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JEE • MATHEMATICS • PROBABILITY

Probability practice · Medium

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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 241-270 of 285 questions
241 Medium
241 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 is needed, is
View answer and explanation
Correct option: B
25\frac{2}{5}
Option B (25\frac{2}{5}) is correct because it calculates the probability of needing an even number of tosses using an infinite geometric series.
242 Medium
242 of 285
Application +4 | -1
Two persons AA and BB take turns in throwing a pair of dice. The first person to through 9 from both dice will be awarded the prize. If AA throws first, then the probability that BB wins the game, is
View answer and explanation
Correct option: B
817\frac{8}{{17}}
Option B (8/17) is correct because it calculates the probability of B winning based on A not throwing 9 and B then throwing 9, considering all possible turns.
243 Medium
243 of 285
Analyzing +4 | -1
Consider the polynomial P(x)=x3+ax2+bx+cP(x)=x^3+a x^2+b x+c where a,b,c∈{1,2,……..,10}a, b, c \in\{1,2, \ldots \ldots . ., 10\} then the probability that P(x)P(x) is divisible by x2+x+1x^2+x+1 is
View answer and explanation
Correct option: B
91000\frac{9}{1000}
Option B (91000\frac{9}{1000}) is correct because the probability is calculated by finding the number of favorable outcomes (where P(x) is divisible by x2+x+1x^2+x+1) and dividing it by the total number of possible outcomes.
244 Medium
244 of 285
Application +4 | -1
A sample space consists of 3 sample points with associated probabilities given as 2p,p2,4p−12 p, p^2, 4 p-1. Then the value of pp is -
View answer and explanation
Correct option: A
p=11−3\mathrm{p}=\sqrt{11}-3
Option D (p=11−3\mathrm{p}=\sqrt{11}-3) is correct because the sum of probabilities of all sample points in a sample space must equal 1, leading to a quadratic equation whose solution gives the value of p.
245 Medium
245 of 285
Application +4 | -1
The probability of event AA occurring is 0.50.5 and B occurring is 0.3. If AA and BB are mutually exclusive events, then the probability of neither AA nor BB occurring is
View answer and explanation
Correct option: D
none of these
The correct answer is D (none of these) because the probability of neither A nor B occurring, when A and B are mutually exclusive with probabilities 0.5 and 0.3 respectively, is 0.2.
246 Medium
246 of 285
Application +4 | -1
A fair coin is tossed repeatedly until a head is obtained. If X\mathrm{X} denotes the number of tosses required, then the conditional probability of X≥6X \geq 6, given that X>3X>3 is
View answer and explanation
Correct option: B
14\frac{1}{4}
Option B (14\frac{1}{4}) is correct because it is the conditional probability calculated based on the geometric distribution of coin tosses.
247 Medium
247 of 285
Application +4 | -1
A group consists 2n2 n students including you and your friend. If the group is split into two different sections AA and BB, each containing nn students, then the probability that you and your friend are in the different sections 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 the probability that you and your friend are in different sections is calculated by dividing the number of favorable arrangements by the total number of arrangements.
248 Medium
248 of 285
Analyzing +4 | -1
10 different books and 2 different pens are given to 3 boys so that each gets equal number of things. The probability that the some boy does not receive both the pens is
View answer and explanation
Correct option: A
5/115/11
Option A (5/11) is correct because it calculates the probability that no single boy receives both pens by subtracting the probability that one boy receives both pens from 1.
249 Medium
249 of 285
Application +4 | -1
A five digit mumber is formed by the digits 1, 2, 3, 4, 5,6 and 8. The probability that the number has even digit at both ends is
View answer and explanation
Correct option: C
27\frac{2}{7}
Option C (27\frac{2}{7}) is correct because the probability is calculated by finding the ratio of favorable outcomes (numbers with even digits at both ends) to the total possible outcomes.
250 Medium
250 of 285
Analyzing +4 | -1
There are 7 seats in a row. Three persons take seats at random. The probability that the middle seat is always occupied and no two persons are consecutive is
View answer and explanation
Correct option: C
4/354/35
Option C (4/35) is correct because it calculates the probability of the middle seat being occupied and no two persons sitting consecutively by considering favorable arrangements and total possible arrangements.
251 Medium
251 of 285
Application +4 | -1
The number of ways in which a pack of 52 cards of four different suits can be distributed equally among four players so that each player gets the Ace, King, Queen and Knave of the same suit is
View answer and explanation
Correct option: A
36!4!(9!)4\frac{36 ! 4 !}{(9 !)^4}
Option A (36!4!(9!)4\frac{36 ! 4 !}{(9 !)^4}) is correct because it represents the number of ways to distribute the cards according to the given conditions.
252 Medium
252 of 285
Application +4 | -1
Two cards are drawn from a well shuffled deck of 52 cards. The probability that one is red card and the other is a queen is
View answer and explanation
Correct option: C
50/66350/663
Option C (50663\frac{50}{663}) is correct because it calculates the probability of drawing one red card and one queen by considering the two mutually exclusive cases.
253 Medium
253 of 285
Application +4 | -1
If the papers of 4 students can be checked by any one of the 7 teachers, then the probability that all the 4 papers are checked by exactly 2 teachers is
View answer and explanation
Correct option: B
6/496 / 49
Option B (649\frac{{6}}{{49}}) is correct because it is the probability calculated by finding the number of ways the 4 papers can be checked by exactly 2 teachers, divided by the total number of ways the 4 papers can be checked by any of the 7 teachers.
254 Medium
254 of 285
Analyzing +4 | -1
In the above question the probability that the binary operation is non-commutative is
View answer and explanation
Correct option: B
nn2−nn(n+1)2nn2\frac{{{n^{{n^2}}} - {n^{\frac{{n\left( {n + 1} \right)}}{2}}}}}{{{n^{{n^2}}}}}
Option D (nn2−nn(n+1)2nn2\frac{{{n^{{n^2}}} - {n^{\frac{{n\left( {n + 1} \right)}}{2}}}}}{{{n^{{n^2}}}}}) is correct because it represents the probability of a binary operation being non-commutative.
255 Medium
255 of 285
Application +4 | -1
If EE and FF are events such that P(E)=14,P(F)=12{P}({E})=\frac{1}{4}, {P}({F})=\frac{1}{2} and P(EP(E and F)=18F)=\frac{1}{8}, then,
I. P(E or F)=58{P}({E}\text{ or }{F})=\frac{5}{8}
II. P(not E and not F)=38{P}(\text{not }{E} \text{ and not } {F})=\frac{3}{8}
View answer and explanation
Correct option: C
Both I and II are true
Option C (Both I and II are true) is correct because both the probability of E or F and the probability of not E and not F are calculated correctly using the given probabilities.
256 Medium
256 of 285
Application +4 | -1
The probability that in a year of 22 nd 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 the probability is calculated by considering both leap and non-leap years in the 22nd century and the probability of having 53 Sundays in each case.
257 Medium
257 of 285
Analyzing +4 | -1
Suppose f(x)=x3+ax2+bx+c,f\left( x \right) = {x^3} + a{x^2} + bx + c, where a, b, ca,\,b,\,c are chosen respectively by throwing a dice three times. Then, the probability that f(x)f\left( x \right) is an increasing function, is
View answer and explanation
Correct option: A
49\frac{4}{9}
Option A (49\frac{4}{9}) is correct because it represents the probability that the derivative of the cubic function is always non-negative, ensuring the function is increasing.
258 Medium
258 of 285
Application +4 | -1
The probability that in the random arrangement of the letters of the word 'UNIVERSITY', the two I's does not come together is
View answer and explanation
Correct option: A
0.8
Option A (0.8) is correct because it calculates the probability by subtracting the probability of the two I's being together from 1.
259 Medium
259 of 285
Analyzing +4 | -1
AA and BB are two students. Their chances of sloving a problem correctly are 13\frac{1}{3} and 14\frac{1}{4}, respectively. If the probability of their making a common error is, 120\frac{1}{20} and they obtain the same answer, then the probability of their answer to be correct is
View answer and explanation
Correct option: D
1013\frac{10}{13}
Option D (1013\frac{10}{13}) is correct because it represents the probability of their answer being correct given that they obtained the same answer.
260 Medium
260 of 285
Analyzing +4 | -1
Consider the system of equations px+qy=0,rx+sy=0\mathrm{px}+\mathrm{qy}=0, \mathrm{rx}+\mathrm{sy}=0 where p,q,r,s∈{−1,1}p, q, r, s \in\{-1,1\}
Statement-1: The probability that the system of equation has a unique solution is 12\frac{1}{2}.
Statement-2: The probability that the system of equations no solution is zero.
View answer and explanation
Correct option: B
Statement-1 is true, statement-2 is true and ststement-2 is NOT the correct explanation for statement-1
Statement-1 is true because the probability of a unique solution is 1/2, and Statement-2 is true because the probability of no solution is zero, as the system is homogeneous. Statement-2 is NOT the correct explanation for Statement-1.
261 Medium
261 of 285
Application +4 | -1
Suppose that the probability that an item produced by a particular machine is defective equals 0.2. If 10 items produced from this machine are selected at random, the probability that not more than one defective is found is
View answer and explanation
Correct option: C
3e2\frac{3}{\mathrm{e}^2}
Option C (3e2\frac{{3}}{{\mathrm{{e}}^2}}) is correct because it uses the Poisson approximation to find the probability of not more than one defective item.
262 Medium
262 of 285
Application +4 | -1
A bag contains 6 red and 3 white balls. Four balls are drawn one by one and not replaced. The probability that they are alternatively of different colours, is
View answer and explanation
Correct option: B
542\frac{5}{{42}}
Option B (542\frac{5}{{42}}) is correct because it calculates the probability of drawing four balls alternatively of different colors (Red-White-Red-White or White-Red-White-Red) without replacement.
263 Medium
263 of 285
Application +4 | -1
The letters of the word 'longest' are arranged at random. The probability that the vowels may occupy only odd positions, is
View answer and explanation
Correct option: A
27\frac{2}{7}
Option A (27\frac{2}{7}) is correct because it represents the probability of arranging the vowels in the odd positions out of all possible arrangements.
264 Medium
264 of 285
Application +4 | -1
Three rifle-men take one shot each at the same target. The probability of the first rifle-man hitting the target is 0.4 , the probability of the second rifle-man hitting the target is 0.5 and the probability of the third rifle-man hitting the target is 0.8 . The Probability that exactly two of them hit the target, is
View answer and explanation
Correct option: B
0.44
Option B (0.44) is correct because it calculates the probability of exactly two rifle-men hitting the target by considering all possible combinations of two successes and one failure.
265 Medium
265 of 285
Application +4 | -1
A man takes a step forward with probability 0.40.4 and backwards with probability 0.6. The probability that at the end of eleven steps he is just one step away from the starting point, is
View answer and explanation
Correct option: C
11C5(0.4)5(0.6)5{ }^{11} C_5(0.4)^5(0.6)^5
Option C (11C5(0.4)5(0.6)6{ }^{11} C_5(0.4)^5(0.6)^6) is correct because it's the probability that the man is one step away, considering both forward and backward scenarios.
266 Medium
266 of 285
Application +4 | -1
Fifteen coupons are numbered 1 to 15. Seven coupons are selected at random, one at a time with replacement. The probability that the largest number appearing on a selected coupon be 9, is
View answer and explanation
Correct option: C
(35)7{\left( {\frac{3}{5}} \right)^7}
Option D is incorrect. Option C (35)7{\left( {\frac{{3}}{{5}}} \right)^7} is correct because it represents the probability of selecting a number less than or equal to 9 in each of the 7 trials with replacement.
267 Medium
267 of 285
Application +4 | -1
A group of 6 boys and 6 girls is randomly divided into two equal groups. The probability that each group contains 3 boys and 3 girls is
View answer and explanation
Correct option: D
100/231100 / 231
Option D (100 / 231) is correct because it is calculated by dividing the number of ways to form a group with 3 boys and 3 girls by the total number of ways to form a group of 6 from 12 people.
268 Medium
268 of 285
Application +4 | -1
AA is a set containing n 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 are disjoint sets, is
View answer and explanation
Correct option: D
(34)n\left(\frac{3}{4}\right)^n
Option D ((34)n\left(\frac{{3}}{{4}}\right)^n) is correct because it calculates the probability of P and Q being disjoint by considering the choices for each element in A.
269 Medium
269 of 285
Application +4 | -1
The probabilities that a student will obtain grades A, B, C   or   DA,\,B,\,C\;\,\mathrm{or}\;\,D are 0.30, 0.35, 0.20 and 0.15 respectively. The probability that he will receive atleast CC grade, is
View answer and explanation
Correct option: D
0.2
Option B (0.85) is correct because it is the sum of the probabilities of receiving a grade of A, B, or C.
270 Medium
270 of 285
Application +4 | -1
India plays two matches each with the West Indies and Australia. In any match, the probabilities of India getting 0,1 and 2 points are 0.45,0.05 and 0.50, respectively. Assuming that the outcomes are independent, the probability of India getting at least 7 points, is
View answer and explanation
Correct option: B
780\frac{7}{80}
Option B (780\frac{7}{80}) is correct because it calculates the probability of India getting at least 7 points by considering the cases of getting exactly 7 or 8 points in the matches.
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