Answer:
P ( B0 / D0 ) = 0.59877
P ( B1 / D0 ) = 0.25793
P ( B2 / D0 ) = 0.14329
Step-by-step explanation:
Given:
- 0 be the event that the batch has 0 defectives = (0
) = 0.52
- 1 be the event that the batch has 1 defectives = (1
) = 0.28
- 2 be the event that the batch has 2 defectives = (2
) = 0.2
- Two components are selected
Find:
What are the probabilities associated with 0, 1, and 2 defective components being in the batch under each of the following conditions?
(a) Neither tested component is defective.
Solution:
Let 0 be the event that neither selected component is defective.
- The event 0 can happen in three different ways:
(i) Our batch of 10 is perfect, and we get no defectives in our sample of two;
P(i) = P(B0) = 0.52
(ii) Our batch of 10 has 1 defective, but our sample of two misses them;
P ( no defect / B1 ) = P ( no defect ) / P ( B 1 )
= 9C2 / 10C2 = 0.8
P ( ii ) = 0.28*0.8 = 0.224
(iii) Our batch has 2 defective, but our sample misses them.
P ( no defect / B2 ) = P ( no defect ) / P ( B 2 )
= 8C2 / 10C2 = 56/90
P ( iii ) = 0.2*56/90 = 0.124444
- Then,
P(Do) = P(i) + P(ii) + P(iii)
P(Do) = 0.52 + 0.224 + 0.124444 = 977/1125
We use the general conditional probability formula:
P ( B0 / D0 ) = P ( B0 & D0 ) / P( D0 )
P ( B0 / D0 ) = 0.52*1125 / 977 = 0.59877
P ( B1 / D0 ) = P ( B1 & D0 ) / P( D0 )
P ( B1 / D0 ) = 0.224*1125 / 977 = 0.25793
P ( B2 / D0 ) = P ( B2 & D0 ) / P( D0 )
P ( B2 / D0 ) = 0.12444*1125 / 977 = 0.14329
