Question

when two coins are thrown sinultaneously the probability of getting one head and one tail up is​

Answer 1

one head:

Let E1= event of getting at least 1 head. Then,

E1= {HT, TH, HH} and, therefore, n(E1) = 3.

Therefore, P(getting at least 1 head) = P(E1) = n(E1)/n(S) = ¾.

one tail:

Let E2= event of getting at least 1 tail. Then,

E2= {TH, HT, TT} and, therefore, n(E2) = 3.

Therefore, P(getting at least 1 tail) = P(E2) = n(E2)/n(S) = ¾.