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JEE • MATHEMATICS • PROBABILITY

Probability practice · Easy

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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 751-780 of 822 questions
751 Easy
751 of 822
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%).
752 Easy
752 of 822
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.
753 Easy
753 of 822
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.
754 Easy
754 of 822
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.
755 Easy
755 of 822
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.
756 Easy
756 of 822
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.
757 Easy
757 of 822
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.
758 Easy
758 of 822
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.
759 Easy
759 of 822
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
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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.
760 Easy
760 of 822
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.
761 Easy
761 of 822
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.
762 Easy
762 of 822
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.
763 Easy
763 of 822
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.
764 Easy
764 of 822
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.
765 Easy
765 of 822
Application +4 | -1
For a binomial variate XX with n=6n = 6, if P(X=2)=9P(X=4)P\left( {X = 2} \right) = 9P\left( {X = 4} \right), then its variance is
View answer and explanation
Correct option: C
98\frac{9}{8}
Option C (9/8) is correct because the variance is calculated using the binomial distribution formula npq after finding p and q.
766 Easy
766 of 822
Application +4 | -1
If XX has a binomial distribution, B(n,p)B(n,p)with parameters nn and pp such that P(X=2)=P(X=3),P\left( {X = 2} \right) = P\left( {X = 3} \right), then E(X),E\left( X \right), the mean of variable X,X, is
View answer and explanation
Correct option: B
3−p3 - p
Option B (3−p3 - p) is correct because, given P(X=2)=P(X=3)P(X=2) = P(X=3), we can derive the relationship np=3−pnp = 3 - p for a binomial distribution, and E(X)=npE(X) = np.
767 Easy
767 of 822
Application +4 | -1
If EE and FF are two events associated with a random experiment for which P(E)=0.6,P(E∪F)P(E)=0.6, P(E \cup F) =0.85=0.85 and P(E∩F)=0.42P(E \cap F)=0.42, then P(F)P(F) is
View answer and explanation
Correct option: B
0.67
Option B (0.67) is correct because it is the probability of event F, calculated using the formula for the probability of the union of two events.
768 Easy
768 of 822
Application +4 | -1
To open a lock, a key is taken out from a collection of nn keys at random. If the lock is not opened with this key, it is put back into the collection and another key is tried. The process is repeated again and again. If it is given that with only one key in the collection, the lock can be opened, then the probability that the lock will open in nn trials, is
View answer and explanation
Correct option: C
1−(n−1n)n1 - {\left( {\frac{{n - 1}}{n}} \right)^n}
Option C (1−(n−1n)n1 - {(\frac{{n - 1}}{n})^n}) is correct because it calculates the probability of opening the lock within n trials by subtracting the probability of not opening the lock in any of the n trials from 1.
769 Easy
769 of 822
Application +4 | -1
The probability that in the toss of two dice, we obtain the sum 7 or 11, is
View answer and explanation
Correct option: C
29\frac{2}{9}
Option C (29\frac{2}{9}) is correct because it is the probability of obtaining a sum of 7 or 11 when tossing two dice, calculated by dividing the number of favorable outcomes by the total number of possible outcomes.
770 Easy
770 of 822
Application +4 | -1
The probability that randomly selected calculator from a store is of brand rr is proportional to rr, r=1,2,…,6r=1,2, \ldots, 6. Further, the probability of a calculator of brand rr being defective is 7−r21,r=1,2,…,6\frac{7-r}{21}, r=1,2, \ldots, 6. Then the probability that a calculator randomly selected from the store being defective is
View answer and explanation
Correct option: A
863\frac{8}{63}
Option A (8/63) is correct because it calculates the probability of a randomly selected calculator being defective, considering the probability of each brand and its defect rate.
771 Easy
771 of 822
Understanding +4 | -1
There are four machines and it is known that exactly two of them are faulty. They are tested, one by one, in a random order till both the faulty machines are identified. Then the probability that only two tests are needed is
View answer and explanation
Correct option: B
1/61 / 6
Option B (1/6) is correct because there is only one way to choose the two faulty machines in the first two tests out of the six possible combinations.
772 Easy
772 of 822
Application +4 | -1
A random variable XX has the following probability distribution
X            1       2        3        4P(X)   k      2k     3k     4k\begin{array}{l} X\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,4\\ P(X)\,\,\,k\,\,\,\,\,\,2k\,\,\,\,\,3k\,\,\,\,\,4k \end{array}
Then, the mean of XX is
View answer and explanation
Correct option: A
3
Option A (3) is correct because it is the mean of the random variable X calculated using the given probability distribution.
773 Easy
773 of 822
Application +4 | -1
From a box containing 20 tickets marked with numbers 1 to 20, four tickets are drawn one by one. After each draw, the ticket is replaced. The probability that the largest value of tickets drawn is 15 is
View answer and explanation
Correct option: B
27320\frac{27}{320}
The correct answer is 27320\frac{27}{320} because it represents the probability that the largest value of four tickets drawn with replacement is 15.
774 Easy
774 of 822
Application +4 | -1
If P(A∩B)=710P(A \cap B)=\frac{7}{10} and P(B)=1720P(B)=\frac{17}{20}, then P(A/B)P(A / B) equals to
View answer and explanation
Correct option: A
1417\frac{14}{17}
Option A (1417\frac{14}{17}) is correct because it is the conditional probability of A given B, calculated using the formula P(A/B)=P(A∩B)P(B)P(A/B) = \frac{P(A \cap B)}{P(B)}.
775 Easy
775 of 822
Application +4 | -1
In a maths paper there are 3 sections A, B and C. Section A is compulsory. Out of sections BB and CC a student has to attempt any one. Passing in the paper means passing in A\mathrm{A} and passing in B\mathrm{B} or C\mathrm{C}. The probability of the student passing in A,B\mathrm{A}, \mathrm{B} and CC are p,qp, q and 1/21 / 2 respectively. If the probability that the student is successful is 1/21 / 2 then
View answer and explanation
Correct option: D
p=q=1/2\mathrm{p}=\mathrm{q}=1 / 2
Option D (p=q=1/2) is correct because it satisfies the probability condition for the student being successful.
776 Easy
776 of 822
Application +4 | -1
Probability that in the toss of two dice we obtain an even sum or a sum less than 5, is
View answer and explanation
Correct option: D
59\frac{5}{9}
Option D (59\frac{5}{9}) is correct because it's calculated by finding the probability of an even sum, the probability of a sum less than 5, and subtracting the probability of their intersection.
777 Easy
777 of 822
Application +4 | -1
Assuming that for a husband-wife couple the chances of their child being a boy or a girl are the same, the probability of their two children being a boy and a girl is
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (12\frac{1}{2}) is correct because there are two equally likely possibilities: boy then girl, or girl then boy, out of the four possible outcomes (BB, GG, BG, GB).
778 Easy
778 of 822
Application +4 | -1
For k=1,2,3k = 1,2,3 the box Bk{B_k} contains kk red balls and (k+1)\left( {k + 1} \right) white balls,
Let P(B1)=12, P(B2)=13P\left( {{B_1}} \right) = \frac{1}{2},\,P\left( {{B_2}} \right) = \frac{1}{3} and P(B3)=16P\left( {{B_3}} \right) = \frac{1}{6}
A box is selected at random and a ball is drawn from it. If a red ball is drawn, then the probability that it has come from box B2{B_2}, is
View answer and explanation
Correct option: B
1439\frac{{14}}{{39}}
Option B (14/39) is correct because it is calculated using Bayes' theorem to find the probability that the red ball came from box B2 given that a red ball was drawn.
779 Easy
779 of 822
Application +4 | -1
A natural number xx is chosen at random from the first 100 natural numbers. The probability that x+100x>50x + \frac{{100}}{x} > 50
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Correct option: C
1120\frac{{11}}{{20}}
Option C (1120\frac{{11}}{{20}}) is the correct probability, calculated by finding the range of x values that satisfy the inequality and dividing by the total number of natural numbers.
780 Easy
780 of 822
Application +4 | -1
If XX is a poisson variate such that P(X=1)=P(X=2)P\left( {X = 1} \right) = P\left( {X = 2} \right), then P(X=4)P\left( {X = 4} \right) is equal to
View answer and explanation
Correct option: C
23e2\frac{2}{{3{e^2}}}
Option C (2/3e²) is correct because by equating the probabilities P(X=1) and P(X=2), we find λ=2, and then calculate P(X=4) using the Poisson distribution formula.
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