SOLUTION: A bag contains three coins, one of which has a head on both sides while the other two coins are normal. A coin is chosen at random from the bag and tossed three times. The number o

Algebra ->  Probability-and-statistics -> SOLUTION: A bag contains three coins, one of which has a head on both sides while the other two coins are normal. A coin is chosen at random from the bag and tossed three times. The number o      Log On


   

Question 1203802: A bag contains three coins, one of which has a head on both sides while the other two coins are normal. A coin is chosen at random from the bag and tossed three times. The number of heads is a random variable, say X.
(a) Find the discrete pdf of X. (Hint: Use the Law of Total Probability with B1 = a normal coin and B2 = two-headed coin.)
(b) Sketch the discrete pdf and the CDF of X.
Answer by Edwin McCravy(20066) About Me  (Show Source):
You can put this solution on YOUR website!
P(B1,H,H,H) = (1/2)(1/2)(1/2)(1/2) = 1/16, n(H) = 3
P(B1,H,H,T) = (1/2)(1/2)(1/2)(1/2) = 1/16, n(H) = 2
P(B1,H,T,H) = (1/2)(1/2)(1/2)(1/2) = 1/16, n(H) = 2
P(B1,H,T,T) = (1/2)(1/2)(1/2)(1/2) = 1/16, n(H) = 1
P(B1,T,H,H) = (1/2)(1/2)(1/2)(1/2) = 1/16, n(H) = 2
P(B1,T,H,T) = (1/2)(1/2)(1/2)(1/2) = 1/16, n(H) = 1
P(B1,T,T,H) = (1/2)(1/2)(1/2)(1/2) = 1/16, n(H) = 1
P(B1,T,T,T) = (1/2)(1/2)(1/2)(1/2) = 1/16, n(H) = 0
P(B2,H,H,H) = (1/2)( 1 )( 1 )( 1 ) =  1/2, n(H) = 3

X(0) = P(0) = 1/16     = 1/16
X(1) = P(1) = 3/16     = 3/16
X(2) = P(2) = 3/16     = 3/16
X(3) = P(3) = 1/16+1/2 = 9/16 
--------------------------
           Sum = 16/16 = 1

n |  X 
0 | 1/16
1 | 3/16
2 | 3/16
3 | 9/16


     

Edwin