Question

Two coins are tossed simultaneously. Find the probability of getting at least
one head.​

Answer 1

Answer:

(vi) getting at least 1 head:

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

Answer 2

Answer:

The probability is 3/4

Step-by-step explanation:

Let S be the sample space.

Then S= {HH, HT, TH, TT).

n(S) = 4.

Let A be the event of getting at least one head.

Then A={HH, HT, TH}

n(A) = 3.

P(A) = n(A) /n(S)

Ans. The probability is 3/4