Please have the answers step by step.

A1
a) Total number of ways in which 5 cards can be selected without replacement are:
52C5
At most there can be 2 red queens in 2 cards .
Probability that there is atleast one red queen is P(One red queen) +P(2 red queens)
P(One red queen ) = 1772.0
552
450*2

C
C

P(2 red queens) = 0075.
552
350

C
C

Thus the required probability is 0.1847
b) Total number of ways in which 5 cards can be selected with replacement is 52^5.
Here we may have all 5 cards to be one of the red queens as it is with replacement.
Thus the required probability =1-P(none of the cards is a red queen )
P(None of the 5 cards is a red queen) =(50/52)^5 =0.8219
Thus the required probability is 1-0.8219 =0.1781
A2)
Let X be the random variable denoting the number of customers arriving at the check out .
The X~Poi(7)
a) We need P(X=0)+P(X=1)+P(X=2)+P(X=3)
Now P(X=k) =
!
7*)7exp(
k
k

Therefore, P(X=0) =.000911
P(X=1)=.00638 ,P(X=2)=.02234, P(X=3).0522
Thus the required probability is : 0.0817
b)We need P(X>=2)
=1-(P(X=0) +P(X=1))
=(1-(.00091+.00638))
=0.9927
C)We need P(X=5)
=...