JEE Mathematics / Lakshmikanta Satapathy / Questions and Answers on Probability involving Independent events, mutually exclusive events in experiments of drawing colored balls from bags and problem solving probability of students using Addition theorem and Multiplication theorem

1. Physics Helpline
L K Satapathy
Probability QA 5

2. Physics Helpline
L K Satapathy
Q1: A bag contains 5 white balls , 7 red balls and 8 black balls. If four balls are drawn
one by one without replacement , find the probability of getting all white balls.
QA Probability - 5
5 1( )
20 4
P A  
Ans : Let event A = ‘the 1st ball is white’ , B = ‘the 2nd ball is white’ ,
C = ‘the 3rd ball is white’ and D = ‘the 4th ball is white.
We are required to find P(A  B  C  D)
= P(A)  P(B/A)  P(C/A B)  P(D/A  B  C)
Now P(A) is the probability of getting a white ball in the 1st draw. When the 1st ball is
drawn, there are 5 white balls in a total of 20 balls.
After a white ball is drawn in the 1st draw , without replacement ,we are left with 4
white balls in a total of 19 balls for the 2nd draw.
4( )
19
P B A 

3. Physics Helpline
L K Satapathy
QA Probability - 5
After a white ball is drawn in the 2nd draw , without replacement , we are left with 3
white balls in a total of 18 balls for the 3rd draw.
3 1( )
18 6
P C A B  
After a white ball is drawn in the 3rd draw , without replacement , we are left with 2
white balls in a total of 17 balls for the 4th draw.
2( )
17
P D A B C  
( )P A B C D   
( ). ( ). ( ). ( )P A P B A P C A B P D A B C   
1 4 1 2 1
4 19 6 17 969
[ ]Ans    

4. Physics Helpline
L K Satapathy
Q2 : A can solve 90% of the problems given in a book and B can solve 70% . What is
the probability that at least one of them will solve the problem selected at random
from the book?
QA Probability - 5
Ans : Let the events ‘A solves the problem’ = E and ‘B solves the problem’ = F
E and F are independent events.
Now the event ‘at least one of them solve it’ = EF
 We are required to find P(EF)
It is given that P(E) = 90% = 0.9 and P(F) = 70% = 0.7
P(E) = 1 – P(E) = 0.1 and P(F) = 1 – P(F) = 0.3
P(EF) = 1 – P[(EF)] = 1 – P(EF) [ Addition Theorem]
= 1 – P(E).P(F) [ E and F are independent]
= 1 – (0.1)(0.3) = 1 – 0.03 = 0.97 (Ans)

5. Physics Helpline
L K Satapathy
QA Probability - 5
Q3 : Bag A contains 4 red and 5 black balls and bag B contains 3 red and 7 black
balls. One ball is drawn from bag A and two from bag B. Find the probability that out
of the 3 balls drawn, 2 are black and 1 is red.
Ans : Two black balls and one red ball can be drawn from the two bags in
two mutually exclusive ways. We define the events as
E = 1 black ball from bag A and 1 black and 1 red ball from bag B.
F = 1 red ball from bag A and 2 black balls from bag B.
Let E1 = ‘1 black ball from bag A’ and E2 = ‘1 black and 1 red ball from bag B’.
and F1 = ‘1 red ball from bag A’ and F2 = ‘2 black balls from bag B’.
 E = ‘1 black ball from bag A and 1 black and 1 red ball from bag B’ = E1E2
 F = ‘1 red ball from bag A and 2 black balls from bag B’ = F1F2
 ‘1 red ball and 2 black balls’ can be drawn when E or F occurs
 The event ‘2 black balls and 1 red ball’ = EF = 1 2 1 2( ) ( )E E F F  

6. Physics Helpline
L K Satapathy
QA Probability - 5
To find P(E1) : Bag A contains 4 red and 5 black balls.
1
5( )
9
P E 
1 black ball can be chosen from 5 black balls in 5 ways
1 ball can be chosen from a total of 9 balls in 9 ways
 Probability of getting 1 black ball from bag A ,
 The required probability is
1 2 1 2 1 2 1 2[( ) ( )] ( ) ( ) . . . (1)P E E F F P E E P F F      
It may be noted that (i) E1 and E2 are independent
(ii) F1 and F2 are independent
(iii) (E1  E2) and (F1  F2) are mutually exclusive
To find P(E1  E2) and P(F1  F2) we need to find P(E1) , P(E2) , P(F1) and P(F2)
E1 = getting 1 black ball from bag A

7. Physics Helpline
L K Satapathy
QA Probability - 5
To find P(E2) : Bag B contains 3 red and 7 black balls.
2
21 7( )
45 15
P E  
1 black ball can be chosen from 7 black balls in 7 ways
2 balls can be chosen from a total of 10 balls in ways
 Probability of getting 1 black and 1 red ball from bag B ,
1 red ball can be chosen from 3 red balls in 3 ways
1 black and 1 red ball can be chosen from bag B in = 73 = 21 ways
10
2
10 9 45
1 2
C  

E2 = getting 1 black and 1 red ball from bag B

8. Physics Helpline
L K Satapathy
QA Probability - 5
To find P(F1) : Bag A contains 4 red and 5 black balls.
1
4( )
9
P F 
1 red ball can be chosen from 4 red balls in 4 ways
1 ball can be chosen from a total of 9 balls in 9 ways
 Probability of getting 1 red ball from bag A ,
F1 = getting 1 red ball from bag A ,
To find P(F2) : Bag B contains 3 red and 7 black balls.
2 black balls can be chosen from 7 black balls in ways
F2 = getting 2 red ball from bag A ,
7
2
7 6 21
1 2
C  

2 balls can be chosen from a total of 10 balls in ways10
2
10 9 45
1 2
C  

2
21 7( )
45 15
P F   Probability of getting 1 black and 1 red ball from bag B ,

9. Physics Helpline
L K Satapathy
QA Probability - 5
We have obtained
1
5( )
9
P E  2
7( )
15
P E  1
4( )
9
P F  2
7( )
15
P F 
1 2 1 2
5 7 7( ) ( ). ( )
9 15 27
P E E P E P E     
1 2 1 2
4 7 28( ) ( ). ( )
9 15 135
and P F F P F P F    
1 2 1 2[( ) ( )]P E E F F  
7 28 35 28 63 7
27 135 135 135
]
1
[
5
Ans    
Required probability
E1 and E2 are independent
F1 and F2 are independent
(E1  E2) and (F1  F2) are
mutually exclusive1 2 1 2( ) ( )P E E P F F   

10. Physics Helpline
L K Satapathy
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