Question

For a fair coin toss, what is the probability of getting at least one head in two tosses?

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $\frac{3}{4}$

Answer 1

Answer: (C)

$\frac{3}{4}$

Answer 2

The problem asks for the probability of getting at least one head in two tosses of a fair coin.

1. Determine the Sample Space: When a fair coin is tossed two times, the possible outcomes are:

• HH (Head on first toss, Head on second toss)
• HT (Head on first toss, Tail on second toss)
• TH (Tail on first toss, Head on second toss)
• TT (Tail on first toss, Tail on second toss)

The total number of possible outcomes is 4.

2. Identify Favorable Outcomes: We are interested in the event of getting "at least one head". This means the outcome must contain one or more heads. The favorable outcomes are:

• HH (contains two heads)
• HT (contains one head)
• TH (contains one head)

The number of favorable outcomes is 3.

3. Calculate the Probability: The probability of an event is given by the formula: $P(\text{Event}) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}$ In this case, the probability of getting at least one head is: $P(\text{at least one head}) = \frac{3}{4}$

Alternative Method (Using Complementary Event): The complement of "at least one head" is "no heads" (i.e., all tails). The only outcome with no heads is TT. The probability of getting no heads (TT) is: $P(\text{no heads}) = \frac{1}{4}$ The probability of "at least one head" is 1 minus the probability of "no heads": $P(\text{at least one head}) = 1 - P(\text{no heads})$ $P(\text{at least one head}) = 1 - \frac{1}{4}$ $P(\text{at least one head}) = \frac{4-1}{4} = \frac{3}{4}$

Both methods yield the same result.