Page 105 - DMTH404_STATISTICS

Unit 7: Modern Approach to Probability

Notes
n A
We are given n(S) = 1000, n(A) = 600, ( )  400

600 30 400 5 600 30
´
´
´
n
Also, ( ) B  +  200 and ( n A  ) B   180
100 100 100
( n A  ) B 180 9
)
Thus, the required probability is given by ( /P A B   
n ( ) B 200 10
Alternative Method :
Writing the given information in a nine-square table, we have:
B B Total
A 180 420 600
A 20 380 400
Total 200 800 1000

180 9
)
From the above table we can write ( /P A B  
200 10

Example 28: A bag contains 2 black and 3 white balls. Two balls are drawn at random one
after the other without replacement. Obtain the probability that (a) Second ball is black given
that the first is white, (b) First ball is white given that the second is black.
Solution.
First ball can be drawn in any one of the 5 ways and then a second ball can be drawn in any one
of the 4 ways. Therefore, two balls can be drawn in 5 ´ 4 = 20 ways. Thus, n(S) = 20.
(a) Let A be the event that first ball is white and A be the event that second is black. We want
1 c 2
to find P A /A h.
2 1
First white ball can be drawn in any of the 3 ways and then a second ball can be drawn in
any of the 4 ways,  n(A ) = 3 ´ 4 = 12.
1
Further, first white ball can be drawn in any of the 3 ways and then a black ball can be
1 b
drawn in any of the 2 ways,  n A  g  3 2  6 .
´
A
2
( n A  A ) 6 1
Thus, (A 2 /A 1 )  1 2   .
P
( )
n A 1 12 2
c
(b) Here we have to find P A /A h.
1 2
The second black ball can be drawn in the following two mutually exclusive ways :
(i) First ball is white and second is black or
(ii) both the balls are black. n A b A g
Thus, n(A ) = 3´ 2 + 2´ 1 = 8, P A / b 1  2 6 3
2
2 1 A g  n A b g  8  4 .
2

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