Binomial Theorem: Real-World Scenarios (8)

Question

A factory produces $1000$ items per day. The probability that any one item is defective is $0.005$. Use the binomial theorem (or its approximation) to find the probability that exactly $3$ items are defective on a given day.

This connects binomial expansion with binomial probability — a real-world JEE Main pattern.

Solution — Step by Step

Final answer: $P(X = 3) \approx 0.14$.

Why This Works

The binomial PMF $\binom{n}{k}p^k(1-p)^{n-k}$ counts how many ways to pick $k$ defectives out of $n$, weighted by the probability of any one such configuration. The binomial theorem ensures these probabilities sum to $1$ (by setting $x = p$ and $y = 1-p$ in $(x+y)^n$).

For large $n$ and small $p$ with $np$ moderate, the binomial converges to the Poisson distribution. Here $np = 5$, so $P(X=3) \approx e^{-5} 5^3/6 \approx 0.140$ — matching our exact answer.

Alternative Method

Poisson approximation. With $\lambda = np = 5$:

$P(X = 3) \approx \frac{e^{-5} \cdot 5^3}{3!} = \frac{0.0067 \times 125}{6} \approx 0.140$

Faster than the binomial, accurate to within a few percent for $n \geq 20$, $p \leq 0.05$.

Common Mistake

Using $(1-p)^n$ instead of $(1-p)^{n-k}$. The exponent is the number of “successes” we did NOT have, which is $n - k = 997$, not $1000$.