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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 271-300 of 1130 questions

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271 Easy
271 of 1130
Application +4 | -1
If XX follows a binomial distribution with parameters n=100n = 100 and P=13P = \frac{1}{3}, then P(X=r)P\left( {X = r} \right) is maximum when rr is equal to
View answer and explanation
Correct option: C
33
Option C is correct because the maximum probability occurs at r = floor((n+1)p), which is 33 in this case.
272 Medium
272 of 1130
Application +4 | -1
Suppose that a die (with faces marked 1 to 6) is loaded in such a manner that for K=1,2,3,….,6K = 1,2,3, \ldots .,6 the probability of the face marked KK turning up when die is tossed is proportional to KK. The probability of the event that the outcome of a toss of the die will be an even number, is equal to
View answer and explanation
Correct option: A
12\frac{1}{2}
Option A, 12\frac{1}{2} is correct because when the probability is proportional to K, the probability of an even number is simply the sum of probabilities of even numbers divided by the sum of probabilities of all numbers.
273 Medium
273 of 1130
Application +4 | -1
An integer is chosen at random from the integers 1,2,3,…,501,2,3, \ldots, 50. The probability that the chosen integer is a multiple of atleast one of 4,6 and 7 is
View answer and explanation
Correct option: C
2150\frac{21}{50}
The correct answer is C (21/50) because it represents the probability calculated using the principle of inclusion-exclusion for multiples of 4, 6, and 7.
274 Easy
274 of 1130
Application +4 | -1
A die is thrown twice and the sum of the numbers appearing is observed to be 6 . The conditional probability that the number 4 has appeared at least once, is
View answer and explanation
Correct option: B
25\frac{2}{5}
Option B (25\frac{2}{5}) is correct because it calculates the conditional probability of the number 4 appearing at least once, given that the sum of the numbers is 6.
275 Easy
275 of 1130
Application +4 | -1
Two dice are thrown together. If the numbers appearing on the two dice are different, then what is the probability that the sum is 6?
View answer and explanation
Correct option: C
215\frac{2}{{15}}
Option C (215\frac{2}{{15}}) is correct because it calculates the conditional probability of the sum being 6 given that the numbers on the two dice are different.
276 Easy
276 of 1130
Application +4 | -1
A coin is tossed nn times the probability of getting head at least once is greater than 0.8. Then the least value of such nn is
View answer and explanation
Correct option: B
3
Option B (3) is correct because it is the smallest integer value of n for which the probability of getting at least one head is greater than 0.8.
277 Medium
277 of 1130
Analyzing +4 | -1
If XX is a binomial variate with parameters nn and pp, where 0<p<10 < p < 1 such that P(X=r)P(X=n−r)\frac{P(X=r)}{P(X=n-r)} is independent of nn and rr, then pp equals
View answer and explanation
Correct option: A
1/21 / 2
Option A (1/21/2) is correct because the condition P(X=r)P(X=n−r)\frac{P(X=r)}{P(X=n-r)} being independent of nn and rr implies that pp must be equal to 1/21/2.
278 Easy
278 of 1130
Understanding +4 | -1
Let Ω\Omega be the sample space and A⊆ΩA \subseteq \Omega be an event.

Given below are two statements:

(S1) : If P(A)=0P(A)=0, then A=∅A=\varnothing

(S2) : If P(A)=1P(A)=1, then A=ΩA=\Omega

Then
View answer and explanation
Correct option: D
Both (S1) and (S2) are true
Option D is correct. If P(A) = 0, A is almost surely empty, and if P(A) = 1, A is almost surely the entire sample space.
279 Medium
279 of 1130
Application +4 | -1
If X\mathrm{X} follows a binomial distribution with parameters n=100\mathrm{n}=100 and p=1/3\mathrm{p}=1 / 3, find r\mathrm{r} for which P(X=r)\mathrm{P}(\mathrm{X}=\mathrm{r}) is maximum
View answer and explanation
Correct option: C
33
Option C (33) is correct because the value of r for which P(X=r) is maximum is the mode of the binomial distribution, which is the integer part of (n+1)p = (100+1)(1/3) = 33.67, hence r=33.
280 Easy
280 of 1130
Application +4 | -1
A bag contains 4 tickets numbered 1, 2, 3, 4 and another bag contains 6 tickets numbered 2, 4, 6, 7, 8, 9. One bag is chosen and a ticket is drawn. The probability that the ticket bears the number 4 is
View answer and explanation
Correct option: C
May-24
Option D (May-24) is correct because the probability is calculated by considering both bags and the chance of drawing a 4 from each.
281 Easy
281 of 1130
Application +4 | -1
If AA and BB are mutually exclusive events with P(B)≠1P(B) \neq 1, then P(A/Bˉ)=P(A / \bar{B})=
View answer and explanation
Correct option: D
P(A)1−P(B)\frac{P(A)}{1-P(B)}
Option D (P(A)/(1-P(B))) is correct because it applies the conditional probability formula and uses the fact that A and B are mutually exclusive.
282 Medium
282 of 1130
Application +4 | -1
An urn contains 6 white and 9 black balls. Two successive draws of 4 balls are made without replacement. The probability that the first draw gives all white balls and second draw gives all black balls is
View answer and explanation
Correct option: D
3715\frac{3}{715}
Option D (3715\frac{3}{715}) is correct because the probability of drawing 4 white balls then 4 black balls without replacement is calculated as 6C415C4⋅9C411C4=3715\frac{{}^6C_4}{{}^{15}C_4} \cdot \frac{{}^9C_4}{{}^{11}C_4} = \frac{3}{715}.
283 Hard
283 of 1130
Understanding +4 | -1
If AA and BB are independent events such that P(A)>0,P(B)>0,P\left( A \right) > 0,P\left( B \right) > 0, then
View answer and explanation
Correct option: D
P(A/B)+P(Aˉ/B)=1P\left( {A/B} \right) + P\left( {\bar A/B} \right) = 1
Option D (P(A/B)+P(Aˉ/B)=1P(A/B) + P(\bar A/B) = 1) is correct because P(A/B)+P(Aˉ/B)P(A/B) + P(\bar A/B) represents the sum of conditional probabilities of an event and its complement given B, which always equals 1.
284 Easy
284 of 1130
Application +4 | -1
AA and BB are two events such that odds against AA are 2:12: 1 odds in favour of A∪BA \cup B are 3:13: 1. If x≤P(B)≤yx \leq P(B) \leq y, then the ordered pair (x,y)(x, y) is
View answer and explanation
Correct option: A
(5/12,3/4)(5 / 12,3 / 4)
Option A ((512,34)\left(\frac{{5}}{{12}}, \frac{{3}}{{4}}\right)) is correct because it determines the range of P(B) based on the given probabilities and the properties of probability.
285 Easy
285 of 1130
Application +4 | -1
A coin is tossed 2n2 n times. The chance that the number of times one gets head is not equal to the number of times one gets tail, is
View answer and explanation
Correct option: C
1−2nCn(14n)1-{ }^{2 n} C_n\left(\frac{1}{4^n}\right)
The correct answer is 1−2nCn(14n)1-{ }^{2 n} C_n\left(\frac{1}{4^n}\right) because it represents 1 minus the probability of getting an equal number of heads and tails.
286 Easy
286 of 1130
Analyzing +4 | -1
AA is a set containing nn elements. A subset PP of AA is chosen at random. The set AA is reconstructed by replacing the elements of PP. A subset QQ is again chosen at random. The probability that PP and QQ have equal number of elements, is
View answer and explanation
Correct option: B
2nCn4n\frac{{ }^{2 n} C_n}{4^n}
Option B (2nCn4n\frac{{{{ }^{2 n} C_n}}}{{4^n}}) is correct because it represents the probability that P and Q have an equal number of elements, considering all possible choices for sets P and Q.
287 Easy
287 of 1130
Application +4 | -1
A company has two plants A and B to manufacture motorcycles. 60%60 \% motorcycles are manufactured at plant A and the remaining are manufactured at plant B. 80%80 \% of the motorcycles manufactured at plant A are rated of the standard quality, while 90%90 \% of the motorcycles manufactured at plant B are rated of the standard quality. A motorcycle picked up randomly from the total production is found to be of the standard quality. If p is the probability that it was manufactured at plant B, then 126p126 \mathrm{p} is
View answer and explanation
Correct option: A
54
Option A (54) is correct because it is the value of 126p, where p is the probability that the motorcycle was manufactured at plant B given that it is of standard quality, calculated using Bayes' theorem.
288 Easy
288 of 1130
Application +4 | -1
It is 8:58: 5 against a person who is now 40 years old living till he is 70 and 4:34: 3 against a person now 50 living till he is 80 . The probability that at least one of these persons will be alive 30 years hence, is
View answer and explanation
Correct option: C
5991\frac{59}{91}
Option A (27\frac{2}{7}) is correct because it calculates the probability that the vowels occupy only odd positions when arranging the letters of the word 'longest'.
289 Easy
289 of 1130
Application +4 | -1
Three coins are tossed together, then the probability of getting at least one head is
View answer and explanation
Correct option: D
78\frac{7}{8}
Option D (78\frac{7}{8}) is correct because it represents the probability of getting at least one head, which is the complement of getting all tails.
290 Medium
290 of 1130
Analyzing +4 | -1
For any two independent events E1{E_1} and E2,  P{(E1∪E2)∩(Eˉ1)∩(Eˉ2)}{E_2},\;P\left\{ {\left( {{E_1} \cup {E_2}} \right) \cap \left( {{{\bar E}_1}} \right) \cap \left( {{{\bar E}_2}} \right)} \right\}
View answer and explanation
Correct option: A
≤1/4\le 1/4
Option A (≤1/4\le 1/4) is correct because the expression simplifies to 0, which is less than or equal to 1/4.
291 Easy
291 of 1130
Analyzing +4 | -1
Out of 3n3 n consecutive integers, three are selected at random. Find the probability that their sum is divisible by 3 , is
View answer and explanation
Correct option: C
3n2−3n+2(3n−1)(3n−2)\frac{3 n^2-3 n+2}{(3 n-1)(3 n-2)}
Option C (3n2−3n+2(3n−1)(3n−2)\frac{{3 n^2-3 n+2}}{{(3 n-1)(3 n-2)}}) is correct because it represents the probability calculated by dividing the number of ways to choose three integers whose sum is divisible by 3 by the total number of ways to choose three integers.
292 Easy
292 of 1130
Application +4 | -1
Two events AA and BB are such that P(B)=0.55P(B)=0.55 and P(AB′)=0.15P\left(A B^{\prime}\right)=0.15. The probability of occurrence of at least one of event is
View answer and explanation
Correct option: A
0.7
Option A (0.7) is correct because it calculates the probability of at least one of the events occurring using the formula P(A∪B) = P(AB') + P(B).
293 Easy
293 of 1130
Application +4 | -1
A bag contains 3 red and 7 black balls. Two balls are selected at random without replacement. If the second selected ball is given to be red, the Probability that the first selected ball is also red, is
View answer and explanation
Correct option: B
29\frac{2}{9}
Option B (29\frac{2}{9}) is correct because it represents the conditional probability that the first ball is red, given that the second ball is red.
294 Easy
294 of 1130
Application +4 | -1
A fair die is tossed repeatedly until a six is obtained. Let XX denote the number of tosses required. Then, the probability that X=3X=3 equals
View answer and explanation
Correct option: A
25216\frac{25}{216}
Option A (25/216) is correct because it represents the probability of not getting a six in the first two tosses and then getting a six on the third toss.
295 Medium
295 of 1130
Application +4 | -1
A bag contains 5 white and 3 black balls and 4 balls are successively drawn out and not replaced. The probability that they are alternately of different colours, is
View answer and explanation
Correct option: C
17\frac{1}{7}
Option C (1/7) is correct because it calculates the probability of alternately drawing white and black balls without replacement.
296 Easy
296 of 1130
Application +4 | -1
A bag contains 5 red and 3 bluc balls. If 3 balls are drawn at random without replacement, then the probability of getting exactly one red ball is
View answer and explanation
Correct option: C
1556\frac{15}{56}
Option C (1556\frac{15}{56}) is correct because it is the probability of drawing exactly one red ball out of three, calculated using combinations.
297 Easy
297 of 1130
Application +4 | -1
100 Students appeared for two examinations, 60 passed the first, 50 passed the second and 30 passed both. The Probability that a student selected at random has failed in both examinations, is
View answer and explanation
Correct option: B
0.2
Option B (0.2) is correct because it calculates the probability of a student failing both exams by subtracting the probability of passing at least one exam from 1.
298 Medium
298 of 1130
Application +4 | -1
Two integers rr and ss are drawn one at a time without replacement from the set 1,2,⋯ ,n1,2, \cdots, n. If pk=P(r≤k∣s≤k)p_k=P(r \leq k \mid s \leq k), then 4p74 p_7 is if n=25n=25.
View answer and explanation
Correct option: A
1
Option A (1) is correct because given the condition p_k = P(r <= k | s <= k), and n=25, k=7, then 4p_7 = 1.
299 Medium
299 of 1130
Application +4 | -1
Bag A contains 4 green and 3 red balls and Bag B\mathrm{B} contains 4 red and 3 green balls. One bag is taken at random and a ball is noted it is green. The probability that it comes from Bag  B\mathrm{Bag} \,\, \mathrm{B} is -
View answer and explanation
Correct option: C
37\frac{3}{7}
Option C (37\frac{{3}}{{7}}) is correct because the probability is calculated using Bayes' theorem, considering the probabilities of drawing a green ball from each bag.
300 Easy
300 of 1130
Application +4 | -1
A number is chosen at random among the first 120 natural numbers. The probability of the number chosen being a multiple of 5 or 15 is
View answer and explanation
Correct option: B
15\frac{1}{5}
Option B (15\frac{1}{5}) is correct because it calculates the probability of a number chosen from the first 120 natural numbers being a multiple of 5 or 15.
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