Question

Two coins are tossed simultaneously. Find the probability of getting
(i) exactly 1 head (ii) at most 1 head (iii) at least 1 head.

Answer 1

If we say that the coin is fair. Then getting a head or a tail have equal probabilities. So, P(H)=P(T)=1/2

Now, P(at most 1 head)=P(getting 0 heads or 1 head) = P(0 heads) + P(1 head) = P(T)*P(T)+P(H)*P(T)+P(T)*P(H)= 3/4

getting at least 1 head:

Let E₆ = event of getting at least 1 head. Then,
E₆ = {HT, TH, HH} and, therefore, n(E₆) = 3.
Therefore, P(getting at least 1 head) = P(E₆) = n(E₆)/n(S) = ¾.

exactly 1 head
E = event of getting 1 head = {HT , TH} = 2
p(getting 1 head) = 2/4 = 1/2