed by one bullet is 0.6 . The Probability that dacoit is still alive is
View answer and explanation
Correct option: B
0.004096
Option B (0.004096) is correct because it represents the probability that the dacoit survives all six shots, calculated as (1 - probability of being killed by one bullet) raised to the power of 6.
388 Easy
388 of 1130
Application +4 | -1
Four fair dice D1,D2,D3D_1, D_2, D_3 atd D4D_4, each having six faces numbered 1,2,3,4,5 and 6 are rolled simultaneously. The probability that D4D_4 shows a number on one of D1,D2D_1, D_2 and D3D_3 is
View answer and explanation
Correct option: A
91216\frac{91}{216}
Option A (91/216) is correct because it calculates the probability that D4 shows a number that is present on at least one of the other three dice.
389 Easy
389 of 1130
Analyzing +4 | -1
The probability that the 13th day of a randomly chosen month is a Friday, is
View answer and explanation
Correct option: C
184\frac{1}{{84}}
Option C (184\frac{1}{{84}}) is correct because it represents the probability of a randomly chosen month having its 13th day as a Friday, considering the probability of selecting a month and the conditional probability of the 13th being a Friday.
390 Easy
390 of 1130
Analyzing +4 | -1
Let AA and BB be two events such that P(B∣A)=25,P(A∣B)=17P(B \mid A)=\frac{2}{5}, P(A \mid B)=\frac{1}{7} and P(A∩B)=19P(A \cap B)=\frac{1}{9}, Consider (S1)P(A′∪B)=56,(S2)P(A′∩B′)=118(S 1) P\left(A^{\prime} \cup B\right)=\frac{5}{6},(S 2) P\left(A^{\prime} \cap B^{\prime}\right)=\frac{1}{18}. Then
View answer and explanation
Correct option: A
Both (S1)(S 1) and (S2)(S 2) are true
Option A (Both (S1) and (S2) are true) is correct because both statements are verified to be true based on the given probabilities and conditional probabilities.
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