Question

A box contains 11 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 ​

Answer 1

The probability that all four selected transistors are defective is 1/330, and the probability that none of the transistors are defective is 35/330.

Step-by-step explanation:

To determine the probability that all 4 selected transistors are defective from a box containing 11 transistors of which 4 are defective, we can use the hypergeometric distribution. The probability of all 4 being defective is calculated as:

• The number of ways to choose 4 defective transistors out of 4: C(4,4) = 1
• The number of ways to choose 0 non-defective transistors out of 7: C(7,0) = 1
• The number of ways to choose any 4 transistors out of 11: C(11,4)
• The probability is then P(all defective) = (C(4,4) * C(7,0)) / C(11,4)

Similarly, to find the probability that none are defective:

• The number of ways to choose 0 defective transistors out of 4: C(4,0) = 1
• The number of ways to choose 4 non-defective transistors out of 7: C(7,4)
• The probability is then P(none defective) = (C(4,0) * C(7,4)) / C(11,4)

Using combination formulas, we find:

1. P(all defective) = (1 * 1) / C(11,4) = 1 / 330
2. P(none defective) = (1 * C(7,4)) / C(11,4) = 35 / 330

Thus, the probabilities are:

• For all defective: 1/330 or approximately 0.00303
• For none defective: 35/330 or approximately 0.10606

Answer 2

The probability that all are defective and the probability that none are defective is 0.01747 and 0.9825 respectively.

Using the parameters given for our Calculation;

• Total = 11
• Number selected = 4

1.)

P(All are defective)

• P(defective) = 4/11 = 0.3636

P(All are defective) = 0.3636 × 0.3636 × 0.3636 × 0.3636 = 0.01747

P(Non are defective) = 1 - P(All are defective)

P(Non are defective) = 1 - 0.1747 = 0.9825