Question

A lot of 106 semiconductor chips contains 29 that are defective. Round your answers to four decimal places (e.g. 98.7654).

a) Two are selected, at random, without replacement, from the lot. Determine the probability that the second chip selected is defective.

b) Three are selected, at random, without replacement, from the lot. Determine the probability that all are defective.

Answer 1

number of defective chips = 29

number of non defective chips = 106 - 29 = 77

a) P ( second chip selected is defective)

= P( first chip is non degective)* P ( second chip is defective) + P ( first chip is defective) * P ( second chip is defective)

= (77/106) * (29/105) + (29/106) * (28/105)

=0.2736

b) P ( all three are defective )= (29/106) * (28/105) * (27/ 104)

=0.0189

And i am exremely sorry for the wrong answer actually i didnt saw that " without replacement" thing.