P(X=2) = ⁵C₂ p² q³. Binomial distribution basics.

Bernoulli Trials — Probability of Exactly 2 Successes in 5 Trials

Question

A fair coin is tossed 5 times. Find the probability of getting exactly 2 heads.

(This is a standard Bernoulli trials problem. The same setup appears with different values — dice, defective items, multiple choice guessing — across CBSE and JEE Main papers.)

Solution — Step by Step

A “Bernoulli trial” means each attempt has exactly two outcomes — success or failure — and trials are independent with the same probability each time.

Here: $n = 5$ (trials), success = getting a head, $p = \frac{1}{2}$ , $q = 1 - p = \frac{1}{2}$ .

For $X \sim B(n, p)$ , the probability of exactly $r$ successes is:

P(X = r) = \binom{n}{r} \, p^r \, q^{n-r}

We need $P(X = 2)$ , so plug in $n = 5$ , $r = 2$ .

\binom{5}{2} = \frac{5!}{2! \cdot 3!} = \frac{5 \times 4}{2 \times 1} = 10

Why $\binom{5}{2}$ ? Because we’re choosing which 2 of the 5 tosses land heads — the order of the remaining tails doesn’t matter for counting arrangements.

P(X = 2) = 10 \times \left(\frac{1}{2}\right)^2 \times \left(\frac{1}{2}\right)^3

= 10 \times \frac{1}{4} \times \frac{1}{8} = 10 \times \frac{1}{32} = \frac{10}{32} = \frac{5}{16}

Answer: $P(X = 2) = \dfrac{5}{16}$

Why This Works

The formula $\binom{n}{r} p^r q^{n-r}$ is doing two separate jobs simultaneously. The $p^r q^{n-r}$ part calculates the probability of one specific sequence — say HHTTТ — where exactly 2 heads appear. Since trials are independent, we just multiply individual probabilities.

But HHTTT is not the only way to get exactly 2 heads. THTHT, TTHHH — there are $\binom{5}{2} = 10$ such sequences in total. Each has the same probability $p^2 q^3$ , so we multiply by 10.

This is the beauty of the binomial setup: it handles both “what’s the probability of this exact sequence” and “how many sequences have this pattern” in one clean formula.

Alternative Method

You can verify by listing all sequences and counting — but that’s impractical for large $n$ . For small $n = 5$ , you could write out the full binomial expansion of $(p + q)^5$ and pick the coefficient of $p^2 q^3$ :

(p + q)^5 = q^5 + 5pq^4 + 10p^2q^3 + 10p^3q^2 + 5p^4q + p^5

The $p^2 q^3$ term has coefficient 10 — same as $\binom{5}{2}$ . Substituting $p = q = \frac{1}{2}$ gives $\frac{10}{32} = \frac{5}{16}$ .

For CBSE board exams, showing the full $(p+q)^5$ expansion can earn method marks even if you make an arithmetic error at the end. Write it out explicitly.

Common Mistake

The most frequent error is writing $P(X = 2) = p^2 q^3$ and stopping — forgetting the $\binom{5}{2}$ term entirely. This gives $\frac{1}{32}$ instead of $\frac{10}{32}$ .

The logic behind the mistake: students think “2 heads, 3 tails, so multiply their probabilities.” That gives the probability of one specific arrangement (like HHTTT), not all arrangements with 2 heads. Always ask yourself — how many ways can this pattern occur? That count is your $\binom{n}{r}$ .

Quick Reference

X \sim B(n, p) \implies P(X = r) = \binom{n}{r} p^r q^{n-r}

Mean: $\mu = np$
Variance: $\sigma^2 = npq$
$p + q = 1$ always

For this problem: Mean $= 5 \times \frac{1}{2} = 2.5$ heads on average.

In JEE Main, this formula appears both as a direct calculation (like this problem) and inside expectation/variance questions where you need $\sum r \cdot P(X = r)$ . Get the basic formula automatic first — the harder variants follow naturally.