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JEE Main Previous Year Questions (2025): Probability
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Page 1
JEE Main Previous Year Questions
(2025): Probability
Question1: Three distinct numbers are selected randomly from the set {?? ,?? ,?? ,…,???? }.
If the probability, that the selected numbers are in an increasing G.P., is
?? ?? ,?????? (?? ,?? )=?? , then ?? +?? is equal to ____ .
JEE Main 2025 (Online) 2nd April Morning Shift
Answer: 2477
Solution:
Common ratio Last triplet Total
r=2 10, 20, 40 10
r=3 4, 12, 36 4
r=4 2, 8, 32 2
r=5 1, 5, 25 1
r=6 1, 6, 36 1
Total = 18
Total choices =
40
?? 3
=9880
Required probability =
18
9880
=
9
4940
=
?? ??
?? +?? =4949
Question2: A card from a pack of 52 cards is lost. From the remaining 51 cards, ?? cards
are drawn and are found to be spades. If the probability of the lost card to be a spade
is
????
????
, then ?? is equal to ____ -
Answer: 2
Solution:
?? (
Lost (spade
n cards are spade
)
=
?? (
?? ?? ?? ?? )?? (?? ?? )
?? (
?? ?? ?? ?? )?? (?? ?? )+?? (
?? ?? ???
?? )?? (???
?? )
Page 2
JEE Main Previous Year Questions
(2025): Probability
Question1: Three distinct numbers are selected randomly from the set {?? ,?? ,?? ,…,???? }.
If the probability, that the selected numbers are in an increasing G.P., is
?? ?? ,?????? (?? ,?? )=?? , then ?? +?? is equal to ____ .
JEE Main 2025 (Online) 2nd April Morning Shift
Answer: 2477
Solution:
Common ratio Last triplet Total
r=2 10, 20, 40 10
r=3 4, 12, 36 4
r=4 2, 8, 32 2
r=5 1, 5, 25 1
r=6 1, 6, 36 1
Total = 18
Total choices =
40
?? 3
=9880
Required probability =
18
9880
=
9
4940
=
?? ??
?? +?? =4949
Question2: A card from a pack of 52 cards is lost. From the remaining 51 cards, ?? cards
are drawn and are found to be spades. If the probability of the lost card to be a spade
is
????
????
, then ?? is equal to ____ -
Answer: 2
Solution:
?? (
Lost (spade
n cards are spade
)
=
?? (
?? ?? ?? ?? )?? (?? ?? )
?? (
?? ?? ?? ?? )?? (?? ?? )+?? (
?? ?? ???
?? )?? (???
?? )
=
12
?? ??
51
?? ?? ×
1
4
12
?? ??
51
?? ?? ×
1
4
+
3
4
×
13
?? ??
51
?? ?? =
1
1+3·
13
?? ??
12
?? ?? =
13-?? 52-??
?
13-?? 52-?? =
11
50
??? =2
Question3: Two balls are selected at random one by one without replacement from a
bag containing 4 white and 6 black balls. If the probability that the first selected ball is
black, given that the second selected ball is also black, is
?? ?? , where ?????? (?? ,?? )=?? ,
then ?? +?? is equal to :
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
A. 4
B. 14
C. 11
D. 13
Answer: B
Solution:
In a bag containing 4 white and 6 black balls, two balls are selected one by one without
replacement. We need to find the probability that the first ball is black, given that the second
ball is black. Let's define the events:
A: The first ball selected is black.
B: The second ball selected is black.
The probability ?? (?? |?? ) is given by the formula:
?? (?? |?? )=
?? (?? n?? )
?? (?? )
Let's compute each part separately:
Probability of both balls being black ?? (?? n?? ) :
The chance that the first ball is black is
6
10
. Given that one black ball is removed, the probability
that the second is also black is
5
9
.
?? (?? n?? )=
6
10
×
5
9
=
30
90
=
1
3
Probability that the second ball is black ?? (?? ) :
This can occur if either the first ball was white or black:
First ball white, second black:
4
10
×
6
9
Both balls black:
6
10
×
5
9
Page 3
JEE Main Previous Year Questions
(2025): Probability
Question1: Three distinct numbers are selected randomly from the set {?? ,?? ,?? ,…,???? }.
If the probability, that the selected numbers are in an increasing G.P., is
?? ?? ,?????? (?? ,?? )=?? , then ?? +?? is equal to ____ .
JEE Main 2025 (Online) 2nd April Morning Shift
Answer: 2477
Solution:
Common ratio Last triplet Total
r=2 10, 20, 40 10
r=3 4, 12, 36 4
r=4 2, 8, 32 2
r=5 1, 5, 25 1
r=6 1, 6, 36 1
Total = 18
Total choices =
40
?? 3
=9880
Required probability =
18
9880
=
9
4940
=
?? ??
?? +?? =4949
Question2: A card from a pack of 52 cards is lost. From the remaining 51 cards, ?? cards
are drawn and are found to be spades. If the probability of the lost card to be a spade
is
????
????
, then ?? is equal to ____ -
Answer: 2
Solution:
?? (
Lost (spade
n cards are spade
)
=
?? (
?? ?? ?? ?? )?? (?? ?? )
?? (
?? ?? ?? ?? )?? (?? ?? )+?? (
?? ?? ???
?? )?? (???
?? )
=
12
?? ??
51
?? ?? ×
1
4
12
?? ??
51
?? ?? ×
1
4
+
3
4
×
13
?? ??
51
?? ?? =
1
1+3·
13
?? ??
12
?? ?? =
13-?? 52-??
?
13-?? 52-?? =
11
50
??? =2
Question3: Two balls are selected at random one by one without replacement from a
bag containing 4 white and 6 black balls. If the probability that the first selected ball is
black, given that the second selected ball is also black, is
?? ?? , where ?????? (?? ,?? )=?? ,
then ?? +?? is equal to :
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
A. 4
B. 14
C. 11
D. 13
Answer: B
Solution:
In a bag containing 4 white and 6 black balls, two balls are selected one by one without
replacement. We need to find the probability that the first ball is black, given that the second
ball is black. Let's define the events:
A: The first ball selected is black.
B: The second ball selected is black.
The probability ?? (?? |?? ) is given by the formula:
?? (?? |?? )=
?? (?? n?? )
?? (?? )
Let's compute each part separately:
Probability of both balls being black ?? (?? n?? ) :
The chance that the first ball is black is
6
10
. Given that one black ball is removed, the probability
that the second is also black is
5
9
.
?? (?? n?? )=
6
10
×
5
9
=
30
90
=
1
3
Probability that the second ball is black ?? (?? ) :
This can occur if either the first ball was white or black:
First ball white, second black:
4
10
×
6
9
Both balls black:
6
10
×
5
9
?? (?? )=
4
10
×
6
9
+
6
10
×
5
9
=
24
90
+
30
90
=
54
90
=
3
5
Now, using these probabilities:
?? (?? |?? )=
1
3
3
5
=
1
3
×
5
3
=
5
9
Here,
5
9
is in its simplest form (gcd(5,9)=1), so ?? =5 and ?? =9.
Therefore, ?? +?? =5+9=14.
Question4: A coin is tossed three times. Let ?? denote the number of times a tail
follows a head. If ?? and ?? ?? denote the mean and variance of ?? , then the value of
???? (?? +?? ?? ) is:
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
A. 64
B. 32
C. 51
D. 48
Answer: D
Solution:
Outcome ?? ?? ?? ??
HHH 0
1
8
TTT 0
1
8
HHT 1
1
8
HTH 1
1
8
THH 0
1
8
TTH 0
1
8
THT 1
1
8
HTT 1
1
8
Page 4
JEE Main Previous Year Questions
(2025): Probability
Question1: Three distinct numbers are selected randomly from the set {?? ,?? ,?? ,…,???? }.
If the probability, that the selected numbers are in an increasing G.P., is
?? ?? ,?????? (?? ,?? )=?? , then ?? +?? is equal to ____ .
JEE Main 2025 (Online) 2nd April Morning Shift
Answer: 2477
Solution:
Common ratio Last triplet Total
r=2 10, 20, 40 10
r=3 4, 12, 36 4
r=4 2, 8, 32 2
r=5 1, 5, 25 1
r=6 1, 6, 36 1
Total = 18
Total choices =
40
?? 3
=9880
Required probability =
18
9880
=
9
4940
=
?? ??
?? +?? =4949
Question2: A card from a pack of 52 cards is lost. From the remaining 51 cards, ?? cards
are drawn and are found to be spades. If the probability of the lost card to be a spade
is
????
????
, then ?? is equal to ____ -
Answer: 2
Solution:
?? (
Lost (spade
n cards are spade
)
=
?? (
?? ?? ?? ?? )?? (?? ?? )
?? (
?? ?? ?? ?? )?? (?? ?? )+?? (
?? ?? ???
?? )?? (???
?? )
=
12
?? ??
51
?? ?? ×
1
4
12
?? ??
51
?? ?? ×
1
4
+
3
4
×
13
?? ??
51
?? ?? =
1
1+3·
13
?? ??
12
?? ?? =
13-?? 52-??
?
13-?? 52-?? =
11
50
??? =2
Question3: Two balls are selected at random one by one without replacement from a
bag containing 4 white and 6 black balls. If the probability that the first selected ball is
black, given that the second selected ball is also black, is
?? ?? , where ?????? (?? ,?? )=?? ,
then ?? +?? is equal to :
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
A. 4
B. 14
C. 11
D. 13
Answer: B
Solution:
In a bag containing 4 white and 6 black balls, two balls are selected one by one without
replacement. We need to find the probability that the first ball is black, given that the second
ball is black. Let's define the events:
A: The first ball selected is black.
B: The second ball selected is black.
The probability ?? (?? |?? ) is given by the formula:
?? (?? |?? )=
?? (?? n?? )
?? (?? )
Let's compute each part separately:
Probability of both balls being black ?? (?? n?? ) :
The chance that the first ball is black is
6
10
. Given that one black ball is removed, the probability
that the second is also black is
5
9
.
?? (?? n?? )=
6
10
×
5
9
=
30
90
=
1
3
Probability that the second ball is black ?? (?? ) :
This can occur if either the first ball was white or black:
First ball white, second black:
4
10
×
6
9
Both balls black:
6
10
×
5
9
?? (?? )=
4
10
×
6
9
+
6
10
×
5
9
=
24
90
+
30
90
=
54
90
=
3
5
Now, using these probabilities:
?? (?? |?? )=
1
3
3
5
=
1
3
×
5
3
=
5
9
Here,
5
9
is in its simplest form (gcd(5,9)=1), so ?? =5 and ?? =9.
Therefore, ?? +?? =5+9=14.
Question4: A coin is tossed three times. Let ?? denote the number of times a tail
follows a head. If ?? and ?? ?? denote the mean and variance of ?? , then the value of
???? (?? +?? ?? ) is:
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
A. 64
B. 32
C. 51
D. 48
Answer: D
Solution:
Outcome ?? ?? ?? ??
HHH 0
1
8
TTT 0
1
8
HHT 1
1
8
HTH 1
1
8
THH 0
1
8
TTH 0
1
8
THT 1
1
8
HTT 1
1
8
?? =??? ???? ?? =
1
2
?? 2
=??? ?? 2
?? ?? -?? 2
=
1
2
-
1
4
=
1
4
64(?? +?? 2
)=64[
1
2
+
1
4
]
=64×
3
4
=48
Question5: If ?? and ?? are two events such that ?? (?? n?? )=?? .?? , and ?? (?? |?? ) and
?? (?? |?? ) are the roots of the equation ???? ?? ?? -?? ?? +?? =?? , then the value of
?? (???
????
)
?? (???
n???
)
is:
JEE Main 2025 (Online) 22nd January Evening Shift
Options:
A.
4
3
B.
7
4
C.
9
4
D.
5
3
Answer: C
Solution:
To solve this problem, start by considering the equation given for the probabilities:
12?? 2
-7?? +1=0
The roots of this equation are:
?? =
1
3
,
1
4
Assume ?? (?? |?? )=
1
3
and ?? (?? |?? )=
1
4
.
From the definitions of conditional probability, we have:
?? (?? |?? )=
?? (?? n?? )
?? (?? )
=
1
3
?? (?? |?? )=
?? (?? n?? )
?? (?? )
=
1
4
Given that ?? (?? n?? )=0.1, we can use these equations to find ?? (?? ) and ?? (?? ) :
From
0.1
?? (?? )
=
1
3
, we find ?? (?? )=0.3.
From
0.1
?? (?? )
=
1
4
, we find ?? (?? )=0.4.
Now, calculate ?? (?? ??? ) using the formula for the union of two events:
Page 5
JEE Main Previous Year Questions
(2025): Probability
Question1: Three distinct numbers are selected randomly from the set {?? ,?? ,?? ,…,???? }.
If the probability, that the selected numbers are in an increasing G.P., is
?? ?? ,?????? (?? ,?? )=?? , then ?? +?? is equal to ____ .
JEE Main 2025 (Online) 2nd April Morning Shift
Answer: 2477
Solution:
Common ratio Last triplet Total
r=2 10, 20, 40 10
r=3 4, 12, 36 4
r=4 2, 8, 32 2
r=5 1, 5, 25 1
r=6 1, 6, 36 1
Total = 18
Total choices =
40
?? 3
=9880
Required probability =
18
9880
=
9
4940
=
?? ??
?? +?? =4949
Question2: A card from a pack of 52 cards is lost. From the remaining 51 cards, ?? cards
are drawn and are found to be spades. If the probability of the lost card to be a spade
is
????
????
, then ?? is equal to ____ -
Answer: 2
Solution:
?? (
Lost (spade
n cards are spade
)
=
?? (
?? ?? ?? ?? )?? (?? ?? )
?? (
?? ?? ?? ?? )?? (?? ?? )+?? (
?? ?? ???
?? )?? (???
?? )
=
12
?? ??
51
?? ?? ×
1
4
12
?? ??
51
?? ?? ×
1
4
+
3
4
×
13
?? ??
51
?? ?? =
1
1+3·
13
?? ??
12
?? ?? =
13-?? 52-??
?
13-?? 52-?? =
11
50
??? =2
Question3: Two balls are selected at random one by one without replacement from a
bag containing 4 white and 6 black balls. If the probability that the first selected ball is
black, given that the second selected ball is also black, is
?? ?? , where ?????? (?? ,?? )=?? ,
then ?? +?? is equal to :
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
A. 4
B. 14
C. 11
D. 13
Answer: B
Solution:
In a bag containing 4 white and 6 black balls, two balls are selected one by one without
replacement. We need to find the probability that the first ball is black, given that the second
ball is black. Let's define the events:
A: The first ball selected is black.
B: The second ball selected is black.
The probability ?? (?? |?? ) is given by the formula:
?? (?? |?? )=
?? (?? n?? )
?? (?? )
Let's compute each part separately:
Probability of both balls being black ?? (?? n?? ) :
The chance that the first ball is black is
6
10
. Given that one black ball is removed, the probability
that the second is also black is
5
9
.
?? (?? n?? )=
6
10
×
5
9
=
30
90
=
1
3
Probability that the second ball is black ?? (?? ) :
This can occur if either the first ball was white or black:
First ball white, second black:
4
10
×
6
9
Both balls black:
6
10
×
5
9
?? (?? )=
4
10
×
6
9
+
6
10
×
5
9
=
24
90
+
30
90
=
54
90
=
3
5
Now, using these probabilities:
?? (?? |?? )=
1
3
3
5
=
1
3
×
5
3
=
5
9
Here,
5
9
is in its simplest form (gcd(5,9)=1), so ?? =5 and ?? =9.
Therefore, ?? +?? =5+9=14.
Question4: A coin is tossed three times. Let ?? denote the number of times a tail
follows a head. If ?? and ?? ?? denote the mean and variance of ?? , then the value of
???? (?? +?? ?? ) is:
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
A. 64
B. 32
C. 51
D. 48
Answer: D
Solution:
Outcome ?? ?? ?? ??
HHH 0
1
8
TTT 0
1
8
HHT 1
1
8
HTH 1
1
8
THH 0
1
8
TTH 0
1
8
THT 1
1
8
HTT 1
1
8
?? =??? ???? ?? =
1
2
?? 2
=??? ?? 2
?? ?? -?? 2
=
1
2
-
1
4
=
1
4
64(?? +?? 2
)=64[
1
2
+
1
4
]
=64×
3
4
=48
Question5: If ?? and ?? are two events such that ?? (?? n?? )=?? .?? , and ?? (?? |?? ) and
?? (?? |?? ) are the roots of the equation ???? ?? ?? -?? ?? +?? =?? , then the value of
?? (???
????
)
?? (???
n???
)
is:
JEE Main 2025 (Online) 22nd January Evening Shift
Options:
A.
4
3
B.
7
4
C.
9
4
D.
5
3
Answer: C
Solution:
To solve this problem, start by considering the equation given for the probabilities:
12?? 2
-7?? +1=0
The roots of this equation are:
?? =
1
3
,
1
4
Assume ?? (?? |?? )=
1
3
and ?? (?? |?? )=
1
4
.
From the definitions of conditional probability, we have:
?? (?? |?? )=
?? (?? n?? )
?? (?? )
=
1
3
?? (?? |?? )=
?? (?? n?? )
?? (?? )
=
1
4
Given that ?? (?? n?? )=0.1, we can use these equations to find ?? (?? ) and ?? (?? ) :
From
0.1
?? (?? )
=
1
3
, we find ?? (?? )=0.3.
From
0.1
?? (?? )
=
1
4
, we find ?? (?? )=0.4.
Now, calculate ?? (?? ??? ) using the formula for the union of two events:
?? (?? ??? )=?? (?? )+?? (?? )-?? (?? n?? )=0.4+0.3-0.1=0.6
The task is to find the value of:
?? (???
????
)
?? (???
n???
)
Using De Morgan's laws and the complements:
?? (???
????
)=?? (?? n?? )=1-?? (?? n?? )=1-0.1=0.9
?? (???
n???
)=1-?? (?? ??? )=1-0.6=0.4
Finally, compute the ratio:
?? (???
????
)
?? (???
n???
)
=
0.9
0.4
=
9
4
Question6: One die has two faces marked 1 , two faces marked 2 , one face marked 3
and one face marked 4 . Another die has one face marked 1 , two faces marked 2 , two
faces marked 3 and one face marked 4. The probability of getting the sum of numbers
to be 4 or 5 , when both the dice are thrown together, is
JEE Main 2025 (Online) 23rd January Morning Shift
Options:
A.
2
3
B.
3
5
C.
4
9
D.
1
2
Answer: D
Solution:
a= number or dice 1
b= number on dice 2
(?? ,?? )=(1,3),(3,1),(2,2),(2,3),(3,2),(1,4),(4,1)
Required probability
=
2
6
×
2
6
+
1
6
×
1
6
+
2
6
×
2
6
+
2
6
×
2
6
+
1
6
×
2
6
+
2
6
×
1
6
+
1
6
×
2
6
=
18
36
=
1
2
Read MoreFAQs on JEE Main Previous Year Questions (2025): Probability
| 1. What is the basic definition of probability in the context of JEE Main exams? |  |
Ans. Probability is a branch of mathematics that deals with the likelihood or chance of an event occurring. It is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes in a given scenario. For example, if you roll a fair six-sided die, the probability of getting a 4 is 1/6, since there is one favorable outcome (rolling a 4) out of six possible outcomes (1, 2, 3, 4, 5, 6).
| 2. How are different types of probability problems typically categorized in JEE Main exams? | |
Ans. Probability problems in JEE Main are generally categorized into three main types: 1. <b>Classical Probability</b>: Problems that involve equally likely outcomes, such as dice rolls or coin tosses. 2. <b>Empirical Probability</b>: Problems based on experimental data or historical frequency of events. 3. <b>Subjective Probability</b>: Problems that involve estimation based on personal judgment or experience rather than exact calculations.
| 3. What are some common formulas used in probability that students should remember for JEE Main? | |
Ans. Some essential probability formulas include: 1. The probability of an event A: P(A) = Number of favorable outcomes / Total number of outcomes. 2. The probability of the complement of an event A: P(A') = 1 - P(A). 3. For independent events A and B: P(A and B) = P(A) × P(B). 4. For mutually exclusive events A and B: P(A or B) = P(A) + P(B).
| 4. Can you explain the concept of conditional probability and provide a formula? | |
Ans. Conditional probability refers to the probability of an event occurring given that another event has already occurred. It is denoted as P(A|B), which is read as "the probability of A given B." The formula for conditional probability is: P(A|B) = P(A and B) / P(B), provided that P(B) > 0. This helps in understanding how the occurrence of one event affects the probability of another.
| 5. What strategies can students employ to solve probability problems effectively in exams? | |
Ans. Students can employ several strategies to tackle probability problems effectively: 1. <b>Understand the Problem</b>: Carefully read the problem statement to identify what is being asked and the relevant events. 2. <b>List Outcomes</b>: For simple problems, listing all possible outcomes can clarify the situation. 3. <b>Use Diagrams</b>: Venn diagrams or tree diagrams can help visualize complex relationships between events. 4. <b>Apply Formulas</b>: Familiarize yourself with key formulas and apply them appropriately. 5. <b>Practice Regularly</b>: Solve a variety of problems to gain confidence and improve problem-solving speed.
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