Question

Three groups $\mathrm{A},\mathrm{B},\mathrm{C}$ are competing for positions on the Board of Directors of a company. The probabilities of their winning are $0.5,0.3,0.2$ respectively. If the group $\mathrm{A}$ wins, the probability of introducing a new product is $0.7$ and the corresponding probabilities for groups $\mathrm{B}$ and $\mathrm{C}$ are $0.6$ and $0.5$ respectively. The probability that the new product will be introduced is

• A. $0.18$
• B. $0.35$
• C. $0.10$
• D. $0.63$

Answer 1

The correct option is D

$0.63$

The explanation for the correct answer.

Given: Probability of winning of $\mathrm{A},\mathrm{B},\mathrm{C}$ are $0.5,0.3,0.2$ respectively

If $A$ wins, then the probability of introducing a new product is $0.7$

If $B$ wins, then the probability of introducing a new product is $0.6$

If $C$ wins, then the probability of introducing a new product is $0.5$

Let $X$ be the event of introducing a new product

$P(X/A)=0.7,P(X/B)=0.6,P(X/C)=0.5$

Required probability $=\mathrm{P}\left(\mathrm{A}\right)·\mathrm{P}\left(\mathrm{X}\right|\mathrm{A})+\mathrm{P}(\mathrm{B})·\mathrm{P}(\mathrm{X}\left|\mathrm{B}\right)+\mathrm{P}\left(\mathrm{C}\right)·\mathrm{P}\left(\mathrm{X}\right|\mathrm{C})$

$$\begin{gathered} =0.5(0.7)+0.3(0.6)+0.2(0.5) \\ =0.35+0.18+0.10 \\ =0.63 \end{gathered}$$

Hence option D is the correct answer.