Two successive tosses of a fair coin. Find the probabilities of (1) exactly one head and (2) at least one head (express as ordered pair).

Introduction / Context:
With two independent fair coin tosses, we compute probabilities for two events: exactly one head, and at least one head.

Given Data / Assumptions:

• Outcomes: HH, HT, TH, TT with equal probability 1/4 each.
• Event A: exactly one head (HT or TH).
• Event B: at least one head (HH, HT, TH).

Concept / Approach:
Count favorable outcomes for each event and divide by 4 (total outcomes). Use complement for “at least one head” if desired.

Step-by-Step Solution:
P(exactly one head) = 2 / 4 = 1/2.P(at least one head) = 1 − P(TT) = 1 − 1/4 = 3/4.

Verification / Alternative check:
Direct counting confirms 2 favorable for A and 3 favorable for B.

Why Other Options Are Wrong:
Other ordered pairs do not match the correct counts; 2/3 or 4/5 are not attainable with a four-element uniform sample space in this context.

Common Pitfalls:
Including HH in “exactly one head” or excluding it from “at least one head”.

Final Answer:
1/2 , 3/4