Conditional Probability - Cell Phone Problem

Question

The screens used for a certain type of cell phone are manufactured by 3 companies, A, B. and C. The proportions of screens supplied by A. B. and C are 0.5, 0.3, and 0.2. respectively, and their screens are defective with probabilities 0.01, 0.02, and 0.03, respectively. Given that the screen on such a phone is defective, what is the probability that Company A manufactured it?

Would it be that:

P(A) = 1/2 because 0.5 of the screens are made by Company A

and

P(B) = 1/100 because its defective 0.01 of the time?

Answer 1

Hint. The probability that a screen is defective is $$P(D)=0.5\cdot 0.01+0.3\cdot 0.02+0.2\cdot 0.03=0.017.$$ The probability that a screen is made by Company A and it defective is $$P(A\cap D)=0.5\cdot 0.01=0.005.$$

Are you able to determine $P(A|D)$?

Answer 2

Formulating this problem in terms of Bayes' Theorem:

$$p(A\vert D) = p(D\vert A)\dfrac{p(A)}{p(D)}$$

You will need to decompose the probability of D as $$\sum_{i \in \{ A,B,C \}} p(D \vert i.)$$

Then, you know all the appropriate probabilities.

Answer 3

This problem is problem 8 from chapter 2 exercises of Introduction to Probability Joseph K. Blitzstein and Jessica Hwang

Let

A = Event screen supplied/manufactured by company A
B = Event screen supplied/manufactured by company B
C = Event screen supplied/manufactured by company C
D = Event screen on phone is defective

Given

$P(A)=0.5$
$P(B)=0.3$
$P(C)=0.2$

$P(D\vert A)=0.01$
$P(D\vert B)=0.02$
$P(D\vert C)=0.03$

Question is asking you to determine the following probability

$P(A\vert D) =\space ?$

Solving using Bayes theorem and LOTP

$P(A\vert D) = \dfrac{P(A,D)}{P(D)} = \dfrac{P(D\vert A)P(A)}{P(D)}$

Both $P(D\vert A)=0.01$ and $P(A)=0.5$ are given so need to find $P(D)$ using LOTP

$P(D) =\sum_{i \in \{ A,B,C \}} P(D \vert i.)P(i) = P(D\vert A)P(A) + P(D\vert B)P(B) + P(D\vert C)P(C)$

Therefore

$P(A\vert D) = \dfrac{0.01 * 0.5}{0.01 * 0.5 + 0.02 * 0.3 + 0.03 *0.2} = \dfrac{5}{17} \approx 29.4\% $