Question

Three companies X, Y and Z supply computers to a university. The percentage of computers supplied by them and the probability of those being defective are tabulated below

Company	% of computers supplied	probability of being defective
X	60%	0.01
Y	30%	0.02
Z	10%	0.03

Give that a computer is defective, the probability that it was supplied by Y is

• A. 0.1
• B. 0.2
• C. 0.3
• D. 0.4

Answer 1

The correct option is D 0.4

Let S and D be the events supply by y and defective respectively.

$\therefore RequiredProbability=P\left(S/D\right)$

$=\frac{P(S\cap D)}{P\left(D\right)}$

$P(S\cap D)=P\left(S\right)\times P\left(D\right)=0.3\times 0.02=0.006$
By total probability,
$P\left(D\right)=0.6\times 0.01+0.3\times 0.02+0.1\times 0.03$
= 0.015

$P\left(S/D\right)=\frac{0.006}{0.015}=0.4$

Note: We can also solve above equation using Baye's theorem.