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JEE • MATHEMATICS • PROBABILITY

Probability practice · Mixed difficulty

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Practice format and chapter quiz

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 1021-1050 of 1130 questions
1021 Easy
1021 of 1130
Application +4 | -1
A number xx is chosen at random from the set {1,2,3,…,100}\{1,2,3, \ldots, 100\}. Let p=p= probability that xx is divisible by 19 and q=q= probability that xx is divisible by 31, then p+qp+q is equal to
View answer and explanation
Correct option: D
0.8
Option D (0.08) is correct because it represents the sum of the probabilities that x is divisible by 19 and 31, calculated by dividing the number of multiples of each number within the set by the total number of elements in the set.
1022 Easy
1022 of 1130
Analyzing +4 | -1
Three-digit numbers are formed using the digits 0,0 , 1,2,3,4,51,2,3,4,5 without repetition of digits. If a number is chosen at random, then the probability that the digits either increase or decrease, is
View answer and explanation
Correct option: C
310\frac{3}{10}
Option C (310\frac{3}{10}) is correct because it represents the ratio of numbers with increasing or decreasing digits to the total number of three-digit numbers formed.
1023 Medium
1023 of 1130
Application +4 | -1
The probability that in a year of 22 nd century chosen at random, there will be 53 Sundays, is
View answer and explanation
Correct option: D
528\frac{5}{28}
Option D (528\frac{5}{28}) is correct because the probability is calculated by considering both leap and non-leap years in the 22nd century and the probability of having 53 Sundays in each case.
1024 Easy
1024 of 1130
Application +4 | -1
If n(A)=8n\left( A \right) = 8 and n(A∩B)=2,n\left( {A \cap B} \right) = 2, then n[(A∩B)′∩A]n\left[ {{{\left( {A \cap B} \right)}^\prime } \cap A} \right] is equal to
View answer and explanation
Correct option: A
6
Option A (6) is correct because n[(A∩B)′∩A]=n(A)−n(A∩B)=8−2=6n[ (A ∩ B)' ∩ A ] = n(A) - n(A ∩ B) = 8 - 2 = 6.
1025 Medium
1025 of 1130
Analyzing +4 | -1
Suppose f(x)=x3+ax2+bx+c,f\left( x \right) = {x^3} + a{x^2} + bx + c, where a, b, ca,\,b,\,c are chosen respectively by throwing a dice three times. Then, the probability that f(x)f\left( x \right) is an increasing function, is
View answer and explanation
Correct option: A
49\frac{4}{9}
Option A (49\frac{4}{9}) is correct because it represents the probability that the derivative of the cubic function is always non-negative, ensuring the function is increasing.
1026 Easy
1026 of 1130
Application +4 | -1
Each of the persons AA and BB independently tosses three fair coins. The probability that both of them get the same number of heads is
View answer and explanation
Correct option: C
516\frac{5}{{16}}
Option C (5/16) is correct because it calculates the probability of both persons getting the same number of heads by considering all possible cases (0, 1, 2, or 3 heads) and summing their probabilities.
1027 Medium
1027 of 1130
Application +4 | -1
The probability that in the random arrangement of the letters of the word 'UNIVERSITY', the two I's does not come together is
View answer and explanation
Correct option: A
0.8
Option A (0.8) is correct because it calculates the probability by subtracting the probability of the two I's being together from 1.
1028 Easy
1028 of 1130
Analyzing +4 | -1
For the three events A, BA,\,B and CC, PP (exactly one of the events AA or BB occurs) = P= \,P (exactly one of the events BB or CC occurs) −P- P (exactly one of the events CC or AA occurs) = p= \,p and PP ( all the three events occur simultaneously) =p2,  = {p^2},\; where 0<p<120 < p < \frac{1}{2}. Then, the probability of at least one of the three events AA, BB and CC occuring, is
View answer and explanation
Correct option: A
3p+2p22\frac{{3p + 2{p^2}}}{2}
Option A (3p+2p22\frac{{3p + 2{p^2}}}{2}) is correct because it represents the probability of at least one of the three events occurring, derived from the given probabilities of exactly one event occurring and all three events occurring simultaneously.
1029 Easy
1029 of 1130
Application +4 | -1
The probability that the same number appear on throwing three dice simultaneously, is
View answer and explanation
Correct option: B
Jan-36
Option C (0.06) is correct because it is the product of the probability of receiving a bonus (30%) and the probability of being skilled given a bonus (20%).
1030 Easy
1030 of 1130
Application +4 | -1
Two dice are thrown. If it is known that the sum of the numbers on the dice is less than 6 , the probability of getting a sum 3 is
View answer and explanation
Correct option: C
15\frac{1}{5}
Option C (15\frac{1}{5}) is correct because it is the conditional probability of getting a sum of 3 given that the sum is less than 6.
1031 Medium
1031 of 1130
Analyzing +4 | -1
AA and BB are two students. Their chances of sloving a problem correctly are 13\frac{1}{3} and 14\frac{1}{4}, respectively. If the probability of their making a common error is, 120\frac{1}{20} and they obtain the same answer, then the probability of their answer to be correct is
View answer and explanation
Correct option: D
1013\frac{10}{13}
Option D (1013\frac{10}{13}) is correct because it represents the probability of their answer being correct given that they obtained the same answer.
1032 Easy
1032 of 1130
Application +4 | -1
If three six-faced fair dice are thrown together, the probability that the sum of the numbers appearing on the dice is 16 is
View answer and explanation
Correct option: A
136\frac{1}{36}
Option A (136\frac{1}{36}) is correct because it is the probability of the sum of three dice being 16, calculated by dividing the number of favorable outcomes by the total number of outcomes.
1033 Easy
1033 of 1130
Application +4 | -1
AA and BB are two independent events such that P(A)12P\left( A \right)\frac{1}{2} and P(B)=13,  P\left( B \right) = \frac{1}{3},\; then PP (neither AA nor BB) is equal to
View answer and explanation
Correct option: D
13\frac{1}{3}
Option D (13\frac{1}{3}) is correct because it calculates the probability of neither A nor B occurring using the formula for the complement of the union of two independent events.
1034 Easy
1034 of 1130
Application +4 | -1
5 persons A,B,C,DA, B, C, D and EE are in a queue of a shop. The probability that AA and EE always occur together, is
View answer and explanation
Correct option: C
25\frac{2}{5}
Option C (25\frac{2}{5}) is correct because it is the probability of A and E always occurring together, calculated by considering them as a single unit.
1035 Medium
1035 of 1130
Analyzing +4 | -1
Consider the system of equations px+qy=0,rx+sy=0\mathrm{px}+\mathrm{qy}=0, \mathrm{rx}+\mathrm{sy}=0 where p,q,r,s∈{−1,1}p, q, r, s \in\{-1,1\}
Statement-1: The probability that the system of equation has a unique solution is 12\frac{1}{2}.
Statement-2: The probability that the system of equations no solution is zero.
View answer and explanation
Correct option: B
Statement-1 is true, statement-2 is true and ststement-2 is NOT the correct explanation for statement-1
Statement-1 is true because the probability of a unique solution is 1/2, and Statement-2 is true because the probability of no solution is zero, as the system is homogeneous. Statement-2 is NOT the correct explanation for Statement-1.
1036 Easy
1036 of 1130
Application +4 | -1
A coin is tossed 10 times. The probability of getting exactly six heads is
View answer and explanation
Correct option: B
105512\frac{{105}}{{512}}
Option B (105512\frac{{105}}{{512}}) is correct because it calculates the probability of getting exactly six heads when a coin is tossed 10 times using the binomial distribution.
1037 Easy
1037 of 1130
Application +4 | -1
Two dice are tossed once. The probability of getting even number at the first die or a total of 8 is
View answer and explanation
Correct option: D
59\frac{5}{9}
Option D (59\frac{5}{9}) is correct because it calculates the probability of getting an even number on the first die or a total of 8 by considering the union of the two events and accounting for their intersection.
1038 Medium
1038 of 1130
Application +4 | -1
Suppose that the probability that an item produced by a particular machine is defective equals 0.2. If 10 items produced from this machine are selected at random, the probability that not more than one defective is found is
View answer and explanation
Correct option: C
3e2\frac{3}{\mathrm{e}^2}
Option C (3e2\frac{{3}}{{\mathrm{{e}}^2}}) is correct because it uses the Poisson approximation to find the probability of not more than one defective item.
1039 Medium
1039 of 1130
Application +4 | -1
A bag contains 6 red and 3 white balls. Four balls are drawn one by one and not replaced. The probability that they are alternatively of different colours, is
View answer and explanation
Correct option: B
542\frac{5}{{42}}
Option B (542\frac{5}{{42}}) is correct because it calculates the probability of drawing four balls alternatively of different colors (Red-White-Red-White or White-Red-White-Red) without replacement.
1040 Hard
1040 of 1130
Analyzing +4 | -1
A number nn is randomly selected from the set {1,2,3,….,1000}\{1,2,3, \ldots ., 1000\}. The probability that ∑i=1ni2∑i=1ni\frac{\sum_{i=1}^{n} i^{2}}{\sum_{i=1}^{n} i} is an integer is
View answer and explanation
Correct option: C
0.334
Option C (0.334) is correct because it represents the probability that the given expression is an integer when n is randomly selected from the set.
1041 Easy
1041 of 1130
Application +4 | -1
An investment consultant predicts that the odds ageinst the price of a certain stock will go up during the next week are 2:12: 1 and the odds in favour of the price remaining the same are 1:31: 3. The probability that the price of the stock will go down during the next week, is
View answer and explanation
Correct option: B
512\frac{5}{12}
Option B (512\frac{5}{12}) is correct because it calculates the probability that the price of the stock will go down, given the odds against the price going up or remaining the same.
1042 Easy
1042 of 1130
Application +4 | -1
In a class of 125 students 70 passed in Mathematics, 55 in Statistic and 30 in both. The probability that a student selected at random from the class, has passed in only one subject is
View answer and explanation
Correct option: A
13/2513/25
Option A (1325\frac{{13}}{{25}}) is correct because it calculates the probability of passing in only one subject by subtracting the probability of passing in both subjects from the probability of passing in each subject individually.
1043 Easy
1043 of 1130
Application +4 | -1
There are five students S1,S2,S3,S4S_1, S_2, S_3, S_4 and S5S_5 in a music class and for them there are five seats R1,R2,R3,R4R_1, R_2, R_3, R_4 and R5R_5 arranged in a row, where initially the seat RiR_i is allotted to the student Si,i=1,2,3,4,5S_i, i=1,2,3,4,5.But, on the examination day, the five students are randomly allotted the five seats. The probability that, on the examination day, the student S1S_1 gets the previously allotted seat R1R_1, and none of the remaining students gets the seat previously allotted to him/her, is
View answer and explanation
Correct option: A
340\frac{3}{40}
Option A (340\frac{{3}}{{40}}) is correct because it calculates the probability of S1 getting R1 and none of the others getting their original seats using derangements.
1044 Medium
1044 of 1130
Application +4 | -1
The letters of the word 'longest' are arranged at random. The probability that the vowels may occupy only odd positions, is
View answer and explanation
Correct option: A
27\frac{2}{7}
Option A (27\frac{2}{7}) is correct because it represents the probability of arranging the vowels in the odd positions out of all possible arrangements.
1045 Easy
1045 of 1130
Application +4 | -1
It has been found that if AA and BB play a game 12 times, AA wins 6 times, BB wins 4 times and they draw twice. AA and BB take part in a series of 3 games. The probability that they will win alternately is
View answer and explanation
Correct option: B
536\frac{5}{36}
Option B (536\frac{5}{36}) is correct because it calculates the probability of A and B winning alternately in a series of 3 games, considering the given win probabilities.
1046 Easy
1046 of 1130
Application +4 | -1
Two cards are drawn successively with replacement from a well shuffled deck of 52 cards, then the mean of the number of aces is
View answer and explanation
Correct option: C
2/132 / 13
The correct answer is 2/132 / 13 because it calculates the mean number of aces when two cards are drawn successively with replacement.
1047 Medium
1047 of 1130
Application +4 | -1
Three rifle-men take one shot each at the same target. The probability of the first rifle-man hitting the target is 0.4 , the probability of the second rifle-man hitting the target is 0.5 and the probability of the third rifle-man hitting the target is 0.8 . The Probability that exactly two of them hit the target, is
View answer and explanation
Correct option: B
0.44
Option B (0.44) is correct because it calculates the probability of exactly two rifle-men hitting the target by considering all possible combinations of two successes and one failure.
1048 Easy
1048 of 1130
Application +4 | -1
The odds against a certain event are 5 :2 and the odds in favour of another independent event are 6 : 5. The probability that at least one of the events will happen, is
View answer and explanation
Correct option: B
5277\frac{{52}}{{77}}
Option B (5277\frac{{52}}{{77}}) is correct because it is calculated by finding the probability that at least one event happens using the complement rule and the given odds.
1049 Medium
1049 of 1130
Application +4 | -1
A man takes a step forward with probability 0.40.4 and backwards with probability 0.6. The probability that at the end of eleven steps he is just one step away from the starting point, is
View answer and explanation
Correct option: C
11C5(0.4)5(0.6)5{ }^{11} C_5(0.4)^5(0.6)^5
Option C (11C5(0.4)5(0.6)6{ }^{11} C_5(0.4)^5(0.6)^6) is correct because it's the probability that the man is one step away, considering both forward and backward scenarios.
1050 Easy
1050 of 1130
Application +4 | -1
A bag contains 10 mangoes out of which 4 are rotten, two mangoes are taken out together. If one of them is found to be good, the probability that other is also good is
View answer and explanation
Correct option: C
518\frac{5}{18}
Option C (5/18) is correct because it accurately calculates the conditional probability that both mangoes are good given that one of them is good.
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