A long-suffering manager has two employees. On any day, Employee A has a 80% cha
ID: 3275264 • Letter: A
Question
A long-suffering manager has two employees. On any day, Employee A has a 80% chance of showing up for work, while Employee B has a 90% chance of showing up for work. The probability that both Employee A and Employee B will show up is 75%. a. Are the events "Employee A will show for work" and "Employee B will show for work" independent? Justify your answer using mathematical reasoning. b. Knowing that only one employee showed for work in a given day, what is the probability that that Employee A showed for work on that day? c. Give the probability distribution for X = number of employees that show up for work in a day.Explanation / Answer
a) P(A) = 0.8 is the probability that employee A show for work
P(B) = 0.9 is the probability that employee B show for work
P(A and B) = 0.75 is the probability that both employees A and B show for work
Now P(A)P(B) = 0.8*0.9 = 0.72 which is not equal to P(A and B)
Therefore A and B are not independent events here.
b) Given that only one employee showed for work , we have to find the probability that it was A that showed for work.
P(A only ) = P(A) - P(A and B) = 0.8 - 0.75 = 0.05
P(B only ) = P(B) - P(A and B) = 0.9 - 0.75 = 0.15
Therefore the required probability that Given that only one employee showed for work , the probability that it was A that showed for work is computed as:
= P(A only) / [ P(A only ) + P(B only) ]
= 0.05/ (0.05 + 0.15)
= 0.05 / 0.2
= 0.25
Therefore 0.25 is the required probability here.
b) X = number of employees that show up for work in a day
P(X = 2) = P(A and B) = 0.75
P(X = 1) = P(A only ) + P(B only ) = 0.05 + 0.15 = 0.2
Therefore, P(X = 0) = 1 - P(X=1) - P(X=2) = 1 - 0.2 - 0.75 = 0.05
Therefore the required probability distribution here is:
P(X=0) = 0.05, P(X=1) = 0.2, P(X=2) = 0.75
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