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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 1-30 of 285 questions

1 Medium

1 of 285

Application +4 | -1

The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05 . What is the probability that out of 5 such bulbs none will fuse after 150 days of use

View answer and explanation

Correct option: B

\left(\frac{19}{20}\right)^5

Option B is correct. The probability that none of the 5 bulbs will fuse is calculated using the binomial probability formula with p = 19/20, representing the probability of a bulb not fusing.

2 Medium

2 of 285

Application +4 | -1

Given that

X

is discrete random variable which takes the values 0, 1, 2 and

P\left( {X = 0} \right) = \frac{{144}}{{169}},\;P\left( {X = 1} \right) = \frac{1}{{169}},

then the value of

P\left( {X = 2} \right)

View answer and explanation

Correct option: B

\frac{{24}}{{169}}

Option B (

\frac{{24}}{{169}}

) is correct because the sum of probabilities for all possible values of a discrete random variable must equal 1.

3 Medium

3 of 285

Analyzing +4 | -1

If an unbiased die, marked with

-2,-1,0,1,2,3

on its faces, is thrown five times, then the probability that product of the outcomes is positive is:

View answer and explanation

Correct option: D

\frac{521}{2592}

Option D (

\frac{{521}}{{2592}}

) is correct because it is the probability calculated by considering the cases where the product of the outcomes is positive.

4 Medium

4 of 285

Application +4 | -1

n

positive integers are taken at random and multiplied together, and

p_n

is the probability that the last digit of the product is

2,4,6

or 8 then

125 p_3-50

is equal to

View answer and explanation

Correct option: B

Option C (6) is correct because the probability that the last digit of the product is 2, 4, 6, or 8 is (4/5)^n - (2/5)^n, and for n=3, 125p_3 - 50 = 6.

5 Medium

5 of 285

Application +4 | -1

A bag contains some white and some black balls, all combinations being equally likely. Total balls are 10. If three are drawn and found black then find probability that the bag contains 1 white and 9 black balls :

View answer and explanation

Correct option: A

\frac{14}{55}

Option A (

\frac{14}{55}

) is correct because it is calculated using Bayes' theorem, considering the probability of drawing three black balls given 1 white and 9 black balls, divided by the total probability of drawing three black balls.

6 Medium

6 of 285

Application +4 | -1

The probability that a teacher will give an unannounced test during any class meeting is 1/5. If a student is absent twice, the probability that he will miss atleast one test, is

View answer and explanation

Correct option: B

9/25

Option B (9/25) is correct because it calculates the probability of missing at least one test by subtracting the probability of missing no tests from 1.

7 Medium

7 of 285

Application +4 | -1

A locker can be opened by dialing a fixed three digit code (between 000 and 999). A stranger who does not know the code tries to open the
locker by dialing three digits at random. The probability that the stranger succeeds at the

k^{\text {th }}

trial is

View answer and explanation

Correct option: B

\frac{k}{1000}

Option D (

a=-\frac{{5 \pi}}{{4}}, b=

any constant) is correct because it satisfies the given integral equation.

8 Medium

8 of 285

Application +4 | -1

A six faced die is a biased one. It is thrice more likely to show an odd number than show an even number. It is thrown twice. The probability that the sum of the numbers in the two throws is even, is

View answer and explanation

Correct option: B

5/8

Option B (

5/8

) is correct because it is calculated by considering the probabilities of getting either two odd numbers or two even numbers in two throws of the biased die.

9 Medium

9 of 285

Application +4 | -1

A

and

B

are events of the same experiments with

P\left( A \right) = 0.2,\;P\left( B \right) = 0.5

, then maximum value of

P\left( {A' \cap B} \right)

View answer and explanation

Correct option: B

0.5

Option B (0.5) is correct because the maximum value of P(A' ∩ B) occurs when B is a subset of A', making P(A' ∩ B) equal to P(B).

10 Medium

10 of 285

Application +4 | -1

Three dice are thrown simultaneously, then probability of throwing a total greater than 4 is

View answer and explanation

Correct option: B

\frac{{53}}{{54}}

Option B (

\frac{{53}}{{54}}

) is correct because it is calculated by subtracting the probability of the complement event (sum less than or equal to 4) from 1.

11 Medium

11 of 285

Application +4 | -1

Suppose

E

and

F

are two independent events of a random experiment. If the probability of occurrence of

E

\frac{1}{5}

and the probability of occurrence of

F

given

E

\frac{1}{{10}}

,then the probability of non-occurrence of at least one of the events

E

and

F

View answer and explanation

Correct option: C

\frac{{49}}{{50}}

Option C (

\frac{{49}}{{50}}

) is correct because it calculates the probability of non-occurrence of at least one of the events E and F using the complement rule and the independence of the events.

12 Medium

12 of 285

Application +4 | -1

n biscuits are distributed among

N

beggars at random. The probability that a particular beggar gets

r\left( {\, < n} \right)

biscuits, is

View answer and explanation

Correct option: A

{ }^n C_r \left(\frac{1}{N}\right)^r\left(\frac{N-1}{N}\right)^{n-r}

Option A (

{ }^n C_r \left(\frac{{1}}{{N}}\right)^r\left(\frac{{N-1}}{{N}}\right)^{{n-r}}

) is correct because it represents the probability of a particular beggar getting r biscuits out of n, with each biscuit having a 1/N chance of going to that beggar.

13 Medium

13 of 285

Analyzing +4 | -1

Let

S

be the sample space of the random experiment of throwing simultaneously two unbiased dice with six faces (numbered 1 to 6) and let

{E_k} = \left\{ {\left( {a,\;b} \right) \in S:ab = k} \right\}\;

for

k \ge 1.

{p_k} = P\left( {{E_k}} \right)

for

k \ge 1,\;

then the correct among the following, is

View answer and explanation

Correct option: A

{p_1} < {p_{30}} < {p_4} < {p_6}

Option D (

p_{36} < p_{11} < p_6 < p_4

) is incorrect. Option A (

p_1 < p_{30} < p_4 < p_6

) is correct because it correctly orders the probabilities of the product of two dice rolls equaling specific values.

14 Medium

14 of 285

Application +4 | -1

A committee of the five is to be chosen from a group of people. The probability that a certain married couple will either serve together or not at all is

View answer and explanation

Correct option: B

\frac{4}{9}

Option B (

\frac{4}{9}

) is correct because it is the probability that a married couple will either serve together or not at all, calculated by dividing favorable combinations by total combinations.

15 Medium

15 of 285

Application +4 | -1

In a bolt factory, machines A, B and C manufacture respectively

20 \%, 30 \%

and

50 \%

of the total bolts. Of their output 3, 4 and 2 percent are respectively defective bolts. A bolts is drawn at random from the product. If the bolt drawn is found the defective, then the probability that it is manufactured by the machine C is

View answer and explanation

Correct option: A

\frac{5}{14}

Option A (5/14) is correct because it applies Bayes' theorem to find the probability that the defective bolt was manufactured by machine C.

Explore Other Chapters in MATHEMATICS

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs COMPLEX NUMBERS 980 MCQs DIFFERENTIAL EQUATIONS 977 MCQs

16 Medium

16 of 285

Application +4 | -1

a bag contains three coins one of which has head on both sides, another is a biased coin that shows up heads

90 \%

of the time and thnbiased coin a coin is taken oute third one is an unbiased from the bag at random and tossed if it shows up a head then the probablilty tht it is the unbiased coin is

View answer and explanation

Correct option: C

\frac{5}{24}

Option C (

\frac{5}{24}

) is correct because it is calculated using Bayes' theorem, considering the probabilities of each coin type and the probability of getting a head.

17 Medium

17 of 285

Application +4 | -1

A pair of fair dice is rolled together till a sum of either 5 or 7 is obtained. If

p

denotes the probability that 7 comes before 5 , then

5 p

is equal

View answer and explanation

Correct option: C

Option C (3) is correct because it is derived by calculating the probability of getting a 7 before a 5, using the probabilities of each event and the sum of an infinite geometric series.

18 Medium

18 of 285

Application +4 | -1

There are

n

different objects, 1, 2, 3,

\cdots, n

, distributed at random in

n

places marked

1,2,3

\cdots, n

. If

p

is the probability that at least three of the objects occupy places corresponding to their number, then

6 p

is equal to

View answer and explanation

Correct option: A

Option A (1) is correct because the probability that at least three objects occupy places corresponding to their number is 1/6, and 6p = 6*(1/6) = 1.

19 Medium

19 of 285

Application +4 | -1

Two events

\mathrm{E}

and

\mathrm{F}

are independent. If

\mathrm{P}(\mathrm{E})=0.3

P(E \cup F)=0.5

, then

P(E \mid F)-P(F \mid E)

equals

View answer and explanation

Correct option: C

\frac{1}{70}

Option C is correct because the difference between conditional probabilities simplifies to the difference between individual probabilities due to independence, and the value is correctly calculated.

20 Medium

20 of 285

Application +4 | -1

If the letters of the word "ATTEMPT" are written down at random then the probability that all the T's are consecutive is

View answer and explanation

Correct option: C

\frac{1}{7}

Option C (

\frac{{1}}{{7}}

) is correct because it represents the probability of all the T's being consecutive in a random arrangement of the letters in \"ATTEMPT\".

21 Medium

21 of 285

Application +4 | -1

There are three bags

X, Y

and

Z.

Bag

X

contains 5 one-rupee coins and 4 five-rupee coins; Bag

Y

contains 4 one-rupee coins and 5 five-rupee coins and Bag

{Z}

contains 3 one-rupee coins and 6 five-rupee coins. A bag is selected at random and a coin drawn from it at random is found to be a one-rupee coin. Then the probability, that it came from bag

{Y},

is:

View answer and explanation

Correct option: A

\frac{1}{3}

Option A (

\frac{1}{3}

) is correct because it represents the probability that the one-rupee coin came from bag Y, calculated using conditional probability.

22 Medium

22 of 285

Application +4 | -1

A

and

B

toss a coin alternately on the understanding that the first to obtain head win the toss. The probability that

A

wins the toss

View answer and explanation

Correct option: B

\frac{2}{3}

Option B (

\frac{2}{3}

) is correct because it calculates the probability of A winning by summing the probabilities of A winning on the first toss, third toss, fifth toss, and so on.

23 Medium

23 of 285

Analyzing +4 | -1

From a group of 10 men and 5 women, four member committees are to be formed each of which must contain at least one woman. Then the probability for these committees to have more women than men, is :

View answer and explanation

Correct option: C

\frac{1}{11}

Option C (

\frac{1}{11}

) is correct because it is the probability of having more women than men in the committee, given that the committee has at least one woman.

24 Medium

24 of 285

Application +4 | -1

The letters of word 'SOCIETY' are placed at random in a row. The Probability that three vowels come together, is

View answer and explanation

Correct option: B

\frac{1}{7}

Option B (

\frac{1}{7}

) is correct because it represents the probability of arranging the letters in 'SOCIETY' such that the three vowels are together.

25 Medium

25 of 285

Application +4 | -1

If the letters of the word MATHEMATICS are arranged arbitrarily, the probability that C comes before E, E before H, H before I and I before S is

View answer and explanation

Correct option: C

\frac{1}{120}

Option C (

\frac{{1}}{{120}}

) is correct because the probability is calculated by dividing the number of favorable arrangements by the total number of arrangements.

26 Medium

26 of 285

Analyzing +4 | -1

Suppose

A

and

B

are two events such that

0 < P(B) < 1

. If

P\left(A \mid B^{\prime}\right)=\frac{P(A)}{1-P(B)}

then which one of the following is not true?

View answer and explanation

Correct option: B

P(A \cap B)=P(A \cup B)

Option B (

P(A \cap B)=P(A \cup B)

) is not true because given the condition,

P(A \cap B) = 0

and

P(A \cup B) = P(A) + P(B)

, which are not generally equal unless either P(A) or P(B) is zero.

27 Medium

27 of 285

Analyzing +4 | -1

A

is a set containing

n

elements. A subset

P

A

is chosen at random. The set

A

is reconstructed by replacing the elements of

P

. A subset

Q

is again chosen at random. The probability that

Q

is a subset of

P

, is

View answer and explanation

Correct option: B

\left(\frac{3}{4}\right)^n

Option B (

\left(\frac{{3}}{{4}}\right)^n

) is correct because it represents the probability that Q is a subset of P, considering all possible subsets P and Q of set A.

28 Medium

28 of 285

Application +4 | -1

The coefficients

a,\,b

and

c

of the quadratic equation,

a{x^2} + bx + c = 0

are obtained by throwing a dice three times. The probability that this equation has equal roots is

View answer and explanation

Correct option: B

\frac{5}{{216}}

Option B (5/216) is correct because it calculates the probability of the quadratic equation having equal roots based on the possible outcomes of rolling a dice three times.

29 Medium

29 of 285

Application +4 | -1

Two cards are drawn at random from a pack of 52 cards. The probability of getting at least a spade and an ace is

View answer and explanation

Correct option: C

Jan-26

Option B (8/221) is correct because it calculates the probability of getting at least one spade and one ace by considering the different combinations.

30 Medium

30 of 285

Application +4 | -1

One ticket is selected at random from 50 tickets numbered 00, 01, 02, …., 49. Then, the probability that the sum of the digits on the selected ticket is 8, given that the product of these digits is zero equals

View answer and explanation

Correct option: A

\frac{1}{{14}}

Option C (

\frac{1}{4}

) is correct because there are two favorable outcomes (HTH, THT) out of a total of eight possible outcomes when tossing a coin three times.

Revision notes and key takeaways

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

Key Takeaways

✓ This set contains 285 single-correct-answer MCQs on PROBABILITY.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

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