JEE Advanced Previous Year Questions (2021-2023): Probability | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


JEE Advanced Previous Year 
Questions (2 1 2 0 -2023): Probability 
Q1. Let ?? be the set of all five digit numbers formed using 1,2,2,2,4,4,0. For 
example, 22240 is in ?? while 02244 and 44422 are not in ?? . Suppose that 
each element of ?? has an equal chance of bein chosen. Let ?? be the 
conditional probability that an element chosen at random is a multiple of 20 
given that it is a multiple of 5 . Then the value of ?????? is equal to :   [JEE 
Advanced 2023 Paper 2 Online] 
Ans: 31 
Explanation: Currently no explanation available 
 
 
Q2. Consider the ?? × ?? square in the figure. Let ?? ?? , ?? ?? , … , ?? ????
 be the points of 
intersections (dots in the picture) in some order. We say that ?? ?? and ?? ?? are 
friends if they are adjacent along a row or along a column. Assume that each 
point ?? ?? has an equal chance of being chosen. 
 
Let ?? ?? be the probability that a randomly chosen point has ?? many friends, 
?? = ?? , ?? , ?? , ?? , ?? . Let ?? be a random variable such that for ?? = ?? , ?? , ?? , ?? , ?? , the 
probability ?? (?? = ?? ) = ?? ?? . Then the value of ???? (?? ) is :     [JEE Advanced 2023 
Paper 2 Online] 
Page 2


JEE Advanced Previous Year 
Questions (2 1 2 0 -2023): Probability 
Q1. Let ?? be the set of all five digit numbers formed using 1,2,2,2,4,4,0. For 
example, 22240 is in ?? while 02244 and 44422 are not in ?? . Suppose that 
each element of ?? has an equal chance of bein chosen. Let ?? be the 
conditional probability that an element chosen at random is a multiple of 20 
given that it is a multiple of 5 . Then the value of ?????? is equal to :   [JEE 
Advanced 2023 Paper 2 Online] 
Ans: 31 
Explanation: Currently no explanation available 
 
 
Q2. Consider the ?? × ?? square in the figure. Let ?? ?? , ?? ?? , … , ?? ????
 be the points of 
intersections (dots in the picture) in some order. We say that ?? ?? and ?? ?? are 
friends if they are adjacent along a row or along a column. Assume that each 
point ?? ?? has an equal chance of being chosen. 
 
Let ?? ?? be the probability that a randomly chosen point has ?? many friends, 
?? = ?? , ?? , ?? , ?? , ?? . Let ?? be a random variable such that for ?? = ?? , ?? , ?? , ?? , ?? , the 
probability ?? (?? = ?? ) = ?? ?? . Then the value of ???? (?? ) is :     [JEE Advanced 2023 
Paper 2 Online] 
Ans:  24 
Explanation: Currently no explanation available 
 
Q3. Consider the ?? × ?? square in the figure. Let ?? ?? , ?? ?? , … , ?? ????
 be the points of 
intersections (dots in the picture) in some order. We say that ?? ?? and ?? ?? are 
friends if they are adjacent along a row or along a column. Assume that each 
point ?? ?? has an equal chance of being chosen. 
 
Two distinct points are chosen randomly out of the points ?? ?? , ?? ?? , … , ?? ????
. Let ?? 
be the probability that they are friends. Then the value of ???? is :   
[JEE Advanced 2023 Paper 2 Online] 
Ans: 0.50 
Explanation: Currently no explanation available 
 
Q4. In a study about a pandemic, data of 900 persons was collected. It was 
found that 
190 persons had symptom of fever, 
220 persons had symptom of cough, 
220 persons had symptom of breathing problem, 
330 persons had symptom of fever or cough or both, 
350 persons had symptom of cough or breathing problem or both, 
340 persons had symptom of fever or breathing problem or both, 
30 persons had all three symptoms (fever, cough and breathing problem). 
If a person is chosen randomly from these 900 persons, then the probability 
that the person has at most one symptom is   [JEE Advanced 2022 Paper 1 
Online] 
Page 3


JEE Advanced Previous Year 
Questions (2 1 2 0 -2023): Probability 
Q1. Let ?? be the set of all five digit numbers formed using 1,2,2,2,4,4,0. For 
example, 22240 is in ?? while 02244 and 44422 are not in ?? . Suppose that 
each element of ?? has an equal chance of bein chosen. Let ?? be the 
conditional probability that an element chosen at random is a multiple of 20 
given that it is a multiple of 5 . Then the value of ?????? is equal to :   [JEE 
Advanced 2023 Paper 2 Online] 
Ans: 31 
Explanation: Currently no explanation available 
 
 
Q2. Consider the ?? × ?? square in the figure. Let ?? ?? , ?? ?? , … , ?? ????
 be the points of 
intersections (dots in the picture) in some order. We say that ?? ?? and ?? ?? are 
friends if they are adjacent along a row or along a column. Assume that each 
point ?? ?? has an equal chance of being chosen. 
 
Let ?? ?? be the probability that a randomly chosen point has ?? many friends, 
?? = ?? , ?? , ?? , ?? , ?? . Let ?? be a random variable such that for ?? = ?? , ?? , ?? , ?? , ?? , the 
probability ?? (?? = ?? ) = ?? ?? . Then the value of ???? (?? ) is :     [JEE Advanced 2023 
Paper 2 Online] 
Ans:  24 
Explanation: Currently no explanation available 
 
Q3. Consider the ?? × ?? square in the figure. Let ?? ?? , ?? ?? , … , ?? ????
 be the points of 
intersections (dots in the picture) in some order. We say that ?? ?? and ?? ?? are 
friends if they are adjacent along a row or along a column. Assume that each 
point ?? ?? has an equal chance of being chosen. 
 
Two distinct points are chosen randomly out of the points ?? ?? , ?? ?? , … , ?? ????
. Let ?? 
be the probability that they are friends. Then the value of ???? is :   
[JEE Advanced 2023 Paper 2 Online] 
Ans: 0.50 
Explanation: Currently no explanation available 
 
Q4. In a study about a pandemic, data of 900 persons was collected. It was 
found that 
190 persons had symptom of fever, 
220 persons had symptom of cough, 
220 persons had symptom of breathing problem, 
330 persons had symptom of fever or cough or both, 
350 persons had symptom of cough or breathing problem or both, 
340 persons had symptom of fever or breathing problem or both, 
30 persons had all three symptoms (fever, cough and breathing problem). 
If a person is chosen randomly from these 900 persons, then the probability 
that the person has at most one symptom is   [JEE Advanced 2022 Paper 1 
Online] 
Ans: 0.79 to 0.81 
Let person had symptom of fever = ?? (?? ) = 190 
Person had symptom of cough = ?? (?? ) = 220 
Person had symptom of breathing = ?? ( ?? ) = 220 
 
Person had symptom of fever or cough 
= ?? ( ?? ? ?? ) = 330 
Person had symptom of cough or breathing 
= ?? (?? ? ?? ) = 350 
Person had symptom of fever or breathing 
= ?? ( ?? ? ?? ) = 340 
Person had symptom of fever, cough and breathing = ?? ( ?? n ?? ? ?? ) = 30 
So ?? ( ?? n ?? ) = ?? ( ?? ) + ?? (?? ) - ?? ( ?? ? ?? ) 
  = 190 + 220- 330   = 80  
and 
 ?? ( ?? n ?? ) = ?? ( ?? ) + ?? ( ?? ) - ?? ( ?? ? ?? )   = 190 + 220- 340   = 70  
  ?????? ?? (?? n ?? ) = ?? (?? ) + ?? ( ?? ) - ?? (?? n ?? )   = 220 + 220- 350   = 90  
Number of people having at most one symptom 
= 70 + 80 + 90 + 480 = 720 
 
Required probability =
720
900
= 0.80 
 
Q5. A number of chosen at random from the set {?? , ?? , ?? , … . , ???????? }. Let ?? be the 
probability that the chosen number is a multiple of 3 or a multiple of 7 . 
Page 4


JEE Advanced Previous Year 
Questions (2 1 2 0 -2023): Probability 
Q1. Let ?? be the set of all five digit numbers formed using 1,2,2,2,4,4,0. For 
example, 22240 is in ?? while 02244 and 44422 are not in ?? . Suppose that 
each element of ?? has an equal chance of bein chosen. Let ?? be the 
conditional probability that an element chosen at random is a multiple of 20 
given that it is a multiple of 5 . Then the value of ?????? is equal to :   [JEE 
Advanced 2023 Paper 2 Online] 
Ans: 31 
Explanation: Currently no explanation available 
 
 
Q2. Consider the ?? × ?? square in the figure. Let ?? ?? , ?? ?? , … , ?? ????
 be the points of 
intersections (dots in the picture) in some order. We say that ?? ?? and ?? ?? are 
friends if they are adjacent along a row or along a column. Assume that each 
point ?? ?? has an equal chance of being chosen. 
 
Let ?? ?? be the probability that a randomly chosen point has ?? many friends, 
?? = ?? , ?? , ?? , ?? , ?? . Let ?? be a random variable such that for ?? = ?? , ?? , ?? , ?? , ?? , the 
probability ?? (?? = ?? ) = ?? ?? . Then the value of ???? (?? ) is :     [JEE Advanced 2023 
Paper 2 Online] 
Ans:  24 
Explanation: Currently no explanation available 
 
Q3. Consider the ?? × ?? square in the figure. Let ?? ?? , ?? ?? , … , ?? ????
 be the points of 
intersections (dots in the picture) in some order. We say that ?? ?? and ?? ?? are 
friends if they are adjacent along a row or along a column. Assume that each 
point ?? ?? has an equal chance of being chosen. 
 
Two distinct points are chosen randomly out of the points ?? ?? , ?? ?? , … , ?? ????
. Let ?? 
be the probability that they are friends. Then the value of ???? is :   
[JEE Advanced 2023 Paper 2 Online] 
Ans: 0.50 
Explanation: Currently no explanation available 
 
Q4. In a study about a pandemic, data of 900 persons was collected. It was 
found that 
190 persons had symptom of fever, 
220 persons had symptom of cough, 
220 persons had symptom of breathing problem, 
330 persons had symptom of fever or cough or both, 
350 persons had symptom of cough or breathing problem or both, 
340 persons had symptom of fever or breathing problem or both, 
30 persons had all three symptoms (fever, cough and breathing problem). 
If a person is chosen randomly from these 900 persons, then the probability 
that the person has at most one symptom is   [JEE Advanced 2022 Paper 1 
Online] 
Ans: 0.79 to 0.81 
Let person had symptom of fever = ?? (?? ) = 190 
Person had symptom of cough = ?? (?? ) = 220 
Person had symptom of breathing = ?? ( ?? ) = 220 
 
Person had symptom of fever or cough 
= ?? ( ?? ? ?? ) = 330 
Person had symptom of cough or breathing 
= ?? (?? ? ?? ) = 350 
Person had symptom of fever or breathing 
= ?? ( ?? ? ?? ) = 340 
Person had symptom of fever, cough and breathing = ?? ( ?? n ?? ? ?? ) = 30 
So ?? ( ?? n ?? ) = ?? ( ?? ) + ?? (?? ) - ?? ( ?? ? ?? ) 
  = 190 + 220- 330   = 80  
and 
 ?? ( ?? n ?? ) = ?? ( ?? ) + ?? ( ?? ) - ?? ( ?? ? ?? )   = 190 + 220- 340   = 70  
  ?????? ?? (?? n ?? ) = ?? (?? ) + ?? ( ?? ) - ?? (?? n ?? )   = 220 + 220- 350   = 90  
Number of people having at most one symptom 
= 70 + 80 + 90 + 480 = 720 
 
Required probability =
720
900
= 0.80 
 
Q5. A number of chosen at random from the set {?? , ?? , ?? , … . , ???????? }. Let ?? be the 
probability that the chosen number is a multiple of 3 or a multiple of 7 . 
Then the value of 500 is 
[JEE Advanced 2021 Paper 2 Online] 
Ans: 214 
 
Given, set = {1,2,3, … ,2000 } 
Let ?? 1
= Event that it is a multiple of 3 = {3,6,9, … ,1998} 
? ?? (?? 1
) = 666 
and ?? 2
= Event that it is a multiple of 7 = {7,14, … ,1995} 
? ?? (?? 2
) = 285 
?? 1
n ?? 2
= multiple of 21 = {21,42, … . ,1995} 
?? (?? 1
n ?? 2
) = 95 
? ?? (?? 1
? ?? 2
) = ?? (?? 1
) + ?? (?? 2
) - ?? (?? 1
n ?? 2
) 
?? (?? 1
? ?? 2
) =
666+285-95
2000
=
856
2000
= ?? (given) 
Hence, 500 ?? = 500×
856
2000
=
856
4
= 214 
 
Q6. The probability that the minimum of chosen numbers is at most 40. 
The value of 
??????
?? ?? ?? is    [JEE Advanced 2021 Paper 1 Online] 
Ans: 76.25 
 ?? 1
= 1 - ?? ( ?? ?? ?? 3 ?????????????? ?????? = 80)   = 1 - (
80
100
)
3
=
125- 64
125
=
61
125
  
So, 
625?? 1
4
=
625
4
×
61
125
=
5
4
× 61 = 76.25 
 
Q7. the probability that the minimum of chosen numbers is at most 40 . 
The value of 
??????
?? ?? ?? is.   [JEE Advanced 2021 Paper 1 Online] 
Ans: 24.5 
?? 2
= 1 - ?? ( all three numbers are > 40) 
= 1 - (
60
100
)
3
= 1 -
27
125
=
98
125
 
So, 
125?? 2
4
=
125
4
×
98
125
=
98
4
= 24.50 
 
?? ?? . Consider an experiment of tossing a coin repeatedly until the outcomes 
of two consecutive tosses are same. If the probability of a random toss 
resulting in head is 
?? ?? , then the probability that the experiment stops with 
head is : 
Page 5


JEE Advanced Previous Year 
Questions (2 1 2 0 -2023): Probability 
Q1. Let ?? be the set of all five digit numbers formed using 1,2,2,2,4,4,0. For 
example, 22240 is in ?? while 02244 and 44422 are not in ?? . Suppose that 
each element of ?? has an equal chance of bein chosen. Let ?? be the 
conditional probability that an element chosen at random is a multiple of 20 
given that it is a multiple of 5 . Then the value of ?????? is equal to :   [JEE 
Advanced 2023 Paper 2 Online] 
Ans: 31 
Explanation: Currently no explanation available 
 
 
Q2. Consider the ?? × ?? square in the figure. Let ?? ?? , ?? ?? , … , ?? ????
 be the points of 
intersections (dots in the picture) in some order. We say that ?? ?? and ?? ?? are 
friends if they are adjacent along a row or along a column. Assume that each 
point ?? ?? has an equal chance of being chosen. 
 
Let ?? ?? be the probability that a randomly chosen point has ?? many friends, 
?? = ?? , ?? , ?? , ?? , ?? . Let ?? be a random variable such that for ?? = ?? , ?? , ?? , ?? , ?? , the 
probability ?? (?? = ?? ) = ?? ?? . Then the value of ???? (?? ) is :     [JEE Advanced 2023 
Paper 2 Online] 
Ans:  24 
Explanation: Currently no explanation available 
 
Q3. Consider the ?? × ?? square in the figure. Let ?? ?? , ?? ?? , … , ?? ????
 be the points of 
intersections (dots in the picture) in some order. We say that ?? ?? and ?? ?? are 
friends if they are adjacent along a row or along a column. Assume that each 
point ?? ?? has an equal chance of being chosen. 
 
Two distinct points are chosen randomly out of the points ?? ?? , ?? ?? , … , ?? ????
. Let ?? 
be the probability that they are friends. Then the value of ???? is :   
[JEE Advanced 2023 Paper 2 Online] 
Ans: 0.50 
Explanation: Currently no explanation available 
 
Q4. In a study about a pandemic, data of 900 persons was collected. It was 
found that 
190 persons had symptom of fever, 
220 persons had symptom of cough, 
220 persons had symptom of breathing problem, 
330 persons had symptom of fever or cough or both, 
350 persons had symptom of cough or breathing problem or both, 
340 persons had symptom of fever or breathing problem or both, 
30 persons had all three symptoms (fever, cough and breathing problem). 
If a person is chosen randomly from these 900 persons, then the probability 
that the person has at most one symptom is   [JEE Advanced 2022 Paper 1 
Online] 
Ans: 0.79 to 0.81 
Let person had symptom of fever = ?? (?? ) = 190 
Person had symptom of cough = ?? (?? ) = 220 
Person had symptom of breathing = ?? ( ?? ) = 220 
 
Person had symptom of fever or cough 
= ?? ( ?? ? ?? ) = 330 
Person had symptom of cough or breathing 
= ?? (?? ? ?? ) = 350 
Person had symptom of fever or breathing 
= ?? ( ?? ? ?? ) = 340 
Person had symptom of fever, cough and breathing = ?? ( ?? n ?? ? ?? ) = 30 
So ?? ( ?? n ?? ) = ?? ( ?? ) + ?? (?? ) - ?? ( ?? ? ?? ) 
  = 190 + 220- 330   = 80  
and 
 ?? ( ?? n ?? ) = ?? ( ?? ) + ?? ( ?? ) - ?? ( ?? ? ?? )   = 190 + 220- 340   = 70  
  ?????? ?? (?? n ?? ) = ?? (?? ) + ?? ( ?? ) - ?? (?? n ?? )   = 220 + 220- 350   = 90  
Number of people having at most one symptom 
= 70 + 80 + 90 + 480 = 720 
 
Required probability =
720
900
= 0.80 
 
Q5. A number of chosen at random from the set {?? , ?? , ?? , … . , ???????? }. Let ?? be the 
probability that the chosen number is a multiple of 3 or a multiple of 7 . 
Then the value of 500 is 
[JEE Advanced 2021 Paper 2 Online] 
Ans: 214 
 
Given, set = {1,2,3, … ,2000 } 
Let ?? 1
= Event that it is a multiple of 3 = {3,6,9, … ,1998} 
? ?? (?? 1
) = 666 
and ?? 2
= Event that it is a multiple of 7 = {7,14, … ,1995} 
? ?? (?? 2
) = 285 
?? 1
n ?? 2
= multiple of 21 = {21,42, … . ,1995} 
?? (?? 1
n ?? 2
) = 95 
? ?? (?? 1
? ?? 2
) = ?? (?? 1
) + ?? (?? 2
) - ?? (?? 1
n ?? 2
) 
?? (?? 1
? ?? 2
) =
666+285-95
2000
=
856
2000
= ?? (given) 
Hence, 500 ?? = 500×
856
2000
=
856
4
= 214 
 
Q6. The probability that the minimum of chosen numbers is at most 40. 
The value of 
??????
?? ?? ?? is    [JEE Advanced 2021 Paper 1 Online] 
Ans: 76.25 
 ?? 1
= 1 - ?? ( ?? ?? ?? 3 ?????????????? ?????? = 80)   = 1 - (
80
100
)
3
=
125- 64
125
=
61
125
  
So, 
625?? 1
4
=
625
4
×
61
125
=
5
4
× 61 = 76.25 
 
Q7. the probability that the minimum of chosen numbers is at most 40 . 
The value of 
??????
?? ?? ?? is.   [JEE Advanced 2021 Paper 1 Online] 
Ans: 24.5 
?? 2
= 1 - ?? ( all three numbers are > 40) 
= 1 - (
60
100
)
3
= 1 -
27
125
=
98
125
 
So, 
125?? 2
4
=
125
4
×
98
125
=
98
4
= 24.50 
 
?? ?? . Consider an experiment of tossing a coin repeatedly until the outcomes 
of two consecutive tosses are same. If the probability of a random toss 
resulting in head is 
?? ?? , then the probability that the experiment stops with 
head is : 
A 
?? ?? 
B 
?? ????
 
 
C 
?? ????
 
D 
?? ??              [JEE Advanced 2023 Paper 2 Online] 
Ans: (B) 
We are given a biased coin with the probability of getting heads (?? ) as 
1
3
 and thus, the 
probability of getting tails (?? ) is 
2
3
 (since the total probability of getting either heads or 
tails should sum to 1). 
The experiment involves tossing this coin repeatedly until we obtain two identical 
outcomes in a row. We are interested in the case where the experiment ends with two 
consecutive heads. 
Let's consider two different scenarios for the sequences of coin tosses : 
Sequences that begin with a head and end with HH (like HH, HTHH, HTHTHH, and so 
on). 
Sequences that begin with a tail and end with HH (like THH, THTHH, THTHTHH, and 
so on). 
Both types of sequences will result in the experiment ending, so we need to consider both 
in our final probability calculation. 
Scenario 1 : Sequences beginning with ?? 
The simplest sequence is getting ???? right away. The probability of this event is (
1
3
)
2
=
1
9
. 
The next sequence is getting a head, then a tail, and then HH (i.e., HTHH). The 
probability of this sequence is 
1
3
·
2
3
· (
1
3
)
2
=
2
81
. 
We can see a pattern in these sequences. Each sequence in this scenario can be described 
as getting one head, followed by some number of "tail, head" pairs, and ending with ???? . 
Hence, the sequences in this scenario form a geometric series, where the ratio between 
consecutive terms is ?? = ?? (???? ) =
1
3
·
2
3
=
2
9
. 
The sum of an infinite geometric series is given by the formula 
?? 1-?? , where ?? is the first 
term of the series and ?? is the common ratio. 
In this scenario, the first term is the probability of getting ???? immediately, which is 
1
9
. 
So, the sum of this series is : 
?? 1
=
1/9
1 - 2/9
=
1/9
7/9
=
1
7
. 
Scenario 2 : Sequences beginning with ?? 
In this scenario, we first get a tail, and then we follow the same pattern as in the first 
scenario. 
The simplest sequence is ?????? , with a probability of 
2
3
· (
1
3
)
2
=
2
27
. 
The next sequence is getting a tail, a head, another tail, and then HH (i.e., THTHH). The 
probability of this sequence is 
2
3
·
1
3
·
2
3
· (
1
3
)
2
=
4
243
.