Question:

The chance of three persons Mr X, Y and Z becoming manager of a company are 4:2:3. The probability that bonus scheme shall be introduced if X, Y and Z become managers are 0.3,0.5,0.8 respectively. Find the probability that the bonus scheme will be introduced. What is the probability that X is appointed as a manager?

Answer:

P(S) = Probability of bonus being introduced.

P(X) = Probability of Mr X being Manager.

P(Y) = Probability of Mr Y being Manager.

P(Z) = Probability of Mr Z being Manager.

P(S/X) = Probability of scheme being introduced when X becomes manager.

P(S/Y) = Probability of scheme being introduced when Y becomes manager.

P(S/Z) = Probability of scheme being introduced when Z becomes manager.

Probability of bonus scheme being introduced

P(S) = P(X).P(S/X) + P(Y).P(S/Y) + P(Z).P(S/Z) = 4(0.3) + 2(0.5) + 3(0.8) = 4.6

Probability that X is appointed as manager

P(S/X) =

= 0.26

Is this Right or Wrong solution......Please let me now ........