SOLUTION: Suppose that you flip a "strange" coin twice, where the probability of two heads and two tails are the same (P(HH)=P(TT)) and the probability of one head and one tail is the same

Question 1032915: Suppose that you flip a "strange" coin twice, where the probability of two heads and two tails are the same (P(HH)=P(TT)) and the probability of one head and one tail is the same in either order (P(HT)=P(TH)) but the probability of HH is 7 times the probability of HT .
a) Find the probability of two heads.
P(HH)=
b) Let A be the event that at least one of the coin flips is a head. What is P(A) ?
P(A)=
c) Let B be the event that the two flips match. What is P(B) ?
P(B)=
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
P(HH) = P(TT) = x
P(HT) = P(TH) = x/7 (since P(HH) = 7*P(HT))

Since HH, TT, HT, TH are disjoint and one of these outcomes must occur, the sum of their probabilities is 1. Hence x + x + (x/7) + (x/7) = 1, or x = 7/16.

a) P(HH) = x = 7/16
b) P(A) = 1 - P(TT) = 9/16
c) P(B) = P(HH) + P(TT) = 7/8
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