Find the probability distribution of number of heads in 4 tosses of fair coin

Question

A fair coin is tossed 4 times. Find the probability distribution of $X$ , where $X$ is the number of heads obtained. Also find the mean and variance.

(NCERT Class 12, Chapter 13 — Probability)

Solution — Step by Step

Each toss is a Bernoulli trial with $p = 1/2$ (probability of head) and $q = 1/2$ . With $n = 4$ trials, $X$ follows a Binomial distribution: $X \sim B(4, 1/2)$ .

P(X = k) = \binom{4}{k} \left(\frac{1}{2}\right)^k \left(\frac{1}{2}\right)^{4-k} = \binom{4}{k} \cdot \frac{1}{16}

$X$ (heads)	$\binom{4}{k}$	$P(X = k)$
0	1	$1/16$
1	4	$4/16 = 1/4$
2	6	$6/16 = 3/8$
3	4	$4/16 = 1/4$
4	1	$1/16$

Verification: $1 + 4 + 6 + 4 + 1 = 16$ , so $\sum P = 16/16 = 1$ ✓

For a binomial distribution: $E(X) = np = 4 \times \frac{1}{2} = \mathbf{2}$

Or by direct calculation:

E(X) = 0 \cdot \frac{1}{16} + 1 \cdot \frac{4}{16} + 2 \cdot \frac{6}{16} + 3 \cdot \frac{4}{16} + 4 \cdot \frac{1}{16} = \frac{0 + 4 + 12 + 12 + 4}{16} = \frac{32}{16} = 2

For a binomial distribution: $\text{Var}(X) = npq = 4 \times \frac{1}{2} \times \frac{1}{2} = \mathbf{1}$

Standard deviation: $\sigma = \sqrt{1} = 1$ .

Why This Works

Each coin toss is independent, with the same probability of success. This is the textbook scenario for a binomial distribution. The probability of getting exactly $k$ heads in $n$ tosses is determined by: how many ways can $k$ heads be arranged among $n$ positions ( $\binom{n}{k}$ ), times the probability of each specific arrangement ( $p^k q^{n-k}$ ).

The mean $np = 2$ makes intuitive sense — on average, half of 4 tosses will show heads. The variance $npq = 1$ tells us the typical deviation from the mean is about 1 head.

Alternative Method — Listing all outcomes

Total outcomes = $2^4 = 16$ . List them: HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT.

Count by number of heads: 0H → 1 way, 1H → 4 ways, 2H → 6 ways, 3H → 4 ways, 4H → 1 way.

For CBSE boards, the question typically asks for the probability distribution table AND the mean. The marking scheme gives 2 marks for the table and 1-2 marks for the mean. Use the binomial formula $E(X) = np$ for the mean — it’s much faster than the direct sum method and gets full marks.

Common Mistake

Students sometimes compute $P(X = 2) = \binom{4}{2} \times (1/2)^2 = 6 \times 1/4 = 3/2$ — a probability greater than 1, which is clearly wrong. The error is forgetting the $(1/2)^{4-k}$ term. The correct computation is $\binom{4}{2} \times (1/2)^2 \times (1/2)^2 = 6/16$ . Always include both the $p^k$ and $q^{n-k}$ factors.

Question

Solution — Step by Step

Why This Works

Alternative Method — Listing all outcomes

Common Mistake

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