Please have the answers step by step.

Question

Robert · Accepted Answer

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) 
  = 1277.

Please have the answers step by step.

Solution

Answer To This Question Is Available To Download

Related Questions & Answers

Submit New Assignment