Question

A shop selling electronic items sells smartphones of only three reputed companies A, B, and C because chances of their manufacturing a defective smartphone are only 5%, 4%, and 2% respectively. In his inventory, he has 25% smartphones from company A, 35% smartphones from company B, and 40% smartphones from company C.
A person buys a smartphone from this shop

(ii) What is the probability that this defective smartphone was manufactured by company B?

Hint:

Remember to use the law of total probability when calculating the overall probability of an event with multiple causes. Bayes' theorem helps reverse the conditioning, making it easier to find the probability of the cause given the observed event.

Answer 1

We are asked to find \( P(B|D) \), the probability that the defective smartphone was from company B. Using Bayes' theorem: \[ P(B|D) = \frac{P(D|B) \cdot P(B)}{P(D)} \] We already know: - \( P(D|B) = 0.04 \)
- \( P(B) = 0.35 \)
- \( P(D) = 0.0345 \)
Substitute the values: \[ P(B|D) = \frac{0.04 \times 0.35}{0.0345} \] \[ P(B|D) = \frac{0.014}{0.0345} \] \[ P(B|D) \approx 0.4058 \] Thus, the probability that the defective smartphone was manufactured by company B is approximately 40.58%.