Question

1) Two coins are to be flipped. The first coin will land on heads with probability .6, the second with probability .7. Assume that the results of the flips are independent, and let X equal the total number of heads that result. (a) Find P X = 1, (b) Determine E[X].

Answer 1

Answer:

(a) P(X=1)=0.46

(b) E[X]=1.3

Step-by-step explanation:

(a)

Let A be the event that first coin will land on heads and B be the event that second coin will land on heads.

According to the given information

$P(A)=0.6$

$P(B)=0.7$

$P(A')=1-P(A)=1-0.6=0.4$

$P(B')=1-P(B)=1-0.7=0.3$

P(X=1) is the probability of getting exactly one head.

P(X=1) = P(1st heads and 2nd tails ∪ 1st tails and 2nd heads)

          = P(1st heads and 2nd tails) + P(1st tails and 2nd heads)

Since the two events are disjoint, therefore we get

$P(X=1)=P(A)P(B')+P(A')P(B)$

$P(X=1)=(0.6)(0.3)+(0.4)(0.7)$

$P(X=1)=0.18+0.28$

$P(X=1)=0.46$

Therefore the value of P(X=1) is 0.46.

(b)

Thevalue of E[X] is

$E[X]=\sum_{x}xP(X=x)$

$E[X]=0P(X=0)+1P(X=1)+2P(X=2)$

$E[X]=P(X=1)+2P(X=2)$                      ..... (1)

First we calculate  the value of P(X=2).

P{X = 2} = P(1st heads and 2nd heads)

             = P(1st heads)P(2nd heads)

$P(X=2)=P(A)P(B)$

$P(X=2)=(0.6)(0.7)$

$P(X=2)=0.42$

Substitute P(X=1)=0.46 and P(X=2)=0.42 in equation (1).

$E[X]=0.46+2(0.42)$

$E[X]=1.3$

Therefore the value of E[X] is 1.3.