Question

A fair coin is tossed twice. Given that the first toss resulted in head, then the probability that the second toss also, would result in head is

• A. \(\frac{1}{8}\)
• B. \(\frac{1}{4}\)
• C. \(\frac{3}{8}\)
• D. \(\frac{1}{2}\)
• E. \(\frac{5}{8}\)

Answer 1

The Correct Option is D

We are tossing a fair coin twice. Given that the first toss resulted in head, the outcome of the second toss is independent of the first toss. The probability of getting heads on any fair coin toss is \( \frac{1}{2} \). Therefore, the probability that the second toss will also result in heads is simply \( \frac{1}{2} \).

So, the correct option is (D) : \(\frac{1}{2}\)

Answer 2

We are tossing a fair coin twice. Let H denote heads and T denote tails.

The possible outcomes are {HH, HT, TH, TT}. Since the coin is fair, each outcome has a probability of \(\frac{1}{4}\).

We are given that the first toss resulted in heads. This means that the possible outcomes are now reduced to {HH, HT}.

We want to find the probability that the second toss also results in heads, given that the first toss resulted in heads. In other words, we want to find \(P(\text{second toss is H} | \text{first toss is H})\).

Using the definition of conditional probability, we have:

\(P(\text{second toss is H} | \text{first toss is H}) = \frac{P(\text{both tosses are H})}{P(\text{first toss is H})}\)

The event "both tosses are H" is HH, which has probability \(\frac{1}{4}\).

The event "first toss is H" is {HH, HT}, which has probability \(\frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}\).

Therefore:

\(P(\text{second toss is H} | \text{first toss is H}) = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{4} \cdot \frac{2}{1} = \frac{2}{4} = \frac{1}{2}\)