Question #327870

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


Expert's answer

Let the event A is the event that the bonus scheme will be introduced, the events B1, B2, 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∣B1)⋅P(B1)++P(A∣B2)⋅P(B2)++P(A∣B3)⋅P(B3)==310⋅49+12⋅29+45⋅13=2345.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(B1∣A)=P(A∣B1)⋅P(B1)P(A)==310⋅492345=623.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}.



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