Question

8. Suppose a ball is drawn at random from a box containing three white and three black balls. After a ball is drawn, it is then replaced and another drawn. What is the probability that of the first four balls drawn, exactly two are white?

9. In a bolt manufacturing factory, machines A, B, and C produce 25%, 30% and 45% of the total output, respectively. Of their outputs, 7%, 6% and 4% are defective bolts,
respectively. If a bolt drawn at random from the production is found to be defective,
what is the probability that it was manufactured by machine C?

Answer 1

8.

$$P(WWBB)+P(WBBW)+P(WBWB)+$$ $$+P(BWWB)+P(BWBW)+P(BBWW)=$$

$$=({3 \over 6})({3\over 6})({3 \over 6})({3 \over 6})+({3 \over 6})({3\over 6})({3 \over 6})({3 \over 6})+$$

$$+({3 \over 6})({3\over 6})({3 \over 6})({3 \over 6})+({3 \over 6})({3\over 6})({3 \over 6})({3 \over 6})+$$

$$+({3 \over 6})({3\over 6})({3 \over 6})({3 \over 6})+({3 \over 6})({3\over 6})({3 \over 6})({3 \over 6})+$$

$$+({3 \over 6})({3\over 6})({3 \over 6})({3 \over 6})+({3 \over 6})({3\over 6})({3 \over 6})({3 \over 6})=$$

$$=6({1\over 16})={3\over 8}$$

9. Let 

​$E_1:$ the event that bolt is produced by machine A,

​$E_2:$ the event that bolt is produced by machine B, and 

​$E_3:$ the event that bolt is produced by machine C.

Here $E_1, E_2,$ and $E_3$ are mutually exclusive and exhaustive events.

We have

$$P(E_1)=0.25,P(E_2)=0.3,P(E_3)=0.45$$

Let $D:$ the event that bolt chosen is found to be defective.

Then

$$P(D|E_1)=0.07,P(D|E_2)=0.06,P(D|E_3)=0.04$$

From Bayes' Theorem

$$P(E_3|D)={P(D|E_3)P(E_3)\over P(D| E_1)P(E_1)+P(D|E_2)P(E_2)+P(D|E_3)P(E_3) }=$$

$$={0.04(0.45)\over 0.07(0.25)+0.06(0.3)+0.04(0.45) }={36\over 107}\approx0.3364$$