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JEE • MATHEMATICS • PROBABILITY

Probability practice · Easy

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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 301-330 of 822 questions

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301 Easy
301 of 822
Understanding +4 | -1
Sixteen players S1,S2,…S16S_1, S_2, \ldots S_{16} play in a tournament. They are divided into eight pairs at random. From each pair a winner is decided on the basis of a game played between the two players of the pair. Assuming that all the palyers are to equal strength. The probability that the players S1S_1 is among the eight winners is
View answer and explanation
Correct option: A
12\frac{1}{2}
Option A (12\frac{1}{2}) is correct because each player is of equal strength, so the probability that S1S_1 wins their first game is simply 1/2.
302 Easy
302 of 822
Understanding +4 | -1
A card is drawn at random from a pack of 100 cards numbered 1 to 100 . The probability of drawing a number, which is a square, is-
View answer and explanation
Correct option: C
1/101 / 10
Option C (1/101 / 10) is correct because there are 10 perfect squares between 1 and 100, so the probability is 10/100.
303 Easy
303 of 822
Application +4 | -1
Suppose XX follows a binomial distribution with parameters n=6n=6 and pp. If 9P(X=4)=9 P(X=4)= P(X=2)P(X=2), then 4p4 p is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because, given the binomial distribution with n=6 and the equation 9P(X=4) = P(X=2), solving for p gives 4p = 1.
304 Easy
304 of 822
Application +4 | -1
An unbiased coin is tossed to get 2 points for turning up a head and one point for the tail. If three unbiased coins are tossed simultaneously, then the probability of getting a total of odd number of points
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Correct option: A
12\frac{1}{2}
Option A (12\frac{1}{2}) is correct because the probability of getting an odd number of points is the same as getting an even number of points when tossing unbiased coins.
305 Easy
305 of 822
Application +4 | -1
Out of 15 persons 10 can speak Hindi and 8 can speak English. If two persons are chosen at random, then the probability that one person speaks Hindi only and the other speaks both Hindi and English is
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Correct option: C
15\frac{1}{5}
Option C (15\frac{1}{5}) is correct because it calculates the probability of selecting one person who speaks only Hindi and another who speaks both languages.
306 Easy
306 of 822
Application +4 | -1
If XX is a poisson variate with P(X=0)=0.8,P\left( {X = 0} \right) = 0.8, then the variance of XX is
View answer and explanation
Correct option: C
log⁡e54{\log _e}\frac{5}{4}
Option C (log⁡e54{\log _e}\frac{5}{4}) is correct because the variance of a Poisson distribution is equal to its mean, which is found using the given probability P(X=0).
307 Easy
307 of 822
Application +4 | -1
In a box, there are 20 cards, out of which 10 are labelled as AA and the remaining 10 are labelled as BB. Cards are drawn at random, one after the other and with replacement, till a second AA - card is obtained. The probability that the second AA - card appears before the third BB - card is:
View answer and explanation
Correct option: D
1116\frac{{11}}{{16}}
Option D is correct because it accurately calculates the probability of the second A-card appearing before the third B-card by considering all possible sequences.
308 Easy
308 of 822
Application +4 | -1
Let AA and BB be two events associated to a random experiment such that PP (Exactly one occurs) =25,P(A∪B)=12=\frac{2}{5}, P(A \cup B)=\frac{1}{2}. Then, P(A∩B)P(A \cap B) is
View answer and explanation
Correct option: A
110\frac{1}{10}
Option A (110\frac{1}{10}) is correct because it uses the given probabilities to find P(A ∩ B) using the formula for the probability of exactly one event occurring.
309 Easy
309 of 822
Application +4 | -1
6 boys and 6 girls sit in a row randomly. The probability that all 6 girls sit together, is
View answer and explanation
Correct option: C
1132\frac{1}{{132}}
Option C (1132\frac{1}{{{{132}}}}) is correct because it is calculated by considering the 6 girls as one unit and then finding the ratio of favorable arrangements to total arrangements.
310 Easy
310 of 822
Analyzing +4 | -1
If P(A∩B)=1/3,P(A∪B)=5/6P(A \cap B)=1 / 3, P(A \cup B)=5 / 6 and P(A)=1/2P(A)=1 / 2, then which one of the following is correct?
View answer and explanation
Correct option: A
AA and BB are independent events
Option A (A and B are independent events) is correct because P(A∩B)=P(A)P(B)P(A \cap B) = P(A)P(B) holds true, indicating independence.
311 Easy
311 of 822
Application +4 | -1
Cards are drawn from a pack of 52 playing cards one by one. The probability that exactly 10 cards will be drawn before the first ace is
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Correct option: C
1644165\frac{164}{4165}
Option C (1644165\frac{164}{4165}) is correct because it represents the probability of drawing 10 non-ace cards followed by an ace from a standard deck of 52 cards.
312 Easy
312 of 822
Application +4 | -1
The probability that in a year of 22nd22 n d century chosen at random, there will be 53 Sundays is
View answer and explanation
Correct option: D
528\frac{5}{28}
Option D (528\frac{{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.
313 Easy
313 of 822
Application +4 | -1
A die is thrown 100 times. If getting an even number is considered a success, the variance of the number of successes, is
View answer and explanation
Correct option: B
25
Option B (25) is correct because the variance of a binomial distribution is calculated as npq, where n=100 and p=1/2.
314 Easy
314 of 822
Application +4 | -1
There are 9999 tickets bearing numbers 0001, 0002, …., 9999. If one ticket is selected from these tickets at random, the probability that the number on the ticket will consists of all different digits, is
View answer and explanation
Correct option: A
50409999\frac{{5040}}{{9999}}
Option A (50409999\frac{{5040}}{{9999}}) is correct because it represents the probability that the number on the ticket consists of all different digits, calculated by dividing the number of favorable outcomes by the total number of outcomes.
315 Easy
315 of 822
Application +4 | -1
Let a biased coin be tossed 5 times. If the probability of getting 4 heads is equal to the probability of getting 5 heads, then the probability of getting atmost two heads is:
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Correct option: D
4664\frac{{46}}{{{6^4}}}
Option D (4664\frac{{46}}{{{6^4}}}) is correct because it is the probability of getting at most two heads in 5 tosses of a biased coin, calculated using binomial probability with the given condition.
316 Easy
316 of 822
Analyzing +4 | -1
The probability that a randomly chosen 2×22 \times 2 matrix with all the entries from the set of first 10 primes, is singular, is equal to :
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Correct option: C
19103\frac{19}{10^3}
Option C (19103\frac{{19}}{{10^3}}) is correct because it represents the probability of a 2x2 matrix with entries from the first 10 primes being singular, calculated by considering cases where rows are proportional.
317 Easy
317 of 822
Application +4 | -1
If birth to a male child and birth to a female child are equal-probable, then what is the probability that at least one of the three children born to a couple is male?
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Correct option: B
78\frac{7}{8}
Option B (78\frac{7}{8}) is correct because it calculates the probability of at least one male child by considering all possible gender combinations for three children.
318 Easy
318 of 822
Understanding +4 | -1
The probability that an ordinary or a non-leap year has 53 sunday, is
View answer and explanation
Correct option: B
17\frac{1}{7}
Option B (17\frac{1}{7}) is correct because in a non-leap year, there is one extra day beyond 52 weeks, giving a 1/7 probability for that day to be a Sunday.
319 Easy
319 of 822
Application +4 | -1
One hundred identical coins, each with probability pp or showing up heads are tossed once. If 0<p<10 < p < 1 and the probability of heads showing on 50 coins is equal to that of heads showing on 51 coins, then the value of pp is
View answer and explanation
Correct option: D
51101\frac{{51}}{{101}}
Option D (51101\frac{{51}}{{101}}) is correct because it is the value of p that satisfies the condition P(X=50) = P(X=51) in a binomial distribution.
320 Easy
320 of 822
Application +4 | -1
A bag contains four tickets numbered 00,01,1000,01,10 and 11. Four tickets are chosen at random with replacement, then the probability that sum of the numbers on the tickets is 23 , is
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Correct option: A
332\frac{3}{32}
Option A is correct because the probability is calculated by finding the coefficient of x23 in the generating function and dividing by the total number of outcomes.
321 Easy
321 of 822
Application +4 | -1
AA and BB are two independent events such that their probabilities are 310\frac{3}{{10}} and 25\frac{2}{{5}} respectively. The probability of exactly one of the events happening is
View answer and explanation
Correct option: A
23/5023/50
Option A (23/5023/50) is correct because it is calculated using the probabilities of independent events and the formula for the probability of exactly one event happening.
322 Easy
322 of 822
Understanding +4 | -1
Two contestants play a game as follows: each is asked to select a digit from 1 to 9. If the two digits match they both win a prize. The probability that they will win a prize in a single trial is
View answer and explanation
Correct option: C
19\frac{1}{9}
Option C (19\frac{{1}}{{9}}) is correct because there are 9 ways the two contestants can select the same digit out of a total of 81 possible outcomes.
323 Easy
323 of 822
Application +4 | -1
Out of 13 applicants for a job, there are 5 women and 8 men. It is desired to select 2 persons for the job. The probability that at least one of the selected persons will be a women is
View answer and explanation
Correct option: A
25/3925/39
Option A (25/39) is correct. The probability of at least one woman is calculated by considering the cases where one woman and one man are selected, or two women are selected.
324 Easy
324 of 822
Application +4 | -1
In a box containing 100 bulbs, 10 are defective. What is the probability that out of a sample of 5 bulbs none is defective?
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Correct option: A
(910)5\left(\frac{9}{10}\right)^5
Option A ((910)5\left(\frac{{9}}{{10}}\right)^5) is correct because it calculates the probability of selecting 5 non-defective bulbs in a row.
325 Easy
325 of 822
Application +4 | -1
The mean and variance of a binomial distribution are 4 and 3 respectively, then the probability of getting exactly six successes in this distribution is
View answer and explanation
Correct option: B
  16C6(14)6(34)10{\;^{16}}{C_6}{\left( {\frac{1}{4}} \right)^6}{\left( {\frac{3}{4}} \right)^{10}}
Option B (  16C6(14)6(34)10{\;^{16}}{C_6}{\left( {\frac{1}{4}} \right)^6}{\left( {\frac{3}{4}} \right)^{10}}) is correct because it uses the binomial probability formula with the correct parameters derived from the mean and variance.
326 Easy
326 of 822
Application +4 | -1
In a school there are 100 students 60 of them don't like Chocolate and 50 don't like Biscuit and 10 of them like none then how many of them like both?
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Correct option: D
None of these
Option D (None of these) is correct because the calculated number of students who like both is 0, which is not among the given options.
327 Easy
327 of 822
Application +4 | -1
From a pack of cards two are accidently dropped. Probability that they are of opposite shade is
View answer and explanation
Correct option: C
2651\frac{{26}}{{51}}
Option C (2651\frac{{26}}{{51}}) is correct because it calculates the probability of drawing two cards of opposite shades by considering the number of ways to choose one red and one black card.
328 Easy
328 of 822
Application +4 | -1
Three dice are thrown together. The probability that the sum of the numbers appearing on them is 9 , is
View answer and explanation
Correct option: C
754\frac{7}{54}
Option B (25512\frac{{25}}{{512}}) is correct because it calculates the probability that the sum of the numbers on the five tickets drawn is 24.
329 Easy
329 of 822
Application +4 | -1
An unbiased cubical die is thrown 5 times. The probability that the maximum number appearing on the die is 4 is
View answer and explanation
Correct option: B
1023/651023 / 6^5
Option B (1023/651023 / 6^5) is correct because it represents the probability that the maximum number appearing on the die is 4 after 5 throws.
330 Easy
330 of 822
Application +4 | -1
A parents has two children. If one of them is boy, then the probability that other is, also a boy, is
View answer and explanation
Correct option: C
13\frac{1}{3}
Option C (13\frac{1}{3}) is correct because it's the conditional probability of having two boys given that one of them is a boy.
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