Question

Items are inspected for flaws by two quality inspectors. Both inspectors inspect every item and the probability that an item has a flaw is 0.1. If a flaw is present,it will be detected by the first inspector with probability 0.92, and by the second inspector with probability 0.7. If an item does not have a flaw, it will be passed by the first inspector with probability 0.95 and by the second inspector with probability 0.8. Assume the inspectors function independently.
(a) If an item has a flaw, what is the probability that it will be found by at least one of the two inspectors?
(b) If an item is passed by the first inspector, what is the probability that it actually has a flaw?
(c) What is the probability that the two inspectors draw different conclusions on the same item?
(d) If an item is passed by both inspectors, what is the probability that it actually has a flaw?

Answer 1

Answer:

a)0.976

b)0.00926

c)0.2402

d)0.35

Step-by-step explanation:

Let $X_i$ be an item passed by inspector i

Let Y be the event that there is a fault in an item

The probability that an item has a flaw is 0.1 i.e. P(Y)=0.1

If a flaw is present ,it will be detected by the first inspector with probability 0.92 i.e.$P(\bar{X_1}|Y)=0.92$

So, $P(X_1|Y)=1-0.92=0.08$

If a flaw is present ,it will be detected by the second inspector with probability 0.7 i.e.$P(\bar{X_2}|Y)=0.7$

So,$P(X_2|Y)=1-0.7=0.3$

If an item does not have a flaw, it will be passed by the first inspector with probability 0.95 i.e. $P(X_1|\bar{Y}) = 0.95$

So, $P(\bar{X_1}|\bar{Y}) = 1-0.95=0.05$

If an item does not have a flaw, it will be passed by the second inspector with probability 0.8 i.e. $P(X_2|\bar{Y}) = 0.8$

So, $P(\bar{X_2}|\bar{Y}) = 1-0.8=0.2$

a)P(found by atleast one inspector | It has flaw )=1-P(found by none inspector | It has flow )

P(found by atleast one inspector | It has flaw )=$1-P(X_1|Y) P(X_2|Y)$

P(found by atleast one inspector | It has flaw )=$1-0.08 \times 0.3$

P(found by atleast one inspector | It has flaw )=0.976

Hence the probability that it will be found by at least one of the two inspectors if it has flaw is 0.976

b)$P(Y|X_1)=\frac{P(X_1|Y) P(Y)}{P(X_1|Y) P(Y)+P(X_1|\bar{Y}) P(\bar{Y})}$

$P(Y|X_1)=\frac{0.08 \times 0.1}{0.08 \times 0.1+0.95 \times 0.9}=0.00926$

C)P( two inspectors draw different conclusions on the same item)=$P(X_1 \cap \bar{X_2} \cap Y)+P(\bar{X_1} \cap X_2 \cap Y)+P(X_1 \cap \bar{X_2} \cap \bar{Y})+P(\bar{X_1} \cap X_2 \cap \bar{Y})$

P( two inspectors draw different conclusions on the same item)=0.2402

D)

$P(Y|(X_1 \cap X_2))=\frac{P(Y \cap X_1 \cap X_2)}{P(X_1 \cap X_2)}\\P(Y|(X_1 \cap X_2))=\frac{P(Y \cap X_1 \cap X_2)}{P(X_1 \cap X_2 \cap Y)+P(X_1 \cap X_2 \cap \bar{Y})}\\P(Y|(X_1 \cap X_2))=0.35$