Question

. A cell phone company has three different production sites. Five percent of the cars from Site 1, 7% from Site 2, and 9% from Site 3 have been recalled due to unexpected shutdown issue. Suppose that 60% of the phones are produced at Site 1, 30% at Site 2, and 10% at Site 3. If a randomly selected cell phone has been recalled, what is the probability that it came from Site 3 (write it up to second decimal place)

Answer 1

Answer:

0.15 = 15% probability that it came from Site 3

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

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

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Being recalled

Event B: Recalled from site 3.

Probability of being recalled:

5% of 60%(from site 1)

7% of 30%(from site 2)

9% of 10%(from site 3).

So

$P(A) = 0.05*0.6 + 0.07*0.3 + 0.09*0.1 = 0.06$

Probability of being recalled, being from site 3.

9% of 10%.

$P(A \cap B) = 0.09*0.1 = 0.009$

If a randomly selected cell phone has been recalled, what is the probability that it came from Site 3?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.009}{0.06} = 0.15$

0.15 = 15% probability that it came from Site 3