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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 481-510 of 1130 questions

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481 Easy
481 of 1130
Application +4 | -1
In a party 23 persons take their seats at a round table. The odds against two particular persons sitting together are:
View answer and explanation
Correct option: A
10:110: 1
Option A (10:110: 1) is correct because it's the odds against two particular persons sitting together, calculated using probability.
482 Medium
482 of 1130
Application +4 | -1
Suppose AA and BB are two mutually exclusive events such that P(A)=13(3p+1)P(A)=\frac{1}{3}(3 p+1) and P(B)=14(1−p)P(B)=\frac{1}{4}(1-p), then the set of possible values of pp lies in the interval:
View answer and explanation
Correct option: C
[−13,59]\left[-\frac{1}{3}, \frac{5}{9}\right]
Option D ([−79,49]\left[-\frac{7}{9}, \frac{4}{9}\right]) is incorrect. The correct answer is C ([−13,59]\left[-\frac{1}{3}, \frac{5}{9}\right]) because it considers the constraints on probabilities for mutually exclusive events.
483 Easy
483 of 1130
Application +4 | -1
A man is known to speak truth in 75% cases. If he throws an unbiased die and tells his friend that it is a six, then the probability that it is actually a six, is
View answer and explanation
Correct option: D
3/83/8
Option D (3/8) is correct because it is the result of applying Bayes' theorem to calculate the probability that it was actually a six given that the man reported it was a six.
484 Easy
484 of 1130
Application +4 | -1
Seven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently equals
View answer and explanation
Correct option: B
715\frac{7}{15}
Option B (7/15) is correct because it represents the probability of arranging the balls such that no two black balls are adjacent, calculated using combinations.
485 Easy
485 of 1130
Application +4 | -1
An article manufactured by a company consists of two parts X{X} and Y{Y}. In the process of manufacture of part X{X}, 9 out of 104 parts may be defective. Similarly, 5 out of 100 are likely to be defective in the manufacture of the part YY. The probability that the assembled product will not be defective, is
View answer and explanation
Correct option: B
361416\frac{361}{416}
Option B (361416\frac{361}{416}) is correct because it represents the probability of both parts X and Y not being defective when assembled.
486 Easy
486 of 1130
Application +4 | -1
AA and BB are two events such that P(A∪B)=34P(A \cup B)=\frac{3}{4}, P(A∩B)=14,P(Aˉ)=23P(A \cap B)=\frac{1}{4}, P(\bar{A})=\frac{2}{3}, then P(Aˉ∩B)=P(\bar{A} \cap B)=
View answer and explanation
Correct option: A
512\frac{5}{12}
Option A (512\frac{5}{12}) is correct because it calculates the probability of the event (Aˉ∩B\bar{A} \cap B) using the given probabilities of (A ∪ B), (A ∩ B), and (Aˉ\bar{A}).
487 Easy
487 of 1130
Application +4 | -1
A pack of cards contains 4 aces, 4 kings, 4 queens and 4 jacks. Two cards are drawn at random. The probability that atleast one of them is an ace is
View answer and explanation
Correct option: C
920\frac{9}{20}
Option C (920\frac{{9}}{{20}}) is correct because it is the probability of at least one ace being drawn, calculated by considering both cases: both cards are aces, or only one is an ace.
488 Hard
488 of 1130
Application +4 | -1
The probability that the roots of the equation x2+nx+12+n2=0{x^2} + nx + \frac{1}{2} + \frac{n}{2} = 0 are real where n∈Nn \in N such that n≤5,n \le 5, is
View answer and explanation
Correct option: D
04-May
Option D (04-May) is correct because it calculates the probability of the roots being real based on the possible values of n and the discriminant condition.
489 Easy
489 of 1130
Analyzing +4 | -1
Box-I contains 30 cards numbered 1 to 30 and Box - II contains cards numbered 31 to 50 . A box is selected at random and a card is drawn from it. The number on the card is found to be a non-prime number. The probability that the card was drawn from Box - I is
View answer and explanation
Correct option: C
817\frac{8}{17}
Option C (817\frac{8}{17}) is correct because it uses Bayes' theorem to find the probability that the card was drawn from Box-I given that it's a non-prime number.
490 Easy
490 of 1130
Application +4 | -1
In a binomial distribution B(n,p)B(n, p), the sum and the product of the mean and the variance are 5 and 6 respectively, then 6(n+p−q)6(n+p-q) is equal to
View answer and explanation
Correct option: A
52
Option A (52) is correct because it is the value obtained by solving the equations derived from the given sum and product of the mean and variance of the binomial distribution.
491 Medium
491 of 1130
Analyzing +4 | -1
An unbiased cubical die is thrown nn times. If xi=x_i= outcome on the ith i^{\text {th }} throw and X=gcd⁡(x1,x2,…..,xn)X=\operatorname{gcd}\left(x_1, x_2, \ldots . ., x_n\right), then E(X)E(X) is equal to
View answer and explanation
Correct option: C
16n(6n+3n+2n+1+8)\frac{1}{6^n}\left(6^n+3^n+2^{n+1}+8\right)
Option D (16n(6n+3n+2n+2+8)\frac{1}{6^n}\left(6^n+3^n+2^{n+2}+8\right)) is correct because it represents the expected value of the greatest common divisor of n throws of a die.
492 Easy
492 of 1130
Application +4 | -1
If XX follows a binomial distribution with parameters n=6n = 6 and p.p. If 4(P(X=4))=P(X=2),4\left( {P\left( {X = 4} \right)} \right) = P\left( {X = 2} \right), then p=p =
View answer and explanation
Correct option: D
1/31/3
Option D (1/3) is correct because solving the equation 4P(X=4)=P(X=2)4P(X=4) = P(X=2) for a binomial distribution with n=6 yields p=1/3.
493 Easy
493 of 1130
Application +4 | -1
The digits 1,2,3,4,5,6,7,81,2,3,4,5,6,7,8 and 9 are written in random order to form a nine digit number. The probability that this number is divisible by 4, is
View answer and explanation
Correct option: C
29\frac{2}{9}
Option C (29\frac{2}{9}) is correct because it calculates the probability of a nine-digit number formed by the digits 1-9 being divisible by 4 by considering the possible combinations of the last two digits.
494 Easy
494 of 1130
Application +4 | -1
Out of 40 consecutive natural numbers, two are chosen at random. Probability that the sum of the numbers is odd, is
View answer and explanation
Correct option: B
2039\frac{{20}}{{39}}
Option B (20/39) is correct because the sum is odd only when one number is odd and the other is even, and this probability is calculated using combinations.
495 Easy
495 of 1130
Application +4 | -1
For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability that the candidate can solve any problem is 45\frac{4}{5}, then the probability that he is unable to solve less than two problem is
View answer and explanation
Correct option: C
545(45)49\frac{{54}}{5}{\left( {\frac{4}{5}} \right)^{49}}
Option C (545(45)49\frac{{54}}{5}{\left( {\frac{4}{5}} \right)^{49}}) is correct because it is the probability that the candidate is unable to solve less than two problems, calculated using the binomial distribution.
496 Medium
496 of 1130
Application +4 | -1
Nine balls of the same size and color, numbered 1,2,…..,91,2, \ldots . ., 9, were put into a packet. Now A draws a ball from the packet, noted that it is of number a, and puts back it. Then B also draws a ball from the packet and noted that it is of number bb. Then the probability for the inequality a−2b+a-2 b+ 10>010>0 to hold is
View answer and explanation
Correct option: D
6181\frac{61}{81}
Option D (61/81) is correct because it represents the probability calculated by counting favorable outcomes where a - 2b + 10 > 0 and dividing by the total possible outcomes.
497 Easy
497 of 1130
Application +4 | -1
In a game two players AA and BB take turns in throwing a pair of fair dice starting with player AA and total of scores on the two dice, in each throw is noted. AA wins the game if he throws a total of 6 before BB throws a total of 7 and BB wins the game if he throws a total of 7 before AA throws a total of six. The game stops as soon as either of the players wins. The probability of AA winning the game is
View answer and explanation
Correct option: B
3061\frac{{30}}{{61}}
The correct answer is B (3061\frac{{30}}{{61}}) because it represents the probability of player A winning the game, calculated by considering the probabilities of throwing a 6 or 7 and using an infinite geometric series.
498 Medium
498 of 1130
Application +4 | -1
The probability that in a year of the 22nd century chosen at random there will be 53 Sundays, is
View answer and explanation
Correct option: D
528\frac{5}{{28}}
Option D (5/28) is correct because it calculates the probability of having 53 Sundays in a randomly chosen year of the 22nd century considering both leap and non-leap years.
499 Easy
499 of 1130
Application +4 | -1
A bag contains 4 red and 4 blue balls. Four balls are drawn one by one from the bag, then find the probability that the drawn balls are in alternate colour.
View answer and explanation
Correct option: D
635\frac{6}{35}
Option D (635\frac{{6}}{{35}}) is correct because it calculates the probability of drawing balls in alternating colors by considering both red-blue and blue-red sequences.
500 Easy
500 of 1130
Application +4 | -1
In a book of 500 pages, it is found that there are 250 typing errors. Assume that poisson law holds for the number of errors per page. Then, the probability that a random sample of 2 pages will contain no error, is
View answer and explanation
Correct option: C
e−1{e^{ - 1}}
Option C (e−1{e^{ - 1}}) is correct because it applies the Poisson distribution to calculate the probability of no errors on a random sample of 2 pages.
501 Easy
501 of 1130
Application +4 | -1
Three persons A,BA, B and CC are to speak at a function along with 7 other persons. The probability that A,BA, B, and CC speak together
View answer and explanation
Correct option: B
115\frac{1}{15}
Option B (115\frac{1}{15}) is correct because it represents the probability that A, B, and C speak together in a function with 7 other people.
502 Easy
502 of 1130
Application +4 | -1
A box contain NN coins, mm of which are fair and rest are biased. The probability of getting a head when a fair coin is tossed is 1/21 / 2, while it is 2/32 / 3 when a biased coin is tossed. A coin is drawn from the box at random and is tossed twice. The first time it shows head and the second time it shows tail. The probability that the coin drawn is fair is
View answer and explanation
Correct option: C
9m8N+m\frac{9 m}{8 N+m}
Option C (9m8N+m\frac{{9 m}}{{8 N+m}}) is correct because it represents the probability that the coin drawn is fair, given the observed sequence of head and tail, using Bayes' theorem.
503 Easy
503 of 1130
Application +4 | -1
Three distinct numbers are selected from first 100 natural numbers. The probability that all the three numbers are divisible by 2 and 3 is
View answer and explanation
Correct option: D
41155\frac{4}{1155}
Option D (41155\frac{4}{1155}) is correct because it represents the probability that all three selected numbers from the first 100 natural numbers are divisible by both 2 and 3.
504 Medium
504 of 1130
Application +4 | -1
A random variate XX takes the values 0, 1, 2, 3 and its mean is 1.3. If P(X=3)=2P(X=1)P\left( {X = 3} \right) = 2P\left( {X = 1} \right) and P(X=2)=0.3P\left( {X = 2} \right) = 0.3, then P(X=0)P\left( {X = 0} \right) is equal to
View answer and explanation
Correct option: D
0.4
Option D (0.4) is correct because it satisfies the given mean and probability conditions for the random variable X.
505 Easy
505 of 1130
Application +4 | -1
Suppose two numbers aa and bb are chosen randomly from the set {1,2,3,4,5,6}\{1,2,3,4,5,6\}. The probability that function f(x)=x3+ax2+bxf(x)=x^3+a x^2+b x is strictly increasing function on R\mathbf{R} is:
View answer and explanation
Correct option: C
1315\frac{13}{15}
Option D (1315\frac{13}{15}) is correct because it represents the probability of the function being strictly increasing, calculated by finding favorable outcomes where a2<3ba^2 < 3b and dividing by the total possible outcomes.
506 Easy
506 of 1130
Understanding +4 | -1
AA and BB 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
2425\frac{{24}}{{25}}
Option B (2425\frac{{24}}{{25}}) is correct because the probability of not winning is 1 minus the probability of winning.
507 Easy
507 of 1130
Application +4 | -1
One coin is thrown 100 times. The probability of coming tail in odd number
View answer and explanation
Correct option: A
12\frac{1}{2}
Option A (12\frac{1}{2}) is correct. In a binomial distribution with p=q=1/2, the probability of an odd number of successes is always 1/2.
508 Easy
508 of 1130
Application +4 | -1
A fair coin is tossed a fixed number of times. If the probability of getting 7 heads is equal to probability of getting 9 heads, then the probability of getting 2 heads is.
View answer and explanation
Correct option: A
15213\frac{{15}}{{{2^{13}}}}
The correct answer is A (15213\frac{{15}}{{{2^{13}}}}) because it correctly uses the binomial probability formula and the given condition to find the probability of getting 2 heads.
509 Easy
509 of 1130
Application +4 | -1
The mean and variance of a binomial distribution are 6 and 4. The parameter nn is
View answer and explanation
Correct option: A
18
Option A (18) is correct. Given the mean (np = 6) and variance (npq = 4) of a binomial distribution, we can solve for n by finding p and q first.
510 Medium
510 of 1130
Application +4 | -1
If the letters of the word ‘REGULATIONS’ be arranged at random, the probability that there will be exactly 4 letters between RR and EE is
View answer and explanation
Correct option: A
655\frac{6}{{55}}
Option A is correct because it represents the probability of having exactly 4 letters between R and E in a random arrangement of the word 'REGULATIONS', calculated by considering the arrangements of the letters with the given constraint.
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