Pr(E)=(2/3)x(1/2) + (1/3)x(2/5)= 14/30.
I.
We know that Pr(A)=2Pr(B), and Pr(B)=3Pr(C); hence, Pr(A)=6Pr(C).
But Pr(A)+Pr(B)+Pr(C)=1. Consequently, 6Pr(C)+3Pr(C)+Pr(C)=1. So,
Pr(C)=1/10; Pr(B)=3/10; Pr(A)=6/10.
II.
We know that Pr(1)+Pr(2)+Pr(3)+Pr(4)+Pr(5)=Pr(6)=1; hence, Pr(1)+2Pr(1)+3Pr(1)+4Pr(1)+5Pr(1)+6Pr(1)=1.
So, Pr(1)=1/21. Consequently, Pr(6)=6/21.
III.
Let's construct the tree:
Pr(E)=(2/3)x(1/2) + (1/3)x(2/5)= 14/30.
IV.
Let's construct the tree, where "vaccine A is successful" is As and "vaccine B is successful" is Bs:
Pr(Joe is successfully vaccinated)=Pr(As v Bs)=1/5
+ (4/5)x(2/5) = 13/25.
Another approach is:
Pr(Joe is successfully vaccinated)=Pr(As v Bs)=
1-Pr(-As & -Bs)= 1-[(4/5)x(3/5)]= 13/25
V.
Let's construct the tree:
Pr(R)= (1/3)x(2/5) + (2/3)x(4/10) = 6/15.
Pr(W)= (1/3)x(3/5) + (2/3)x(2/10) = 1/3.
Pr(B)= (2/3)x(4/10) = 4/15.
VI.
Let's construct the tree:
Pr(H)= (1/3)x(1/2)+(1/3)+(1/3)x(1/3) = 11/18.
Pr(T)= 1-Pr(H) = 1-(11/18) = 7/18.
VII.
Let's construct the tree:
1. Pr(H on second flip) = (1/4)x(1/2)x(3/4) + (1/4)x(1/2)x(1/2) + (3/4)x(3/4)x(1/2)
+ (3/4)x(1/4)x(3/4) = 37/64.
2. Pr(T on first flip & T on second flip)= (1/4)x(1/2)x(1/2) +
(3/4)x(1/4)x(1/4) = 7/64.
3. Pr(H on first flip & H on second flip) = (1/4)x(1/2)x(3/4) +
(3/4)x(3/4)x(1/2) = 3/8.