Respuesta :
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
[tex]P(A)=0.6[/tex]
[tex]P(B)=0.7[/tex]
[tex]P(A')=1-P(A)=1-0.6=0.4[/tex]
[tex]P(B')=1-P(B)=1-0.7=0.3[/tex]
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
[tex]P(X=1)=P(A)P(B')+P(A')P(B)[/tex]
[tex]P(X=1)=(0.6)(0.3)+(0.4)(0.7)[/tex]
[tex]P(X=1)=0.18+0.28[/tex]
[tex]P(X=1)=0.46[/tex]
Therefore the value of P(X=1) is 0.46.
(b)
Thevalue of E[X] is
[tex]E[X]=\sum_{x}xP(X=x)[/tex]
[tex]E[X]=0P(X=0)+1P(X=1)+2P(X=2)[/tex]
[tex]E[X]=P(X=1)+2P(X=2)[/tex] ..... (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)
[tex]P(X=2)=P(A)P(B)[/tex]
[tex]P(X=2)=(0.6)(0.7)[/tex]
[tex]P(X=2)=0.42[/tex]
Substitute P(X=1)=0.46 and P(X=2)=0.42 in equation (1).
[tex]E[X]=0.46+2(0.42)[/tex]
[tex]E[X]=1.3[/tex]
Therefore the value of E[X] is 1.3.
