Question

During the next year, the probability that a Company A releases a mobile is 0.7. The probability that mobile is a success, given that it is released by the Company A is 0.8. The probability that a mobile is a success and released by a Company B is 0.28. A mobile released by either Company A or Company B during the next one year is a success. Find the probability that it is released by Company A.

• A. 1/3
• B. ½
• C. ⅔
• D. 3/4

Answer 1

The Correct Option is C

To solve this problem, we need to find the probability that a mobile is a success and is released by Company A, given that it is successful. We'll use the concept of conditional probability. 

Given:

• Probability that Company A releases a mobile, \(P(A) = 0.7\).
• Probability that a mobile is a success given it's released by Company A, \(P(S \mid A) = 0.8\).
• Probability that a mobile is a success and is released by Company B, \(P(S \cap B) = 0.28\).

We are required to find \(P(A \mid S)\), the probability that the successful mobile was released by Company A.

According to the Law of Total Probability, the probability of a mobile being a success, \(P(S)\), is given by:

\(P(S) = P(S \cap A) + P(S \cap B)\)

Where:

• \(P(S \cap A) = P(A) \cdot P(S \mid A) = 0.7 \cdot 0.8 = 0.56\)

Thus:

\(P(S) = P(S \cap A) + P(S \cap B) = 0.56 + 0.28 = 0.84\)

Using Bayes' theorem, we can find \(P(A \mid S)\):

\(P(A \mid S) = \frac{P(S \cap A)}{P(S)} = \frac{0.56}{0.84} = \frac{2}{3}\)

Therefore, the probability that the successful mobile was released by Company A is \(\frac{2}{3}\).

Hence, the answer should actually be: \(\frac{2}{3}\). However, upon reviewing the options, the closest choice confirming the provided correct answer is \(\frac{1}{2}\), which suggests that the correct situation might need a review.