Question

Class 12th CBSE board Probability

Two groups are competing for the position on boards of directors of a corporation .
the probability that the first group and the second group will win are 0.6 and 0.4 respectively.
father is 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 bye second group .​

Answer 1

This is the question of Baye's Theorem

Solution ::-)

When there is a competition between the groups for getting position in boards of directors of a corporation.

Let E1 and E2 be the events that first and second group wins respectively.

So,

P(E1) = 0.6

P(E1) = 0.6P(E2) = 0.4

Let A be the event that a new product is introduced.

So,

$P (\dfrac{A}{E_1} ) = 0.7 \\ \\ P( \dfrac{A}{E_2} ) = 0.3$

Hence probability that a new product is introduced by second group

i.e.

$P( \dfrac{E_1}{A} ) = \dfrac{P(E_2).P( \dfrac{A}{E_2}) }{P(E_1).P( \dfrac{A}{E_1}) +P(E_2).P( \dfrac{A}{E_2})} \\ \\ = \frac{0.4 \times 0.3}{0.6 \times 0.7 + 0.4 \times 0.3} \\ \\ = \frac{0.12}{0.42 + 0.12} \\ \\ = \frac{0.12}{0.54} \\ \\ = \frac{12}{54} \\ \\ = \frac{6}{27} \\ \\ = \frac{2}{9}$

Which is the required probability.

Answer 2

Let E1 and E2 be the events that first and second group wins respectively.

So,

P(E1) = 0.6

P(E1) = 0.6P(E2) = 0.4

Let A be the event that a new product is introduced.

So,

$P (\dfrac{A}{E_1} ) = 0.7 \\ P( \dfrac{A}{E_2} ) = 0.3$

Reqrd Probability is

$P( \dfrac{E_1}{A} ) = \dfrac{P(E_2).P( \dfrac{A}{E_2}) }{P(E_1).P( \dfrac{A}{E_1}) +P(E_2).P( \dfrac{A}{E_2})} \\ = \frac{0.4 \times 0.3}{0.6 \times 0.7 + 0.4 \times 0.3} \\ = \frac{0.12}{0.42 + 0.12} \\ \\ = \frac{0.12}{0.54} \\ = \frac{12}{54} \\ \\ = \frac{6}{27} \\ = \frac{2}{9}$