Question

The probability that a microscope is defective is .17%. Find the probability that one of the first 50 microscopes is defective

Answer 1

Let $X$ be the number of defective microscopes. Because the events (finding that certain microscope is defecrive or not) have equal probability $p = 1.7 \cdot {10^{ - 3}}$ that it will be defecrive, we can conclude that $X$ follows the binomial distribution $X \sim B(n,p)$ (where $n = 50$) with probability mass function

$${p_X}(k) = C_n^k{p^k}{(1 - p)^{n - k}}$$

where $C_n^k = \frac{{n!}}{{k!(n - k)!}}$ is the binomial coefficient.

Under "one of the first" they mean "exactly one", we need to find the probability $P(X = 1)$, thus

$$P(X = 1) = {p_X}(1) = \frac{{50!}}{{1!49!}} \cdot 1.7 \cdot {10^{ - 3}} \cdot {(1 - 1.7 \cdot {10^{ - 3}})^{49}} \approx 0.0782008$$

If they mean "one or more", we need to find the probability $P(X \geqslant 1)$ that can be rewritten as

$$P(X \geqslant 1) = 1 - P(X < 1) = 1 - P(X = 0)$$

and

$$P(X = 0) = {p_X}(0) = \frac{{50!}}{{0!50!}} \cdot {(1.7 \cdot {10^{ - 3}})^0} \cdot {(1 - 1.7 \cdot {10^{ - 3}})^{50}} \approx 0.918446$$

thus

$$P(X \geqslant 1) \approx 1 - 0.918446 \approx 0.081554$$

Answer: $P(X = 1) \approx 0.0782008$.