Question

Suppose we toss a fair coin three times in a row. Let A be the event of exactly 2 tails. Let B be the event that the first 2 tosses are tails. Let C be the event that all three tosses are tails. What is the probability of the union of A, B, and C

Answer 1

Answer: 1/2

Explanation:

Since it's a fair coin, the probability of tossing head, P(H) = probability of tossing tail, P(T) = 1/2.

Hence, for the event of A being exactly 2 tails, then we have a possibility of:

A= [ HTT, THT, TTH]

For the event of B being first two tosses resulting in tail, it becomes:

B= [TTH)

For event of C being that the three tosses are tail, it becomes:

C=[TTT]

Hence, union of Event A, B and C is given as:

AuBuC= [HTT, THT, TTH, TTT]

Prob [AuBuC] = [ (1/2 * 1/2 * 1/2)] * 4

P[AuBuC] = (1/8) *4

P(AuBuC) = 1/2.