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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 451-480 of 822 questions

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451 Easy
451 of 822
Application +4 | -1
The probability of having at least one tail in 4 throws with a coin, is
View answer and explanation
Correct option: A
1516\frac{{15}}{{16}}
Option A (1516\frac{{15}}{{16}}) is correct because it represents the probability of getting at least one tail, which is the complement of getting no tails in 4 throws.
452 Easy
452 of 822
Application +4 | -1
Five persons A,   B,   C,   DA,\,\;B,\,\;C,\;\,D and EE are in queue of a shop. The probability that AA and EE always together, is
View answer and explanation
Correct option: C
25\frac{2}{5}
Option C (25\frac{2}{5}) is correct because it represents the probability of A and E being together, calculated by considering them as a single unit.
453 Easy
453 of 822
Application +4 | -1
Suppose n(≥3)n(\geq 3) persons are arranged in a row. The probability that two particular persons are not together is
View answer and explanation
Correct option: A
1−2n1-\frac{2}{n}
Option A (1−2n1-\frac{2}{n}) is correct because it represents the probability that two particular persons are not together, calculated by subtracting the probability that they are together from 1.
454 Easy
454 of 822
Application +4 | -1
An article manufactured by a company consits of two parts AA and BB. In the processs of manufacutre of part AA, 9 out of 100 are likely to be defective. Similarly 5 out of 100 are likely to be defective in the manufacture of part BB. The probability that the assembled part will not be defective, is
View answer and explanation
Correct option: A
0.8645
Option A (0.8645) is correct because it represents the probability that both parts A and B are non-defective, resulting in a non-defective assembled article.
455 Easy
455 of 822
Application +4 | -1
If a triangle is formed by choosing the vertices of 10 sided regular polygon. Then probability that triangle has exactly one side in common with the polygon, is
View answer and explanation
Correct option: D
12\frac{1}{2}
The correct answer is D (12\frac{{1}}{{2}}) because it is obtained by calculating the probability of forming a triangle with exactly one side common with the polygon.
456 Easy
456 of 822
Application +4 | -1
A fair die is tossed repeatedly until a six is obtained. Let XX denote the number of tosses required. The probability that X≥3X \geq 3 equals
View answer and explanation
Correct option: D
2536\frac{25}{36}
Option D (2536\frac{{25}}{{36}}) is correct because it represents the probability that at least 3 tosses are required to get a six, which is the probability of not getting a six in the first two tosses.
457 Easy
457 of 822
Application +4 | -1
A rifleman is firing at a distant target and has only 10%10 \% chance of hitting it. The least number of rounds, he must fire in order to have more than 50%50 \% chance of hitting it at least once, is
View answer and explanation
Correct option: C
7
Option C (7) is correct because it finds the minimum number of rounds needed for the probability of hitting the target at least once to be greater than 50%.
458 Easy
458 of 822
Application +4 | -1
A fair die is rolled. The probability that the first time 1 occurs at the even throw is
View answer and explanation
Correct option: B
511\frac{5}{{11}}
Option B (5/11) is correct because it represents the sum of an infinite geometric series where the first 1 occurs on an even throw.
459 Easy
459 of 822
Analyzing +4 | -1
If AA and BB are two independent events such that P(Bˉ)=αP(Aˉ)=710,P(Bˉ)=αP(\bar{B})=\alpha P(\bar{A})=\frac{7}{10}, P(\bar{B})=\alpha and P(A∪B)=810P(A \cup B)=\frac{8}{10}, then α\alpha, is
View answer and explanation
Correct option: A
2/72 / 7
Option A (2/7) is correct because it is the value of α\alpha that satisfies the given probability conditions for independent events A and B.
460 Easy
460 of 822
Analyzing +4 | -1
Let A,B,CA, B, C be pairwise independent events with P(C)>0P(C)>0 and P(A∩B∩C)=0P(A \cap B \cap C)=0. Then, P(Ac∩Bc/C)P\left(A^c \cap B^c / C\right) is equal to
View answer and explanation
Correct option: C
P(Ac)−P(B)P\left(A^c\right)-P(B)
Option C (P(Ac)−P(B)P(A^c)-P(B)) is the correct answer because it correctly applies the properties of pairwise independent events and conditional probability to simplify the expression for P(Ac∩Bc/C)P(A^c \cap B^c / C).
461 Easy
461 of 822
Application +4 | -1
′X′'X' speaks truth in 60%60 \% and ' YY ' in 50%50 \% of the cases. The probability that they contradict each other while narrating the some incident, is
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (12\frac{1}{2}) is correct because it calculates the probability of them contradicting each other by considering the cases where one speaks the truth and the other lies.
462 Easy
462 of 822
Application +4 | -1
An urn contains 5 red and 2 green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn; if the drawn ball is red, then a green ball is added to the urn the original ball is not returned to the urn. Now, a second ball is drawn at random from it. The probability that the second ball is red, is
View answer and explanation
Correct option: B
3249\frac{32}{49}
Option B (3249\frac{{32}}{{49}}) is correct because it calculates the probability of drawing a red ball in the second draw considering the two possible scenarios of the first draw.
463 Easy
463 of 822
Application +4 | -1
A box contains 10 mangoes out of which 4 are rotten. 2 mangoes are taken out together. If one of them is found to be good, the probability that the other is also good is
View answer and explanation
Correct option: C
5/135/13
Option C (5/135/13) is the correct probability, calculated using conditional probability considering that one mango is already known to be good.
464 Easy
464 of 822
Application +4 | -1
2n2 n boys are randomly divided into two subgroups containing nn boys each. The probability that the two tallest boys are in different groups is
View answer and explanation
Correct option: A
n2n−1\frac{n}{2 n-1}
Option A (n2n−1\frac{{n}}{{2n-1}}) is correct because it represents the probability that the two tallest boys are in different groups when 2n boys are divided into two groups of n.
465 Easy
465 of 822
Application +4 | -1
If the standard deviation of the binomial distribution (q+p)16(q+p)^{16} is 2, then mean of the distribution is
View answer and explanation
Correct option: B
8
Option B (8) is correct. Given the standard deviation, we can find p and then calculate the mean using the formula mean = np.
466 Easy
466 of 822
Application +4 | -1
If the mean and variance of a binomial variate XX are 2 and 1 respectively, then the probability that XX takes a value greater than 1, is
View answer and explanation
Correct option: D
15/1615 / 16
Option D (15/1615 / 16) is correct because it calculates the probability that X is greater than 1, given the mean and variance of a binomial variate.
467 Easy
467 of 822
Application +4 | -1
While shuffling a pack of 52 playing cards. Unfortunately 4 cards are dropped on the ground it is known that 2 out of 4 are spades. What is the probability that all dropped 4 are spades?
View answer and explanation
Correct option: D
none of these
Option D (none of these) is correct because the provided options do not match the correct conditional probability calculation.
468 Easy
468 of 822
Application +4 | -1
If four positive integers are taken at random and multiplied together, then the probability that the last digit is 1, 3, 7 or 9 is
View answer and explanation
Correct option: D
16625\frac{16}{625}
Option D (16625\frac{16}{625}) is correct because it represents the probability that the last digit of the product of four random positive integers is 1, 3, 7, or 9.
469 Easy
469 of 822
Application +4 | -1
The probability that the birth days of six different persons will fall in exactly two calendar months is
View answer and explanation
Correct option: D
341125\frac{341}{12^5}
Option D (341125\frac{341}{12^5}) is correct because it calculates the probability of six birthdays falling in exactly two months, accounting for the combinations of months and excluding cases where all birthdays are in one month.
470 Easy
470 of 822
Analyzing +4 | -1
A three digit number, which is a multiple of 11, is chosen at random. Probability that the number so chosen is also a multiple of 9, is equal to
View answer and explanation
Correct option: A
19\frac{1}{9}
Option A (1/9) is correct because it represents the probability that a randomly chosen three-digit multiple of 11 is also a multiple of 9.
471 Easy
471 of 822
Application +4 | -1
An integer is chosen at random from the first 200 positive integers. The probability that the integer chosen is divisible by 6 or 8 is
View answer and explanation
Correct option: B
1/41 / 4
Option B (14\frac{1}{4}) is correct because it calculates the probability of an integer being divisible by 6 or 8 by considering the individual probabilities and subtracting the probability of being divisible by both.
472 Easy
472 of 822
Application +4 | -1
Let SS be the sample space of all five digit numbers. It pp is the probability that a randomly selected number from SS, is multiple of 7 but not divisible by 5 , then 9p9 p is equal to
View answer and explanation
Correct option: C
1.02851.0285
Option C (1.02851.0285) is correct because it is the result of calculating the probability of a 5-digit number being a multiple of 7 but not divisible by 5, and then multiplying by 9.
473 Easy
473 of 822
Application +4 | -1
If n( A)=1000,n( B)=500n(\mathrm{~A})=1000, n(\mathrm{~B})=500 and if n( A∩B)≥1n(\mathrm{~A} \cap \mathrm{B}) \geq 1 and n( A∪B)=pn(\mathrm{~A} \cup \mathrm{B})=p, then
View answer and explanation
Correct option: D
1000≤p≤14991000 \leq p \leq 1499
Option D (1000≤p≤14991000 \leq p \leq 1499) is correct because it represents the range of possible values for n(A∪B)n(A \cup B) given the constraints on n(A∩B)n(A \cap B).
474 Easy
474 of 822
Application +4 | -1
A die is thrown. Let AA be the event that the number obtained is greater than 3. Let B be the event that the number obtained is less than 5. Then, P(A∪B)P(A \cup B) is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the union of the events A and B covers all possible outcomes when throwing a die.
475 Easy
475 of 822
Application +4 | -1
A,B,CA, B, C are three mutually exclusive and exhaustive events associated with a random experiment. If P(B){P}({B}) =32P(A)=\frac{3}{2} {P}({A}) and P(C)=12P(B){P}({C})=\frac{1}{2} {P}({B}), then P(A)={P}({A})=
View answer and explanation
Correct option: A
413\frac{4}{13}
Option A (413\frac{{4}}{{13}}) is correct because it's the probability of event A, calculated using the mutually exclusive and exhaustive nature of events A, B, and C.
476 Easy
476 of 822
Analyzing +4 | -1
Six dice are thrown simultaneously. The probability that exactly three of them show the same face and remaining three show different faces, is
View answer and explanation
Correct option: C
(5!)22(66)\frac{(5 !)^2}{2\left(6^6\right)}
Option C ((5!)22(66)\frac{{(5 !)^2}}{{2(6^6)}}) is correct because it represents the probability of exactly three dice showing the same face and the remaining three showing different faces.
477 Easy
477 of 822
Application +4 | -1
A class consists of 80 students, 25 of them are girls. If 10 of the students are rich and 20 of the students are fair complexioned, then the probability of selecting a fair complexioned rich girl from the class (assuming three traits as independent) is
View answer and explanation
Correct option: C
5512\frac{5}{512}
Option C (5512\frac{5}{512}) is correct because it correctly calculates the probability of selecting a fair complexioned rich girl assuming independence.
478 Easy
478 of 822
Application +4 | -1
If AA and BB each toss three coins. The probability that both 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 that both A and B get the same number of heads by summing the probabilities of each getting 0, 1, 2, or 3 heads.
479 Easy
479 of 822
Application +4 | -1
The probability that when 12 balls are distributed among three boxes, the first will contain three balls is,
View answer and explanation
Correct option: B
12C3×29312\frac{{ }^{12} C_3 \times 2^9}{3^{12}}
Option B (12C3×29312\frac{{ }^{12} C_3 \times 2^9}{3^{12}}) is correct because it calculates the probability of the first box containing three balls by considering the combinations of choosing the three balls and the distribution of the remaining balls into the other two boxes.
480 Easy
480 of 822
Analyzing +4 | -1
Let AA be a set consisting of nn elements. The probability of selecting two subsets PP and QQ of set AA such that Q=PˉQ=\bar{P}, is
View answer and explanation
Correct option: B
12n−1\frac{1}{2^n-1}
Option B (12n−1\frac{{1}}{{2^n-1}}) is correct because it represents the probability of selecting two subsets P and Q of set A such that Q is the complement of P.
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