Skip to main content
Back to PROBABILITY

Your Progress

0 / 1130 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • PROBABILITY

Probability practice · Mixed difficulty

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 331-360 of 1130 questions

Preparing practice and saved progress…

331 Easy
331 of 1130
Application +4 | -1
Let 20 distinct balls have been randomly distributed in to 4 distinct boxes, 5 into each. Let 'A' be the event that two specific balls have been put into a particular box. The probability of occurence of event ' A\mathrm{A} ' is :
View answer and explanation
Correct option: A
119\frac{1}{19}
Option A (119\frac{1}{19}) is correct because after placing the first ball, there are 19 remaining spots, and only 4 of them are in the same box, so the probability is 4/19. However, the solution provided calculates the probability of any two balls being in the same box, not two specific balls. The correct probability is 1/19.
332 Easy
332 of 1130
Application +4 | -1
A fair die is tossed until six is obtained on it. Let XX be the number of required tosses, then the conditional probability P(X≥6∣X>2)P\left( {X \ge 6|X > 2} \right) is
View answer and explanation
Correct option: D
2536\frac{{25}}{{36}}
Option D (2536\frac{{25}}{{36}}) is correct because it is the conditional probability of needing at least 6 tosses given that more than 2 tosses are already needed.
333 Easy
333 of 1130
Application +4 | -1
Four fair dice D1,D2,D3D_1, D_2, D_3 and D4D_4 each having six faces numbered 1,2,3,4,51,2,3,4,5 and 6 are rolled simultaneously. The probability that D4D_4 shows a number appearing on one of D1,D2D_1, D_2 and D3D_3, is
View answer and explanation
Correct option: A
91216\frac{91}{216}
Option A (91216\frac{{91}}{{216}}) is correct because it sums the probabilities of the different cases where D4 shows a number appearing on one of D1, D2, and D3.
334 Easy
334 of 1130
Application +4 | -1
An urn contains 4 white and 3 red balls. There balls are drawn with replacement from this urn. Then, the standard deviation of the number of red balls drawn is
View answer and explanation
Correct option: A
67\frac{6}{7}
Option A is correct because the standard deviation is calculated using the binomial distribution with replacement.
335 Medium
335 of 1130
Analyzing +4 | -1
10 mangoes are to be distributed among 5 persons. The probability that at least one of them will receive none, is
View answer and explanation
Correct option: D
125143\frac{125}{143}
Option D (125143\frac{125}{143}) is correct because it calculates the probability that at least one person receives no mangoes by subtracting the probability that each person receives at least one mango from 1.
336 Medium
336 of 1130
Application +4 | -1
There are three events E1,E2{E}_1, {E}_2 and E3,{E}_3, one of which must, and only one can happen. The odds are 7 to 4 against E1{E}_1 and 5 to 3 against E2{E}_2. The odds against E3{E}_3 is
View answer and explanation
Correct option: D
65:2365: 23
Option D (65:2365: 23) is correct because it calculates the odds against E3 by first finding the probability of E3 and then using the odds formula.
337 Easy
337 of 1130
Application +4 | -1
A basket contains 5 apples and 7 oranges and another basket contains 4 apples and 8 oranges. One fruit is picked out from each basket. The probability that the fruits are both apples or both oranges, is
View answer and explanation
Correct option: C
68144\frac{68}{144}
Option C (68144\frac{68}{144}) is correct because it calculates the probability of picking both apples or both oranges from two baskets.
338 Medium
338 of 1130
Application +4 | -1
There are 5 identical red balls and 3 identical yellow balls in a box. Two balls are drawn at random, then the probability that their colour is different is
View answer and explanation
Correct option: C
1528\frac{15}{28}
Option C (1528\frac{15}{28}) is correct because it represents the probability of drawing two balls of different colors, calculated by considering the combinations of red and yellow balls.
339 Medium
339 of 1130
Analyzing +4 | -1
From a set of 100 cards numbered 1 to 100, one card is drawn at random. The probability that the number obtained on the card is divisible by 6 or 8 but not by 24, is
View answer and explanation
Correct option: D
15\frac{1}{5}
Option D (15\frac{1}{5}) is correct because it correctly calculates the number of integers divisible by 6 or 8 but not by 24 and divides it by the total number of cards.
340 Easy
340 of 1130
Application +4 | -1
Let ww be a complex cube root of unity with w≠1w \neq 1. A fair die is thrown three times. If r1,r2r_1, r_2 and r3r_3 are the numbers obtained on the die, then the probability that wr1+wr2+wr3=0w^{r_1}+w^{r_2}+w^{r_3}=0 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 the sum of the cube roots of unity being zero by considering the possible combinations of dice rolls.
341 Medium
341 of 1130
Application +4 | -1
The probabilities of Ramesh using car or scooter or bus or train for going to office are respectively 17,37,27\frac{1}{7}, \frac{3}{7}, \frac{2}{7} and 17\frac{1}{7}. The probabilities of his reaching the office late using these modes of transport are respectively 29,19,49\frac{2}{9}, \frac{1}{9}, \frac{4}{9} and 19\frac{1}{9}, On one day Ramesh reaches his office on time. Then the probability that he used car on that day is
View answer and explanation
Correct option: B
17\frac{1}{7}
Option B (17\frac{1}{7}) is correct because it is calculated using Bayes' theorem, considering the probability of using a car given that Ramesh reached the office on time.
342 Easy
342 of 1130
Analyzing +4 | -1
A pack of plying cards was found to contain only 51 cards. If the first 13 cards which are examined are all red, then the probability that the missing cards is black, is
View answer and explanation
Correct option: B
23\frac{2}{3}
Option B (23\frac{2}{3}) is correct because it is calculated using conditional probability and Bayes' theorem, considering the prior probabilities of losing a black or red card.
343 Medium
343 of 1130
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
16\frac{1}{6}
Option B (16\frac{1}{6}) is correct because it is the probability that the first two machines tested are the faulty ones.
344 Easy
344 of 1130
Application +4 | -1
The probability of selecting integers a∈[−5,30] a \in \left[ { - 5,30} \right]\, such that x2+2(a+4)x−5a+64>0,{x^2} + 2\left( {a + 4} \right)x\mathrm{ } - \mathrm{ }5a\mathrm{ } + \mathrm{ }64\mathrm{ } > \mathrm{ }0,\mathrm{ } for all x∈R,x \in R, is
View answer and explanation
Correct option: B
29\frac{2}{{9}}
Option B (29\frac{2}{{9}}) is correct because it represents the probability of 'a' lying within the interval (-16, 3) given the constraint a ∈ [-5, 30].
345 Easy
345 of 1130
Application +4 | -1
A coin is tossed 3 times by 2 persons. The probability that both get equal number of heads, is
View answer and explanation
Correct option: C
5/165 / 16
The correct answer is 5/165 / 16 because it represents the probability that both persons get an equal number of heads when tossing a coin 3 times each.
346 Medium
346 of 1130
Analyzing +4 | -1
If x=3n,x=3^n, where nn is a positive integral value, then what is the probability that xx will have 3 at unit's place?
View answer and explanation
Correct option: B
14\frac{1}{4}
Option B (14\frac{1}{4}) is correct because the unit's digit of 3n3^n repeats every 4 powers (3, 9, 7, 1).
347 Medium
347 of 1130
Application +4 | -1
For a party 7 guests are invited by a husband and his wife. They sit in a row for dinner. The probability that the husband and his wife sit together, is
View answer and explanation
Correct option: B
29\frac{2}{9}
Option B (29\frac{2}{9}) is correct because it is the probability of the husband and wife sitting together, calculated by considering them as a single unit.
348 Easy
348 of 1130
Application +4 | -1
A die thrown two times and the sum of the scores appearing on the die is observed to be a multiple of 4. Then the conditional probability that the score 4 has appeared at least once is
View answer and explanation
Correct option: D
19\frac{1}{9}
Option D (19\frac{1}{9}) is correct because it is the conditional probability calculated based on the given events.
349 Hard
349 of 1130
Application +4 | -1
In an assembly of 4 persons the probability that at least 2 of them have the same birthday, is
View answer and explanation
Correct option: D
0.016
Option D (0.016) is correct because it approximates the probability that at least two people out of four share the same birthday.
350 Easy
350 of 1130
Application +4 | -1
If AA and BB are two events, such that P(A∪B)=34,  P(A∩B)=14,P(Aˉ)=23P\left( {A \cup B} \right) = \frac{3}{4},\;P\left( {A \cap B} \right) = \frac{1}{4},P\left( {\bar A} \right) = \frac{2}{3}, then P(Aˉ∩B)P\left( {\bar A \cap B} \right) is equal to
View answer and explanation
Correct option: A
512\frac{5}{{12}}
Option C (25216\frac{{25}}{{216}}) is correct because it uses the binomial probability formula to calculate the probability of getting a sum of 7 exactly twice in 4 independent trials.
351 Easy
351 of 1130
Application +4 | -1
Three identical dice are thrown together. The probability that distinct numbers appear on them, is
View answer and explanation
Correct option: B
59\frac{5}{9}
Option A (35324\frac{{35}}{{324}}) is correct because it calculates the probability that the sum of the numbers appearing on four dice is 13.
352 Easy
352 of 1130
Application +4 | -1
In eight throws of a die 1 or 3 is considered a success. Then, the standard deviation of success, is
View answer and explanation
Correct option: C
43\frac{4}{3}
Option C (43\frac{4}{3}) is correct because it is the standard deviation calculated using the formula npq\sqrt{npq} with the given parameters.
353 Medium
353 of 1130
Application +4 | -1
If AA and BB are mutually exclusive events with P(B)≠1P\left( B \right) \ne 1, then P(A∣Bˉ)P(A|\bar B) is equal to (Here, Bˉ\bar B is the complement of the event BB)
View answer and explanation
Correct option: D
P(A)1−P(B)\frac{{P\left( A \right)}}{{1 - P\left( B \right)}}
Option D (P(A)/(1-P(B))) is correct because it correctly applies the conditional probability formula given that A and B are mutually exclusive.
354 Easy
354 of 1130
Application +4 | -1
Two numbers are selected randomly from the set S={1,2,3,4,5,6}S=\{1,2,3,4,5,6\} without replacement one by one. The probability that the minimum of the two numbers is less than 4 is
View answer and explanation
Correct option: D
45\frac{4}{5}
Option D (45\frac{4}{5}) is correct because it calculates the probability that at least one of the two selected numbers is less than 4.
355 Easy
355 of 1130
Application +4 | -1
An unbiased coin is tossed eight times. The probability of obtaining at least one head and at least one tail is :
View answer and explanation
Correct option: B
127128\frac{127}{128}
Option B (127/128) is correct because it calculates the probability of not getting all heads or all tails in eight coin tosses.
356 Easy
356 of 1130
Application +4 | -1
Three groups A,B,CA, B, C are contesting for position on the Board of Directors of a company. The probabilities of their winning are 0.5,0.3,0.20.5,0.3,0.2 respectively. If the group A wins, the probability of introducing a new product is 0.7 and the corresponding probabilities for group BB and CC are 0.6 and 0.5 respectively. The probability that the new product will be introduced, is
View answer and explanation
Correct option: B
0.63
Option B (0.63) is correct because it calculates the overall probability of introducing a new product by considering the probabilities of each group winning and the conditional probabilities of introducing the product given each group wins.
357 Easy
357 of 1130
Application +4 | -1
Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Let XX denote the random variable of numbers of aces obtained in the two drawn cards. Then P(X=1)+P(X=2)P(X=1)+P(X=2) equals
View answer and explanation
Correct option: B
25169\frac{25}{169}
Option B (25169\frac{25}{169}) is correct because it calculates the sum of probabilities P(X=1) and P(X=2) using binomial distribution.
358 Easy
358 of 1130
Application +4 | -1
It is given the events AA and BB are such that P(A)=14,P(A∣B)=12P\left( A \right) = \frac{1}{4},P\left( {A\mathrm{|}B} \right) = \frac{1}{2} and P(B∣A)=23P\left( {B\mathrm{|}A} \right) = \frac{2}{3}. Then, P(B)P\left( B \right) is
View answer and explanation
Correct option: C
13\frac{1}{3}
Option C (13\frac{1}{3}) is correct because it is calculated using the conditional probability formula and the given probabilities.
359 Easy
359 of 1130
Application +4 | -1
Two dice are tossed 6 times. Then the probability that 7 will show an exactly four of the tosses is
View answer and explanation
Correct option: C
12515552\frac{{125}}{{15552}}
Option C (12515552\frac{{125}}{{15552}}) is correct because it is the probability calculated using the binomial probability formula for exactly four successes out of six trials.
360 Easy
360 of 1130
Application +4 | -1
A pair of a dice thrown, if 5. appears on at least one of the dice, then the probability that the sum is 10 or greater, is
View answer and explanation
Correct option: D
112\frac{1}{{12}}
Option D (112\frac{1}{{12}}) is correct because it is calculated by dividing the number of favorable outcomes (sum is 10 or greater with at least one 5) by the total number of outcomes in the sample space (36).
Revision notes and key takeaways

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

Showing 331-360 of 1130 questions

Your practice

0 / 1130 answered · 1130 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.