Browse Questions

Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.

Toolbox:
• Given $E_1, E_2, E_3.....E_n$ are mutually exclusive and exhaustive events, we can find the conditional probability $P(E_i|A)$ for any event A associated w/ $E_i$ using the Bayes theorem as follows: $\;P(E_i/A)$=$\large \frac{P(E_i)P(A/E_i)}{\sum_{i=1}^{n}\;P(E_i)P(A/E_i)}$
Let $E_1$ be the event that the first group wins. P($E_1$) = 0.6
Let $E_2$ be the event that the second group wins. P($E_2$) = 0.4
Let A be the event where a new product is introduced.
P (introducing a new product if the first group wins) = P (A|$E_1$) = 0.7
P (introducing a new product if the second group wins) = P (A|$E_2$) = 0.3
We need to find the probability that the new product was introduced by the second group, i.e,, P($E_2$|A).
We can use Baye's theorem, according to which $P(E_2|A) = \large\frac{P(E_2)(P(A|E_2)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$
P ($E_2$|A) = $\large\frac{0.3 \times 0.4}{0.3 \times 0.4 + 0.7 \times 0.6}$ = $\large\frac{0.012}{0.012+0.042} = \frac{0.012}{0.054} = \frac{2}{9}$
edited Jun 19, 2013