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

Step-by-step 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,

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=(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=(11!)/(3!(11-3)!)=165`

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