Solution for Section 6.1 Question 1a

1a) i) Let H represent a coin showing heads and T represent a coin showing tails. Let E be the event that you win.

Then your sample space is S = {HH, HT, TH, TT}, and there is only one way to win, so the set E contains one element (either HH or TT).
Therefore
P(E) = n(E) / n(S) = 1/4.


ii) In this scenario S = {HH,  TT} as TH and HT are ignored and E = {HH}. Hence 
P(E) = n(E) / n(S) = 1/2.

Back to Section 6.1