Answer to Question #327870 in Statistics and Probability for Keke

Question

The probability of x,y and z becoming managers are 4/9 ,2/9 and 1/3 respectively.The probabilities that the bonus scheme will be introduced if x,y and z becomes managers are 3/10 ,1/2 ,and 4/5 respectively

(a)what is the probability that the bonus scheme will be introduced?

(b)if the bonus scheme has been introduced,what is the probability that the manager appointed was x

Answer 1

Let the event A is the event that the bonus scheme will be introduced, the events B_1, B_{2, }B₃ are the events that x,y and z will become a manager respectively.

(a) Using the law of total probability we can calculate the probability P(A):

$P(A) =P(A|B_1) \cdot P(B_1)+\\ +P(A|B_2) \cdot P(B_2)+\\+P(A|B_3) \cdot P(B_3)=\\ =\cfrac{3} {10} \cdot\cfrac{4} {9}+\cfrac{1} {2} \cdot\cfrac{2} {9}+\cfrac{4} {5} \cdot\cfrac{1} {3}=\cfrac{23}{45}.$

(b) We use Bayes theorem:

$P(B_1|A) =\cfrac{P(A|B_1)\cdot P(B_1) }{P(A)}=\\ =\cfrac{\cfrac{3} {10} \cdot\cfrac{4}{9}} {\cfrac{23}{45}} =\cfrac{6}{23}.$