Answer:
P[X=3,Y=3] = 0.0416
Step-by-step explanation:
Solution:
- X is the RV denoting the no. of customers in line.
- Y is the sum of Customers C.
- Where no. of Customers C's to be summed is equal to the X value.
- Since both events are independent we have:
P[X=3,Y=3] = P[X=3]*P[Y=3/X=3]
P[X=3].P[Y=3/X=3] = P[X=3]*P[C1+C2+C3=3/X=3]
P[X=3]*P[C1+C2+C3=3/X=3] = P[X=3]*P[C1=1,C2=1,C3=1]
P[X=3]*P[C1=1,C2=1,C3=1] = P[X=3]*(P[C=1]^3)
- Thus, we have:
P[X=3,Y=3] = P[X=3]*(P[C=1]^3) = 0.25*(0.55)^3
P[X=3,Y=3] = 0.0416
×
