Question

A batch of 40 injection-molded parts contains 6 parts that have suffered excessive shrinkage (a) If two parts are selected at random, and without replacement, what is the probability 6. that the second part selected is one with excessive shrinkage? (b) If three parts are selected at random, and without replacement, what is the probability that the third part selected is one with excessive shrinkage?

Answer 1

From the given data,

P(Suffered excessive shrinkage) = 6 / 40

P( Do not suffered excessive shrinkage) = 1 - 6/40 = 34/40

a)

P( Second part of two parts suffered excessive shrinkage) = P( first part do not suffered excessive shrinkage) *

p( Second part suffered excessive shrinkage) = (34/40) * (6/39) = 0.85 * 0.153846 = 0.130769

b)

P( third of the three parts suffered excessive shrinkage) = P( first part do not suffered excessive shrinkage) *

p( Second part do not suffered excessive shrinkage) * P( Third part suffered excessive shrinkage)

= (34/40)  * (33/39) * (6/38) = 0.85 * 0.846153 * 0.157894 = 0.113562