Question

Two groups are competing for the position on the board of directors of a corporation. The probability 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 introduce was by the second group.

Answer 1

Let $E_1$ the event that the first group wins and $E_2$ the event that the second group wins.

Then, $E_1$ and $E_2$ are mutually exclusive and exhaustive events.

$therefore$ $P\left(E_1\right)=0.6=\frac{6}{10}$ and $P\left(E_2\right)=0.4=\frac{4}{10}$.

Let E: the event that the new product is introduced.

$P\left(\frac{E_1}{E}\right)$ = P (introducing a new product, if the first group wins)

=0.7=$\frac{6}{10}$

$therefore$ $P\left(\frac{E_2}{E}\right)$ = P (introducing a new product, if the second group wins)

=0.3=$\frac{3}{10}$

The probability that the new product is introduced by the second group is given by $P\left(\frac{E_2}{E}\right)$

By using Baye's theorem, we obtain

$P\frac{E_2}{E}=\frac{P\left(\frac{E}{E_2}\right)P\left(E_2\right)}{P\left(\frac{E}{E_1}\right)P\left(E_1\right)+P\left(\frac{E}{E_2}\right)P\left(E_2\right)}=\frac{\frac{3}{10}\times \frac{4}{10}}{\frac{7}{10}\times \frac{6}{10}+\frac{3}{10}\times \frac{4}{10}}=\frac{12}{42+12}\frac{12}{54}=\frac{2}{9}$