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JEE • MATHEMATICS • PROBABILITY

Probability practice · Medium

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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 211-240 of 285 questions
211 Medium
211 of 285
Application +4 | -1
In a lottery three were 90 tickets numbered 1 to 90. Five tickets were drawn at random. The probability that two of the tickets drawn numbers 15 and 89, is
View answer and explanation
Correct option: A
2/801
Option A (2/801) is correct because it represents the probability of selecting two specific tickets (15 and 89) out of 90 when drawing five tickets at random.
212 Medium
212 of 285
Application +4 | -1
A sportsman's chance of shooting an animal at a distance r>ar > a ("a" is constant) is given to be a2r2\frac{a^2}{r^2}. He fires at r=2ar=2 a and if he misses, then again fires at r=3ar=3 a. He repeats the same process at r=4a,5ar=4 a, 5 a and 6a6 a. When he misses at r=6ar=6 a, the animal escapes into the jungle. If the odds against the sportsman are p:q{p}: {q}, then q−p{q}-{p} is ____.
View answer and explanation
Correct option: A
5
Option A (5) is correct because it is the difference between q and p, where p:q are the odds against the sportsman, calculated from the probability of the animal escaping.
213 Medium
213 of 285
Analyzing +4 | -1
For two events AA and B,B, if P(A)=P(AB)=14   and  P(BA)=12,P\left( A \right) = P\left( {\frac{A}{B}} \right) = \frac{1}{4}\;\,\mathrm{and}\,\,P\left( {\frac{B}{A}} \right) = \frac{1}{2}, then
View answer and explanation
Correct option: C
P(B′A′)=12P\left( {\frac{{B'}}{{A'}}} \right) = \frac{1}{2}
Option D (P(B′A′)=12P(\frac{{B'}}{{A'}}) = \frac{1}{2}) is correct because after calculating the probabilities of the intersections and conditional probabilities, it satisfies the given conditions and independence.
214 Medium
214 of 285
Application +4 | -1
From a well-shuffled pack of 52 cards, four cards are selected at random. The probability of drawing exactly 2 spades and exactly 2 aces is
View answer and explanation
Correct option: A
1494270725\frac{1494}{270725}
Option A (1494270725\frac{1494}{270725}) is correct because it calculates the probability of drawing exactly 2 spades and 2 aces considering both cases: one spade ace and two non-spade aces.
215 Medium
215 of 285
Application +4 | -1
If a∈[−20,0],a \in \left[ { - 20,0} \right], then the probability that the graph of the function y=16  x2+8(a+5)x−7  a−5y = 16\;{x^2} + 8\left( {a + 5} \right)x - 7\;a - 5 is strictly above the xx-axis is
View answer and explanation
Correct option: A
01-Feb
Option A (01-Feb) is correct because the probability is calculated by finding the range of 'a' for which the quadratic is always positive and dividing by the total range of 'a'.
216 Medium
216 of 285
Application +4 | -1
Let E1,E2,E3E_1, E_2, E_3 be three mutually exclusive events such that P(E1)=2+3p6,P(E2)=2−p8P\left(E_1\right)=\frac{2+3 p}{6}, P\left(E_2\right)=\frac{2-p}{8} and P(E3)=1−p2P\left(E_3\right)=\frac{1-p}{2}. If the maximum and minimum values of pp are p1p_1 and p2p_2, then (p1+p2)\left(p_1+p_2\right) is equal to :
View answer and explanation
Correct option: B
53\frac{5}{3}
Option B (53\frac{{5}}{{3}}) is correct because after finding the range of p for each probability and their sum, the intersection gives the valid range, and p1+p2 = 5/3.
217 Medium
217 of 285
Application +4 | -1
An unbiased die is tossed until a number greater than 4 appears. The probability that an even number of tosses needed, is
View answer and explanation
Correct option: B
25\frac{2}{5}
Option B (2/5) is correct because it calculates the probability that an even number of tosses is needed by considering the geometric series of probabilities.
218 Medium
218 of 285
Application +4 | -1
Two non-negative integers are chosen at random. The probability that the sum of the square is divisible by 10 , is
View answer and explanation
Correct option: B
950\frac{9}{50}
Option B (9/50) is correct because there are 18 pairs of digits (0-9) whose squares sum to a multiple of 10, out of 100 possible pairs.
219 Medium
219 of 285
Application +4 | -1
Two dice are rolled together and the numbers x,yx, y on them form complex number z=x+iyz=x+i y. The probability that ∣z∣≤3|z| \leq 3 is
View answer and explanation
Correct option: C
112\frac{1}{12}
Option C (112\frac{1}{12}) is correct because it represents the probability of the complex number's magnitude being less than or equal to 3, calculated by finding the favorable outcomes and dividing by the total possible outcomes when rolling two dice.
220 Medium
220 of 285
Application +4 | -1
A box contains tickets numbered 1 to 20. 3 tickets are drawn from the box with replacement. The probability that the largest number on the tickets is 7 , is
View answer and explanation
Correct option: D
None of these
The correct answer is D (None of these) because the probability is calculated by subtracting the probability of the largest number being less than or equal to 6 from the probability of the largest number being less than or equal to 7.
221 Medium
221 of 285
Application +4 | -1
If a fair coin is tossed 20 times and let we get head nn times, then probability that nn is odd, is
View answer and explanation
Correct option: A
12\frac{1}{2}
Option A (12\frac{1}{2}) is correct because the probability of getting an odd number of heads in 20 tosses of a fair coin is exactly one-half.
222 Medium
222 of 285
Application +4 | -1
The probability that an event A occurs in a single trial of an experiment is 0.6 . In the first three independent trials of the experiment, the probability that A occurs atleast once is
View answer and explanation
Correct option: B
0.936
Option B (0.936) is correct because it calculates the probability of event A occurring at least once in three trials by subtracting the probability of A not occurring in any of the trials from 1.
223 Medium
223 of 285
Application +4 | -1
There are 12 white and 12 red balls in a bag. Balls are drawn one by one with replacement from the bag. The probability that 7th drawn ball is 4th white, is
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (12\frac{1}{2}) is correct because with replacement, the probability of drawing a white ball remains constant at 12\frac{1}{2} for each draw.
224 Medium
224 of 285
Application +4 | -1
India plays two ODI matches each with Australia and Pakistan. The probability of India getting 0, 1, 2 are 0.45, 0.05, 0.50. The probability of India getting at least 7 points in the series is
View answer and explanation
Correct option: C
0.08750.0875
Option C (0.0875) is correct because it calculates the probability of India getting at least 7 points in the series based on the given probabilities.
225 Medium
225 of 285
Application +4 | -1
A five-digit number (having all different digits) is formed using the digits 1,2,3,4,5,6,7,81,2,3,4,5,6,7,8 and 9. The probability that the formed number either begins or ends with an odd digit, is equal to:
View answer and explanation
Correct option: A
56\frac{5}{6}
Option A (56\frac{{5}}{{6}}) is correct because it represents the probability of a five-digit number formed with distinct digits from 1 to 9 either starting or ending with an odd digit, calculated using permutations and combinations.
226 Medium
226 of 285
Application +4 | -1
4 five-rupee coins, 3 two-rupee coins and 2 one-rupee coins are stacked together in a column at random. The probability that the coins of the same denominator are consecutive is
View answer and explanation
Correct option: B
1/2101/210
Option B (1/2101/210) is correct because it calculates the probability of coins of the same denomination being consecutive in a random stack.
227 Medium
227 of 285
Application +4 | -1
A binomial variate X\mathrm{X} has mean =6=6 and variance =2=2 the probability that 5≤X≤75 \leq \mathrm{X} \leq 7 is
View answer and explanation
Correct option: C
46726561\frac{4672}{6561}
Option C is correct because the probability is calculated using the binomial distribution formula with the given mean and variance.
228 Medium
228 of 285
Application +4 | -1
Five horses are in a race. Mr. AA selects two of the horses at random and bets on them. The probability that Mr. AA selected the winning horse, is
View answer and explanation
Correct option: D
25\frac{2}{5}
Option D (2/5) is correct because it calculates the probability of selecting the winning horse by considering the complement: the probability of not selecting the winning horse.
229 Medium
229 of 285
Analyzing +4 | -1
The random variable XX follows binomial distribution B(n,p)B(n, p), for which the difference of the mean and the variance is 1 .

If 2P(X=2)=3P(X=1)2 P(X=2)=3 P(X=1), then n2P(X>1)n^{2} P(X>1) is equal to
View answer and explanation
Correct option: B
11
The correct answer is B (11) because it is derived by using the properties of binomial distribution, solving for n and p, and then calculating the required probability.
230 Medium
230 of 285
Application +4 | -1
Let AA and BB be two events such that P(A)=0.3P(A)=0.3 and P(A∪B)=0.8P(A \cup B)=0.8. If AA and BB are independent events, then P(B)=P(B)=
View answer and explanation
Correct option: A
5/75 / 7
Option A (5/75 / 7) is correct because it is the probability of event B calculated using the independence of events A and B, and the formula for the probability of the union of two events.
231 Medium
231 of 285
Application +4 | -1
In order to get at least once a head with probability ≥0.9,\ge 0.9, the number of times a coin needs to be tossed is
View answer and explanation
Correct option: B
4
Option B (4) is correct because it determines the minimum number of tosses needed to achieve a probability of at least 0.9 of getting at least one head.
232 Medium
232 of 285
Application +4 | -1
AA and BB are two independent witnesses (ie, there is no collusion between them) in a case. The probability that AA will speak the truth is xx and the probability that BB will speak the truth is yy, AA and BB agree in a certain statement. The probability that the statement is true, is
View answer and explanation
Correct option: D
xy1−x−y+2xy\frac{{xy}}{{1 - x - y + 2xy}}
Option D (xy1−x−y+2xy\frac{{xy}}{{1 - x - y + 2xy}}) is correct because it represents the probability that both witnesses are telling the truth given that they agree, calculated using Bayes' theorem.
233 Medium
233 of 285
Understanding +4 | -1
Consider the experiment of rolling a die. Let A{A} be the event 'getting a prime number' and BB be the event 'getting an odd number'.

Then, which of the following is true?

I. A{A} or B=A∪B={1,2,3}{B}={A} \cup {B}=\{1,2,3\}

II. A{A} and B=A∩B={3,5}{B}={A} \cap {B}=\{3,5\}

III. A{A} but not B=A−B={2}{B}={A}-{B}=\{2\}

IV.Not A=A′={1,5,6}{A}={A}^{\prime}=\{1,5,6\}
View answer and explanation
Correct option: C
II and III is true
Option C (II and III is true) is correct because the intersection of A and B is {3, 5}, and A - B is {2}, given A = {2, 3, 5} and B = {1, 3, 5}.
234 Medium
234 of 285
Application +4 | -1
A box conatins 2n2 n cards numbered 20,212^0, 2^1, 22,…,22n−12^2, \ldots, 2^{2 n-1}. Two cards are taken out at random from the box, at random without replacement. Probability that sum of the numbers on the card is divisible by 3 is
View answer and explanation
Correct option: B
n−12n−1\frac{n-1}{2 n-1}
Option B (n−12n−1\frac{{n-1}}{{2n-1}}) is correct because it represents the probability calculated by finding the number of favorable cases where the sum of the numbers on the cards is divisible by 3, divided by the total number of ways to choose two cards.
235 Medium
235 of 285
Application +4 | -1
A man takes a step forward with probability 0.4 and backward with probability 0.6 . The probability that at the end of eleven steps he is one step away from the starting point is
View answer and explanation
Correct option: B
462×(625)5462 \times\left(\frac{6}{25}\right)^5
Option B (462×(625)5462 \times\left(\frac{{6}}{{25}}\right)^5) is correct because it represents the probability of ending one step away after 11 steps, considering both forward and backward movements.
236 Medium
236 of 285
Application +4 | -1
Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house, is
View answer and explanation
Correct option: C
19\frac{1}{9}
Option C (19\frac{1}{9}) is correct because it represents the probability that all three persons apply for the same house out of the three available houses.
237 Medium
237 of 285
Application +4 | -1
Two marbles are drawn in succession from a box
containing 10 red, 30 white, 20 blue and 15 orange
marbles, with replacement being made after each
drawing. Then the probability, that first drawn
marble is red and second drawn marble is white, is
View answer and explanation
Correct option: A
475\frac{4}{75}
Option A (475\frac{4}{75}) is correct because it is the probability of drawing a red marble first and a white marble second, with replacement, calculated as the product of the individual probabilities.
238 Medium
238 of 285
Application +4 | -1
If from each of three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, one ball is drawn at random, then the probability that 2 white and 1 black ball will be drawn, is
View answer and explanation
Correct option: A
1332\frac{{13}}{{32}}
Option A (1332\frac{{13}}{{32}}) is correct because it calculates the probability of drawing 2 white and 1 black ball by considering all possible combinations and their respective probabilities.
239 Medium
239 of 285
Analyzing +4 | -1
Let S={w1,w2,……}S=\left\{w_1, w_2, \ldots \ldots\right\} be the sample space associated to a random experiment. Let P(wn)=P(wn−1)2,n≥2P\left(w_n\right)=\frac{P\left(w_{n-1}\right)}{2}, n \geq 2. Let A={2k+3l:k,l∈N}A=\{2 k+3 l: k, l \in N\} and B={wn:n∈A}B=\left\{w_n: n \in A\right\}. Then P(B)P(B) is equal to
View answer and explanation
Correct option: C
364\frac{3}{64}
Option C (364\frac{3}{64}) is correct because it is the probability of the event B, calculated by finding the probabilities of the individual events and summing them.
240 Medium
240 of 285
Application +4 | -1
Three persons A,B,CA, B, C throw a die in succession. The one getting 'six' wins. If AA starts then the probability of BB winning is
View answer and explanation
Correct option: D
3091\frac{30}{91}
Option D (3091\frac{30}{91}) is correct because it represents the probability of B winning the game, calculated by considering the infinite series of probabilities where B wins on their turn.
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