Understanding the Binomial Distribution
The problem presents a classic scenario for the Binomial Distribution. A binomial distribution applies when you have a fixed number of independent trials, each with only two possible outcomes (success or failure).
The Formula
The probability of getting exactly $k$ successes in $n$ trials is given by: $P(X = k) = \binom{n}{k} p^k q^{n-k}$
Where:
- $n = 4$ (number of bulbs chosen)
- $p = 0.20$ (probability of a bulb being defective)
- $q = 1 - p = 0.80$ (probability of a bulb being good)
- $\binom{n}{k} = \frac{n!}{k!(n-k)!}$
Step-by-Step Solution
(i) Probability of exactly one defective bulb ($k=1$)
$P(X=1) = \binom{4}{1} (0.2)^1 (0.8)^3$ $P(X=1) = 4 \times 0.2 \times 0.512 = 0.4096$
(ii) Probability of zero defective bulbs ($k=0$)
$P(X=0) = \binom{4}{0} (0.2)^0 (0.8)^4$ $P(X=0) = 1 \times 1 \times 0.4096 = 0.4096$
(iii) Probability of at most 2 defective bulbs ($X \le 2$)
This is the sum of probabilities for 0, 1, and 2 defective bulbs: $P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$ We already have $P(0)$ and $P(1)$. Let's find $P(2)$: $P(X=2) = \binom{4}{2} (0.2)^2 (0.8)^2 = 6 \times 0.04 \times 0.64 = 0.1536$
Now, sum them up: $P(X \le 2) = 0.4096 + 0.4096 + 0.1536 = 0.9728$