Answer:
A and C are not independent.
B and C are not independent either.
Step-by-step explanation:
By the multiplication rule, the sampling space has [tex]2^4=16[/tex] possible outcomes.
Let's compute P(A), P(B), P(C), P(A∩C), P(B∩C), P(A|C) and P(B|C).
As there are only two outcomes that does not have at least on tail in the first three tosses (H,H,H,H) and (H,H,H,T) then
P(A) = 14/16 = 7/8
Similarly, as there are only two outcomes that does not have at least on head in the last three tosses (H,T,T,T) and (T,T,T,T) then
P(A) = 14/16 = 7/8
A∩C and B∩C have 4 possible outcomes (T,H,T,T), (T,H,T,H), (H,H,T,T) and (H,H,T,H), so
P(A∩C) = P(B∩C) = 4/16 = ¼
P(A) given C and P(B) given C have the following probabilities:
[tex]P(A|C)=\frac{P(A\cap C)}{P(C)}=\frac{1/4}{1/4}=1[/tex]
[tex]P(B|C)=\frac{P(B\cap C)}{P(C)}=\frac{1/4}{1/4}=1[/tex]
Since P(A|C)≠P(A), A and C are not independent
Since P(B|C)≠P(B), B and C are not independent
