Printed circuit cards are placed in a functional test after being populated with semiconductor chips. A lot contains 110 cards, and 20 are selected without replacement for functional testing.
(a) If 20 cards are defective, what is the probability that at least 1 defective card is in the sample?
(b) If 5 cards are defective, what is the probability that at least 1 defective card appears in the sample?

The answers to the questions are;

(a) P(At least 1 defective)

= 0.9883.

(b) P(At least 1 defective)

= 0.6409.

Step-by-step explanation:

There are 110 cards and 20 defectives.

a) The probability of at least one defective is given by

P(At least 1 defective) = 1 - P(0 defective)

P(0 defective) = 20C0 × (90C0)/(110C20) = 0.0116

1 - 0.0116 = 0.9883

b) For a set of 110 boards that has 5 defective and 105 non-defective

P(At least 1 defective) = 1 - P(0 defective)

P(0 defective) = (20C0)(90C5)/(110C5) = 0.35909

1-0.35909

= 0.6409

See explanation

Step-by-step explanation:

There are so many ways we can rewrite .

We can apply the power rule to obtain:

.

We could also simplify this further because

This implies that:

.

Or

.

All these are equivalent expressions.