JEE Mathematics/ Lakshmikanta Satapathy/ Questions and answers part 7 involving probability distribution and determination of mean and variance of a random variable

1. Physics Helpline
L K Satapathy
Probability QA 7

2. Physics Helpline
L K Satapathy
Q1: Two numbers are selected at random (without replacement) from the first six
positive integers. Let X denote the larger of the two numbers obtained. Find the mean
and variance of X.
QA Probability - 7
6
2
6 5 15
1 2
C ways 

Ans : The first six positive integers are { 1 , 2 , 3 , 4 , 5 , 6 }
We have to select 2 numbers from a total of 6.
This can be done in
Since the numbers are not repeated , the two numbers are not equal
 If X denotes the larger of the two numbers , then X is a random variable
which can take the values 2 , 3 , 4 , 5 or 6.
 Probability of selecting each pair of numbers =
 S = { 12 , 13 , 14 , 15 , 16 , 23 , 24 , 25 , 26 , 34 , 35 , 36 , 45 , 46 , 56 }
1
15

3. Physics Helpline
L K Satapathy
QA Probability - 7
  1( 2) {(1,2)}
15
P X P  
  2( 3) {(1,3),(2,3)}
15
P X P  
  3( 4) {(1,4),(2,4),(3,4)}
15
P X P  
  4( 5) {(1,5),(2,5),(3,5),(4,5)}
15
P X P  
  5( 6) {(1,6),(2,6),(3,6),(4,6),(5,6)}
15
P X P  
X
P(X)
Probability distribution
2 3 4 65
1
15
2
15
3
15
5
15
4
15

4. Physics Helpline
L K Satapathy
QA Probability - 7
i ip x 2 6 12 20 30
15
   
70 14
15 3
 
2
i ip x 4 18 48 100 180
15
   
350 70
15 3
 
2 2 2 70 196 210 196 14
3 9 9 9
[ ]x i i Anp sx       
2
3
4
5
6
Computation Table
ix i ip x
1
15
2
15
3
15
ip 2
i ip x
4
15
5
15
2
15
6
15
12
15
20
15
30
15
4
15
18
15
48
15
100
15
180
15
14
3i ip x  2 70
3i ip x 
14 [
3
]i i Ansp x  

5. Physics Helpline
L K Satapathy
QA Probability - 7
Q2: A fair coin is tossed four times. Let X represent the number of heads obtained in
the four tosses. Find the mean and variance of X.
Ans : The sample space for tossing a coin 4 times is
S = { HHHH , HHHT , HHTH , HHTT , HTHH , HTHT , HTTH , HTTT ,
THHH , THHT , THTH , THTT , TTHH , TTHT , TTTH , TTTT }
Given that X represents the number of heads obtained
 X is a random variable which can take the values 0 , 1 , 2 , 3 or 4.
Number of elements of sample space = 16
 Probability of each outcome
and each of these are equally likely to occur
1
16


6. Physics Helpline
L K Satapathy
QA Probability - 7
1
16
P (X = 0) = P( TTTT )
P (X = 1) = P( HTTT , THTT , TTHT , TTTH )
= P( HTTT ) + P( THTT ) + P( TTHT ) + P( TTTH )
1 1 1 1 4
16 16 16 16 16
    
P (X = 2) = P( HHTT , HTHT , HTTH , THHT , THTH , TTHH )
= P( HHTT ) + P( HTHT ) + P( HTTH ) + P( THHT ) + P( THTH ) + P( TTHH )
1 1 1 1 1 1 6
16 16 16 16 16 16 16
      

7. Physics Helpline
L K Satapathy
QA Probability - 7
P (X = 3) = P( HHHT , HHTH , HTHH , THHH )
= P( HHHT ) + P( HHTH ) + P( HTHH ) + P( THHH )
1 1 1 1 4
16 16 16 16 16
    
1
16
P (X = 4) = P( HHHH )
X
P(X)
Probability distribution
0 1 2 43
1
16
4
16
6
16
1
16
4
16

8. Physics Helpline
L K Satapathy
QA Probability - 7
0
1
2
3
4
Computation Table
ix i ip x
1
16
4
16
6
16
ip 2
i ip x
4
16
1
16
0
4
16
12
16
12
16
4
16
0
4
16
24
16
36
16
16
16
2i ip x  2
5i ip x 
i ip x 4 12 12 4 2
16
   
2
i ip x 4 24 36 16 5
16
   
2 2 2
[1 ]5 4
x i i
An
p x
s
  
  

[ ]2i ix A sp n  

9. Physics Helpline
L K Satapathy
QA Probability - 7
Q3: In a meeting, 70% of the members favour a certain proposal and 30% oppose it. A
member is selected at random and we take X = 0 if he opposed the proposal and X = 1
if he favoured it . Find the mean and variance of X.
Ans : It is given that 30% of the members oppose the proposal for which X = 0
30 3( 0) 0.3
100 10
P X    
and 70% of the members favour the proposal for which X = 1
70 7( 1) 0.7
100 10
P X    
X
P(X)
Probability distribution
0 1
0.3 0.7
0 0.3 1 0.7 0.7i ip x     
2 2 2
0.3 0 0.7 1 0.7i ip x     
0 [.7 ]i i A sp x n  
2 2 2 2
0.7 (0.7) 0.7 0.49 0.21 [ ]x i i Ansp x       

10. Physics Helpline
L K Satapathy
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