Ask Questions, Get Answers

Home  >>  CBSE XII  >>  Math  >>  Probability

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.

Download clay6 mobile app

1 Answer

  • 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} $ = $\large\frac{0.012}{0.012+0.042} = \frac{0.012}{0.054} = \frac{2}{9}$
answered Mar 8, 2013 by poojasapani_1
edited Jun 19, 2013 by balaji.thirumalai

Related questions