Questions: A box contains 16 transistors, 4 of which are defective. If 4 are selected at random, find the probability of the statements below. a. All are defective b. None are defective a. The probability is (Type a fraction. Simplify your answer.)

Solution Steps

To find the probability that all selected transistors are defective, we use the hypergeometric distribution formula:

\[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \]

Substituting the values \(N = 16\), \(K = 4\), \(n = 4\), and \(k = 4\):

\[ P(X = 4) = \frac{\binom{4}{4} \binom{12}{0}}{\binom{16}{4}} = \frac{1 \cdot 1}{1820} = 0.000549 \]

Thus, the probability that all are defective is approximately \(0.0005\).

Next, we calculate the probability that none of the selected transistors are defective. Again, we use the hypergeometric distribution formula:

\[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \]

Substituting the values \(N = 16\), \(K = 4\), \(n = 4\), and \(k = 0\):

\[ P(X = 0) = \frac{\binom{4}{0} \binom{12}{4}}{\binom{16}{4}} = \frac{1 \cdot 495}{1820} = 0.272 \]

Thus, the probability that none are defective is approximately \(0.272\).

Final Answer