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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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751 Easy
751 of 1130
Application +4 | -1
Four positive integers are taken at random and are multiplied together. Then the probability that the product ends in an odd digit other than 5 is
View answer and explanation
Correct option: C
16625\frac{16}{625}
Option C (16625\frac{16}{625}) is correct because the probability of a product ending in an odd digit other than 5 requires each of the four integers to end in 1, 3, 7, or 9.
752 Easy
752 of 1130
Understanding +4 | -1
There are nn letters and nn addressed envelopes, the probability that all the letters are not kept in the right envelope, is
View answer and explanation
Correct option: B
1−1n!1 - \frac{1}{{n!}}
Option B (1−1n!1 - \frac{{1}}{{n!}}) is correct because it represents the probability of at least one letter being in the wrong envelope, which is 1 minus the probability of all letters being in the correct envelope.
753 Easy
753 of 1130
Application +4 | -1
A committee of five is to be chosen from a group of 9 people. The probability that a certain married couple will either serve together or not at all is
View answer and explanation
Correct option: C
49\frac{4}{9}
Option C (49\frac{4}{9}) is correct because it calculates the probability of the married couple either serving together or not at all in the committee.
754 Easy
754 of 1130
Application +4 | -1
The mean and variance of a random variable XX having a binomial distribution are 4 and 2 respectively. Then, P(X>6)P(X > 6) is equal to
View answer and explanation
Correct option: C
9256\frac{9}{{256}}
Option C (9/256) is correct because it is calculated using the binomial distribution formula with the determined values of n, p, and q.
755 Easy
755 of 1130
Application +4 | -1
If P(A)=0.4,P(B′)=0.6P(A)=0.4, P\left(B^{\prime}\right)=0.6 and P(A∩B)P(A \cap B) =0.15=0.15 then the value of P(A/A′∪B′)P\left(A / A^{\prime} \cup B^{\prime}\right) is
View answer and explanation
Correct option: B
517\frac{5}{17}
Option B (517\frac{{5}}{{17}}) is correct because it represents the conditional probability of A given the union of A' and B', calculated using the given probabilities.
756 Easy
756 of 1130
Application +4 | -1
A dice is thrown 5 times, then the probability that an even number will come up exactly 3 times, is
View answer and explanation
Correct option: A
516\frac{5}{{16}}
Option A (5/16) is correct because the probability is calculated using the binomial probability formula with n=5, k=3, and p=1/2.
757 Medium
757 of 1130
Application +4 | -1
If there are 6 girls and 5 boys who sit in a row, then the probability that no two boys sit together is
View answer and explanation
Correct option: C
6  !7  !2  !11  !\frac{{6\;!7\;!}}{{2\;!11\;!}}
Option C (6  !7  !2  !11  !\frac{{6\;!7\;!}}{{2\;!11\;!}}) is correct because it calculates the probability of no two boys sitting together by arranging the girls first and then placing the boys in the gaps.
758 Easy
758 of 1130
Application +4 | -1
Two dice are thrown together. Then, the probability that the sum of numbers appearing on them is a prime number, is
View answer and explanation
Correct option: A
512\frac{5}{{12}}
Option A (512\frac{5}{{12}}) is correct because the probability is calculated by dividing the number of favorable outcomes (sum is a prime number) by the total possible outcomes when two dice are thrown.
759 Easy
759 of 1130
Analyzing +4 | -1
Urn AA contains 6 red and 4 black balls and urn BB contains 4 red and 6 black balls. One ball is drawn at random from urn AA and placed in urn BB. Then one ball is drawn at random from urn BB and placed in urn AA. If one ball is now drawn at random from urn AA, the probability that it is found to be red, is
View answer and explanation
Correct option: A
32/5532 / 55
Option A (32/5532 / 55) is correct because it represents the probability of drawing a red ball from urn A after two transfers, calculated using conditional probabilities.
760 Easy
760 of 1130
Application +4 | -1
If a random variable XX follows the Binomial distribution B(33,p)B\left( {33,p} \right) such that 3P(X=0)=P(X=1),3P\left( {X = 0} \right) = P\left( {X = 1} \right), then the value of P(X=15)P(X=18)−P(X=16)P(X=17)\frac{{P\left( {X = 15} \right)}}{{P\left( {X = 18} \right)}} - \frac{{P\left( {X = 16} \right)}}{{P\left( {X = 17} \right)}} is equal to:
View answer and explanation
Correct option: A
1320
Option A (1320) is correct because after finding the value of p from the given condition, we can calculate the value of the expression P(X=15)P(X=18)−P(X=16)P(X=17)\frac{{P(X = 15)}}{{P(X = 18)}} - \frac{{P(X = 16)}}{{P(X = 17)}} which equals 1320.
761 Easy
761 of 1130
Analyzing +4 | -1
Three numbers are chosen from 1 to 30. The probability that they are not consecutive, is
View answer and explanation
Correct option: B
144145\frac{144}{145}
Option B (144145\frac{{144}}{{145}}) is correct because it calculates the probability of three numbers not being consecutive by subtracting the probability of them being consecutive from 1.
762 Easy
762 of 1130
Application +4 | -1
If a coin be tossed nn times, then probability that the head comes odd times is
View answer and explanation
Correct option: A
1/21/2
Option A (1/21/2) is correct because the probability of getting an odd number of heads in n coin tosses is always 1/2, due to the symmetry of the binomial distribution.
763 Medium
763 of 1130
Understanding +4 | -1
A random variable XX follows binomial distribution with mean α\alpha and variance β\beta. Then
View answer and explanation
Correct option: B
0<β<α0 < \beta < \alpha
Option B (0<β<α0 < \beta < \alpha) is correct because in a binomial distribution, the variance is always less than the mean.
764 Easy
764 of 1130
Understanding +4 | -1
If A   and  BA\;\,\mathrm{and\;}B are two mutually exclusive events, then
View answer and explanation
Correct option: A
P(A)<P(Bˉ)P\left( A \right) < P\left( {\bar B} \right)
Option A (P(A)<P(Bˉ)P(A) < P(\bar B)) is correct because if A and B are mutually exclusive, then A⊆BˉA \subseteq \bar B, which implies P(A)≤P(Bˉ)P(A) \le P(\bar B).
765 Easy
765 of 1130
Application +4 | -1
The mean and standard deviation of a binomial variate XX are 4 and 3\sqrt 3 respectively. Then, P(X≥1)P\left( {X \ge 1} \right) is equal to
View answer and explanation
Correct option: B
1−(34)161 - {\left( {\frac{3}{4}} \right)^{16}}
Option B (1−(34)161 - {\left( {\frac{3}{4}} \right)^{16}}) is correct because it represents the probability of X being greater than or equal to 1, calculated as 1 minus the probability of X being 0.
766 Easy
766 of 1130
Application +4 | -1
If a random variable XX has the following probability distribution values of XX.
X:01234567P(X):0k2k2k3kk22k27k2+k\begin{array}{cccccccccccccccccccc} {X:}&0&1&2&3&4&5&6&7 \\ {P(X):}&0&k&{2k}&{2k}&{3k}&{{k^2}}&{2{k^2}}&{7{k^2} + k} \end{array}
Then, P(X≥6)=P(X \geq 6)=
View answer and explanation
Correct option: A
19100\frac{19}{100}
Option A (19100\frac{19}{100}) is correct because it's the probability calculated after finding the value of k and then summing the probabilities for X >= 6.
767 Medium
767 of 1130
Application +4 | -1
A box contains 2 fifty paise coins, 5 twenty five paise coins and 15 ten paise coins. Five coins are taken out of the box at random. Probability that the value of these five coins is less than one rupee and fifity paise is
View answer and explanation
Correct option: C
1−17022C51-\frac{170}{{}^{22} C_5}
Option C (1−17022C51-\frac{170}{{}^{22} C_5}) is correct because it calculates the probability by subtracting the unfavorable cases (value >= 1.50) from the total probability (1).
768 Easy
768 of 1130
Application +4 | -1
A class consists of 100 students, 25 of them are girls and 75 boys, 20 of them are rich and remaining poor, 40 of them are fair complexioned. The Probability of selecting a fair complexioned rich girl is
View answer and explanation
Correct option: A
0.02
Option A (0.02) is correct because it's the product of the probabilities of selecting a fair-complexioned student, a rich student, and a girl, assuming independence.
769 Medium
769 of 1130
Application +4 | -1
The probability that a rectangle picked up from a chessboard has the area 6 cm26 \mathrm{~cm}^2 where the distance between consecutive parallel lines on the board is 1 cm1 \mathrm{~cm}, is
View answer and explanation
Correct option: D
11108\frac{11}{108}
Option D (11108\frac{11}{108}) is the correct probability, calculated by finding the number of rectangles with area 6 and dividing by the total number of rectangles on the chessboard.
770 Easy
770 of 1130
Application +4 | -1
Two integers are selected at random from the set {1, 2, …… , 11}. Given that the sum of selected numbers is even, the conditional probability that both the number are even is
View answer and explanation
Correct option: A
25\frac{2}{5}
Option A (2/5) is correct because it is the conditional probability of both numbers being even, given that their sum is even, calculated using the conditional probability formula.
771 Medium
771 of 1130
Application +4 | -1
If in a trial the probability of success is twice the probability of failure. In six trials the probability of at least four successes is
View answer and explanation
Correct option: A
496729\frac{{496}}{{729}}
Option A (496729\frac{{496}}{{729}}) is correct because it represents the probability of at least four successes in six trials, calculated using the binomial distribution formula.
772 Easy
772 of 1130
Understanding +4 | -1
The probability of obtaining sum ‘8’ in a single throw of two dice is
View answer and explanation
Correct option: B
536\frac{5}{{36}}
Option B (536\frac{5}{{36}}) is correct because there are 5 combinations that sum to 8: (2,6), (3,5), (4,4), (5,3), and (6,2), out of 36 possible outcomes.
773 Medium
773 of 1130
Analyzing +4 | -1
If three numbers are selected from the set of the first 20 natural numbers, the probability that they are in GPG P, is
View answer and explanation
Correct option: C
111140\frac{11}{1140}
Option C (111140\frac{11}{1140}) is correct because it represents the ratio of the number of possible geometric progressions to the total number of ways to choose three numbers.
774 Easy
774 of 1130
Application +4 | -1
In a certain population 10% of the people are rich, 5% are famous and 3% are rich and famous. The probability that a person picked at random from the population is either famous or rich but not both, is equal to
View answer and explanation
Correct option: C
0.09
Option C (0.09) is correct because it represents the probability of being either rich or famous, but not both, calculated using the given probabilities and the principle of inclusion-exclusion.
775 Easy
775 of 1130
Analyzing +4 | -1
A person puts three cards addresses to three different people in three envelopes with three different addresses without looking. What is the probability that the cards go into their respective envelopes?
View answer and explanation
Correct option: B
16\frac{1}{6}
Option B (1/6) is correct because there is only one way for all three cards to go into their respective envelopes out of the 3! = 6 possible arrangements.
776 Easy
776 of 1130
Understanding +4 | -1
The probability of getting at least one tail in 4 throws of a coin is
View answer and explanation
Correct option: A
1516\frac{{15}}{{16}}
Option A, 1516\frac{{15}}{{16}}, is correct because it calculates the probability of getting at least one tail as the complement of getting all heads.
777 Medium
777 of 1130
Application +4 | -1
From a pack of 52 playing cards, half of the cards are randomly removed without looking at them. From the remaining cards, 3 cards are drawn randomly. The probability that all are king, is :
View answer and explanation
Correct option: A
125.17.13\frac{1}{25.17 .13}
Option A (125.17.13\frac{1}{25.17.13}) is correct because it is the probability of drawing 3 kings from the remaining cards after half the deck is removed.
778 Easy
778 of 1130
Application +4 | -1
At the college entrance examination each candidate is admitted or rejected according to whether he has passed or failed the tests. Of the candidates who are really capable, 80%80 \% pass the tests and of the incapable, 25%25 \% pass the test. Given that 40%40 \% of the candidates are really capable, then the proportion of capable college students is about
View answer and explanation
Correct option: A
68%68 \%
Option A (68%68 \%) is correct because it is the result of applying Bayes' theorem to calculate the conditional probability of a candidate being capable given they passed the test.
779 Easy
779 of 1130
Application +4 | -1
For nn independent events Ai′s,  P(Ai)=1/(1+i),{A'_i}s,\;P\left( {{A_i}} \right) = 1/\left( {1 + i} \right), i=1,  2,...,n.i = 1,\;2,...,n. The probability that atleast one of the events occurs is
View answer and explanation
Correct option: C
n/(n+1)n/\left( {n + 1} \right)
Option C, n/(n+1)n/(n+1), is correct because it represents the probability of at least one event occurring, calculated as 1 minus the probability of none of the events occurring.
780 Medium
780 of 1130
Application +4 | -1
The probability that AA speaks truth is 45  \frac{4}{5}\mathrm{\;} while this probability for BB is 34\frac{3}{4}. The probability that they contradict each other when asked to speak on a fact, is
View answer and explanation
Correct option: C
720\frac{7}{{20}}
Option C (7/20) is correct because the probability of them contradicting each other is the sum of the probabilities of A being truthful and B being untruthful, and vice versa.
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