In a bolt factory, machines A,B and C manufacture respectively \[20 \%, 30 \%\] and \[50 \%\] of the total bolts. Of their output 3, 4 and 2 percent are respectively defective bolts. A bolt is drawn at random from the product. If t...
Solutions · 1
Official solution
Given : P(A)=10020=102P(B)=10030=103P(C)=10050=105
Let E→ Event that the bolt is defective.
P(E/A)=1003,P(BE)=1004,P(CE)=1002
So, P(C/E)=P(AE)×P(A)+P(BE)×P(B)+P(CE)×P(C)P(CE)×P(C)=1003×102+1004×103+1002×105105×1002=6+12+1010=2810=145