Skip to main content
Back to PROBABILITY

Your Progress

0 / 822 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • PROBABILITY

Probability practice · Easy

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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.

Take a chapter quiz →
Showing 541-570 of 822 questions

Preparing practice and saved progress…

541 Easy
541 of 822
Application +4 | -1
A man draws a card from a pack of 52 cards and then replaces it. After shuffling the pack, he again draws a card. This he repeats a number of times. The probability that he will draw a heart for the first time in the third draw is
View answer and explanation
Correct option: A
964\frac{9}{64}
Option A (964\frac{9}{64}) is correct because it calculates the probability of not drawing a heart in the first two draws and then drawing a heart in the third draw, with replacement.
542 Easy
542 of 822
Application +4 | -1
If XX is a binomial variate with the range {0,1,2,3,4,5,6} and P(X=2)=4P(X=4),P\left( {X = 2} \right) = 4P\left( {X = 4} \right), then the parameter pp of XX is
View answer and explanation
Correct option: A
13\frac{1}{3}
Option A (13\frac{1}{3}) is correct because it is the parameter p of the binomial variate X calculated using the given relationship between P(X=2) and P(X=4).
543 Easy
543 of 822
Application +4 | -1
A letter is taken out at random from `ASSISTANT’ and another is taken out from `STATISTICS’. The probability that they are the same letters is
View answer and explanation
Correct option: C
1990\frac{{19}}{{90}}
Option C (1990\frac{{19}}{{90}}) is correct because it sums the probabilities of drawing the same letter (A, I, S, or T) from both words.
544 Easy
544 of 822
Application +4 | -1
Five coins are tossed simultaneously. The probability that at least one head turning up, is
View answer and explanation
Correct option: D
31/3231 / 32
Option D (3132\frac{31}{32}) is correct because it's the probability of at least one head, calculated as 1 minus the probability of no heads.
545 Easy
545 of 822
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.
546 Easy
546 of 822
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.
547 Easy
547 of 822
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.
548 Easy
548 of 822
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.
549 Easy
549 of 822
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.
550 Easy
550 of 822
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.
551 Easy
551 of 822
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.
552 Easy
552 of 822
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.
553 Easy
553 of 822
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.
554 Easy
554 of 822
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.
555 Easy
555 of 822
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.
556 Easy
556 of 822
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).
557 Easy
557 of 822
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.
558 Easy
558 of 822
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.
559 Easy
559 of 822
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.
560 Easy
560 of 822
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.
561 Easy
561 of 822
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.
562 Easy
562 of 822
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.
563 Easy
563 of 822
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.
564 Easy
564 of 822
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.
565 Easy
565 of 822
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.
566 Easy
566 of 822
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.
567 Easy
567 of 822
Understanding +4 | -1
A and B play a game where each is asked to select a number from 1 to 25. If the two numbers match, both of them win a prize. The probability that they will not win a prize in a single trial is
View answer and explanation
Correct option: B
24/2524 / 25
Option B (24/25) is correct because it represents the complement of the probability that they will win a prize, which is 1/25.
568 Easy
568 of 822
Application +4 | -1
A random variable XX takes values 0, 1, 2, 3,… with probability P(X=x)=k(x+1)(15)x,P\left( {X = x} \right) = k\left( {x + 1} \right){\left( {\frac{1}{5}} \right)^x}, where kk is constant, then P(X=0)P\left( {X = 0} \right) is
View answer and explanation
Correct option: D
1625\frac{{16}}{{25}}
Option D (1625\frac{{16}}{{25}}) is correct because it is the value of P(X=0) that satisfies the condition for a valid probability distribution.
569 Easy
569 of 822
Understanding +4 | -1
If A\mathrm{A} and B\mathrm{B} are independent events, then which of the following is not true?
View answer and explanation
Correct option: C
P(A/B)=P(B/A)\mathrm{P}(\mathrm{A} / \mathrm{B})=\mathrm{P}(\mathrm{B} / \mathrm{A})
Option C (P(A/B)=P(B/A)\mathrm{P}(\mathrm{A} / \mathrm{B})=\mathrm{P}(\mathrm{B} / \mathrm{A})) is not true because for independent events, P(A/B) = P(A) and P(B/A) = P(B), and in general, P(A) is not equal to P(B).
570 Easy
570 of 822
Understanding +4 | -1
If AA and BB are two mutually exclusive events, then
View answer and explanation
Correct option: A
P(A)≤P(Bˉ)P(A) \leq P(\bar{B})
Option A is correct because if A and B are mutually exclusive, then A is a subset of the complement of B, meaning P(A) is less than or equal to P(B').
Revision notes and key takeaways

Review PROBABILITY in your textbook or class notes, then use this set to check your understanding.

Showing 541-570 of 822 questions

Your practice

0 / 822 answered · 822 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.