3. P ( x 》2 )
= P (x = 2) + P ( x = 3 )
= 3C2 (1/2)^2 (1/2) + 3C3 (1/2)^3 (1/2)^0
= 3/8 + 1/8
= 1/2
4. Since only 3 coins were tossed , it is impossible to get 5 heads in this event.
So I would say that the possibility of getting 5 heads is 0
5. P (X = 0)
= 3C0 (1/2)^0 (1/2)^3
= 1/8
