Question

A batch of 25 injection-molded parts contains 5 that have suffered excessive shrinkage. Round your answers to four decimal places (e.g. 98.7654). (a) If two parts are selected at random, and without replacement, what is the probability that the second part is one with excessive shrinkage? (b) If three parts are selected at random, and without replacement, what is the probability that the third part is one with excessive shrinkage?

Answer 1

Answer:

(a) The probability that the second part is the one with excessive shrinkage is 0.2000.

(b) The probability that the third part is the one with excessive shrinkage is 0.2000.

Step-by-step explanation:

Let the variable X ₙ denote the nth part that has  suffered excessive shrinkage.

(a)

It is provided that two parts are selected at random.

The parts are selected without replacement.

Now, for the two parts selected it is possible that either both have excessive shrinkage or only the second part has excessive shrinkage.

The probability that the second part is the one with excessive shrinkage is:

P (X₂) = P (X₂ ∩ X₁) + P (X₂ ∩ X₁')

         $=[\frac{4}{24}\times \frac{5}{25}]+[\frac{5}{24}\times \frac{20}{25}]$               (without replacement)

         $=\frac{1}{30}+\frac{1}{6}$

         $=0.2000$

Thus, the probability that the second part is the one with excessive shrinkage is 0.2000.

(b)

It is provided that three parts are selected at random.

The parts are selected without replacement.

Now, for the three parts selected it is possible that either all three have excessive shrinkage or any two of the three has excessive shrinkage or only the third part has excessive shrinkage.

The probability that the third part is the one with excessive shrinkage is:

P (X₃) = P (X₃ ∩ X₂ ∩ X₁) + P (X₃ ∩ X₂ ∩ X₁')

                 + P (X₃ ∩ X₂' ∩ X₁) + P (X₃ ∩ X₂' ∩ X₁')

         $=[\frac{3}{23}\times \frac{4}{24}\times \frac{5}{25}]+[\frac{4}{23}\times \frac{5}{24}\times \frac{20}{25}]+[\frac{4}{23}\times \frac{20}{25}\times \frac{5}{24}]+[\frac{5}{23}\times \frac{19}{24}\times \frac{20}{25}]$  

         $=0.2000$

Thus, the probability that the third part is the one with excessive shrinkage is 0.2000.