Understanding the Binomial Distribution
The problem provided is a classic application of the Binomial Distribution. A binomial experiment occurs when:
- There are a fixed number of trials ($n$).
- Each trial has only two possible outcomes: success or failure.
- The probability of success ($p$) is constant for each trial.
- The trials are independent.
In this scenario, $n = 4$ (the sample size) and the probability of a defective chip is $p = 0.20$. Therefore, the probability of a non-defective chip is $q = 1 - p = 0.80$.
The Binomial Formula is: $P(X = k) = \binom{n}{k} p^k q^{n-k}$ where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.
Step-by-Step Solutions
(i) Probability that no chip is defective ($k=0$)
$P(X=0) = \binom{4}{0} (0.2)^0 (0.8)^4 = 1 \cdot 1 \cdot 0.4096 = 0.4096$
(ii) Probability that one chip is defective ($k=1$)
$P(X=1) = \binom{4}{1} (0.2)^1 (0.8)^3 = 4 \cdot 0.2 \cdot 0.512 = 0.4096$
(iii) Probability that more than one chip is defective ($X > 1$)
It is easier to use the complement rule: $P(X > 1) = 1 - [P(X=0) + P(X=1)]$ $P(X > 1) = 1 - [0.4096 + 0.4096] = 1 - 0.8192 = 0.1808$
Summary
- No defective: $40.96\%$
- Exactly one defective: $40.96\%$
- More than one defective: $18.08\%$