Question

Mr. Smith is looking for his reading glasses. The glasses could be in the kitchen with probability 0.2, in the living room with probability 0.3, and in the bedroom with probability 0.5. If the glasses are in the kitchen, the probability that Mr. Smith will find them in the kitchen is 0.8. If the glasses are in the living room, the probability that Mr. Smith will find them in the living room is 0.6. If the glasses are in the bedroom, the probability that Mr. Smith will find them in the bedroom is 0.4.
a. Find the probability that the reading glasses were in the kitchen, given that Mr. Smith has found them.
b. What is the probability that Mr. Smith will not find his reading glasses in the bedroom, given that the glasses are in the kitchen?
c. Find the probability that the reading glasses are in the bedroom, given that Mr. Smith did not find them in the bedroom?

Answer 1

Answer:

(a) 0.2963

(b)0.6522

(c) 0.50

Step-by-step explanation:

Let,

K = Glasses are in the kitchen,

L = Glasses are in the living room

B = Glasses are in the bedroom

F = Mr. Smith found the glasses

Given:

$P(K) =0.20\\P(L)=0.30\\P(B)=0.50\\P(F|K)=0.80\\P(F|L)=0.60\\P(F|B)=0.40$

Compute the probability that Mr. Smith found the glasses:

$P(F)=P(F|K)P(K)+P(F|L)P(L)+P(F|B)P(B)\\=(0.80\times0.20)+(0.60\times0.30)+(0.40\times0.50)\\=0.54$

(a)

Determine the probability that the glasses were in the  that they Mr. Smith found the glasses :

Use the conditional probability formula:

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$

The probability, P (K|F) as follows:

$P(K|F)=\frac{P(F|K)P(K)}{P(F)}\\=\frac{0.80\times0.20}{0.54}\\=0.2963$

(b)

Determine the probability that Mr. Smith will not find his reading glasses in the bedroom, given that the glasses are in the kitchen:

$P(B^{c}|F^{c})=\frac{P(F^{c}|B^{c})P(B^{c})}{P(F^{c})} \\=\frac{[1-P(F|B)][1-P(B)]}{1-P(F)} \\=\frac{(1-0.40)(1-0.50)}{(1-0.54)} \\=0.6522$

(c)

Determine the probability that the reading glasses are in the bedroom, given that Mr. Smith did not find them in the bedroom:

$P(B|(F|B)^{c})=\frac{(1-P(F|B))P(B)}{1-P(F|B)} \\=\frac{(1-0.40)\times0.50}{(1-0.40)} \\=0.50$