A box contains 10 transistors, 3 of which are defective. If 3 are selected at random, find the probability of the statements below. a. All are defective b. None are defective Answers has to be a fraction

1. 1/120

2. 7/24

Step-by-step explanation:

n = 10

Number of defective = 3

1. Probability of all defective =

(Number of defective/n) x (number of defective -1/n -1)...(number of defective -n/n-1)

=3/10 x 2/9 X 1/8

= 6/720

= 1/120

2. probability of none defective

Since 3 are defective, those that are not defective = 7

The probability is therefore:

7/10 x 6/9 x 5/8

= 210/720

= 7/24.

step-by-step explanation:

do you have a copy of the table?

the answer is c

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