Question

In a certain small town, 3 professional burglars are currently out of prison: Alex, Becky, and Carl. Alex has in the past committed 55% of the burglaries committed by the three, Becky 31%, and Carl the rest. But only 1/3 of Alex’s jobs are burglaries of a residence, while half of Becky’s are, and all of Carl’s are.
a) What is the probability that the next burglary in town (if one of the three did it) is the burglary of a residence?

b)Sure enough, a resident reports a home burglary. If one of the three did it, what is the probability Becky was guilty?

Answer 1

Answer:

(a) The probability that the next burglary in town (if one of the three did it) is the burglary of a residence is 0.4783.

(b) The probability that a burglary of a residence is committed by Becky is 0.3241.

Step-by-step explanation:

Let's denote the events as follows:

A = a burglary was committed by Alex.

B = a burglary was committed by Becky.

C = a burglary was committed by Carl.

R = a burglary is a residence burglary.

Given:

P (A) = 0.55,

P (B) = 0.31,

P (C) = 1 - P (A) - P (B) = 1 - 0.55 - 0.31 = 0.14.

P (R ∩ A) = $0.55\times\frac{1}{3}=0.183$

P (R ∩ B) = $0.31\times\frac{1}{2}=0.155$

P (R ∩ C) = $0.14\times1=0.14$

(a)

The total probability formula is:

$P(X)=P(X\cap Y)+P(X\cap Z)$

Compute the burglary of a residence as follows:

$P(R)=P(R\cap A)+P(R\cap B)+P(R\cap C)\\=0.1833+0.155+0.14\\=0.4783$

Thus, the probability that the next burglary in town (if one of the three did it) is the burglary of a residence is 0.4783.

(b)

Compute the probability that a burglary of a residence is committed by Becky as follows:

$P(B|R)=\frac{P(R\cap B)}{P(R)}=\frac{0.155}{0.4783}=0.321$

Thus, the probability that a burglary of a residence is committed by Becky is 0.3241.