Question

A box contains 14 transistors, 3 of which are defective. If 3 are selected at random, find the probability of the statements below.a. All are defectiveb. None are defective

Answer 1

ANSWERS

a. 1/364

b. 165/364

EXPLANATION

a. There are 11 non-defective transistors and 3 defective transistors in the box. When the first one is selected, there are 3 defective transistors and 14 transistors in total. When the second one is selected, assuming that the first was defective, there are 2 defective and 13 transistors in total. Finally, if the first two were defective, when the third transistor is selected there will be only 1 defective transistor out of 12 transistors in total.

Therefore, the probability that the three transistors selected are defective is,

$P(all\text{ }defective)=\frac{3}{14}\cdot\frac{2}{13}\cdot\frac{1}{12}=\frac{6}{2184}=\frac{1}{364}$

b. Now, we have to find the probability that the three transistors selected are non-defective.

First, we have to find how many ways are to select 3 transistors out of the 14 total transistors,

$_{14}C_3=\frac{14!}{3!(14-3)!}=364$

Also, we have to find how many ways are to select 3 non-defective transistors - note that these must be picked out of the 11 non-defective ones,

$_{11}C_3=\frac{11!}{3!(11-3)!}=165$

Therefore, the probability that the three transistors selected are non-defective is,

$P(none\text{ }defective)=\frac{165}{364}$