Question

Find the mean of number of heads in three tosses of a fair coin.

Hint:

For any binomial experiment (a fixed number of independent trials with two outcomes), the mean is simply the number of trials times the probability of success. It's a quick and powerful shortcut.

Answer 1

This is an example of a binomial distribution.
Let \(X\) be the random variable representing the number of heads.
The number of trials (tosses), \(n = 3\).
The probability of success (getting a head) in a single trial, \(p = \frac{1}{2}\).
The mean, or expected value \(E(X)\), of a binomial distribution is given by the formula \(E(X) = np\).
Substituting the values: \[ E(X) = 3 \times \frac{1}{2} = 1.5 \] The mean number of heads in three tosses of a fair coin is 1.5.
Alternatively, we can list the sample space and calculate the mean directly:
The possible outcomes are: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
Number of heads (X) can be 0, 1, 2, or 3.
P(X=0) [TTT] = 1/8
P(X=1) [HTT, THT, TTH] = 3/8
P(X=2) [HHT, HTH, THH] = 3/8
P(X=3) [HHH] = 1/8
The mean is \(E(X) = \sum x \cdot P(X=x)\): \[ E(X) = (0 \times \frac{1}{8}) + (1 \times \frac{3}{8}) + (2 \times \frac{3}{8}) + (3 \times \frac{1}{8}) \] \[ E(X) = \frac{0 + 3 + 6 + 3}{8} = \frac{12}{8} = 1.5 \]