Question

inspector a visually inspected 1000 ceramic mugs and found flaws in 36 of them. inspector b inspected the same 1000 ceramic mugs and found flaws in 42 of them. a total of 950 mugs were found to be good by both inspectors. one of the 1000 mugs is selected randomly. (a) find the probability that a flaw was found in this mug by at least one of the two inspectors. (b) find the probability that flaws were found in this mug by both inspectors. (c) find the probability that a flaw was found by inspector b but not by inspector a.

Answer 1

Answer:

a = 0.05

b = 0.028

c =0.014

Step-by-step explanation:

Given that;

inspector A visually inspected 1000 ceramic mugs and found flaws in 36 of them.

Let X represent Inspector A

P(X) = $\frac{36}{1000}$

= 0.036

inspector B visually inspected 1000 ceramic mugs and found flaws in 42 of them.

Let Y represent Inspector B

P(Y) = $\frac{42}{1000}$

= 0.042

A total of 950 mugs were found to be good by both inspectors.

i.e (X∩Y)' = 950

P (X∩Y)'  = $\frac{950}{1000}$

= 0.095

We are tasked to find :

a) the probability that a flaw was found in this mug by at least one of the two inspectors [i.e P (X∪Y)]

b) the probability that flaws were found in this mug by both inspectors.

[i.e P ( X∩Y)]

c) the probability that a flaw was found by inspector B but not by inspector A. [i.e P (X) - P ( X∩Y)]

With a);

P (X∪Y) = 1 - P(A∩B)'

P (X∪Y) = 1 - 0.950

P (X∪Y) = 0.05

b)

P(X∩Y) = P(X) + P(Y) - P(X∪Y)

P(X∩Y) = 0.036 + 0.042 + 0.05

P(X∩Y) = 0.028

c)

P (X) - P ( X∩Y) = 0.042 - 0.028

P (X) - P ( X∩Y) = 0.014

Therefore;

the probability that a flaw was found in this mug by at least one of the two inspectors = 0.05

the probability that flaws were found in this mug by both inspectors = 0.028;      &

the probability that a flaw was found by inspector B but not by inspector A = 0.014