Answered: A corporation receives a shipment…

College Algebra

10th Edition

ISBN:9781337282291

Author:Ron Larson

Publisher:Ron Larson

Chapter8: Sequences, Series,and Probability

Section8.7: Probability

Problem 11ECP: A manufacturer has determined that a machine averages one faulty unit for every 500 it produces....

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Question

A corporation receives a shipment having 18 pieces out of which 4 are substandard. If 6 pieces are selected as a random sample, and the shipment is rejected if any of these items proves to be substandard. Calculate the following if possible. If any part of question (you feel) is not possible to calculate give reason:
a. Develop the probability distribution table for the number of substandard pieces.

b. Calculate the expected number of substandard pieces and its variance.

c. What is the probability that less than 3 substandard pieces in the random sample?

d. Calculate the probability of accepting a shipment

Expert Solution

Step 1

The probability of a substandard piece,

$\begin{array}{rcl} P (S) & = & \frac{T o t a l s u b s \tan d a r d p i e c e s f o u n d}{T o t a l p i e c e s c h e c k e d} \\ = & \frac{4}{18} \\ = & \frac{2}{9} \end{array}$
Let X be the number of substandard pieces in the shipment.

Probability for zero substandard pieces out of 6 pieces,

$\begin{array}{rcl} P (X = 0) & = & \frac{{}^{14}C_{6}}{{}^{18}C_{6}} \\ = & \frac{14!}{6! \cdot 8!} \times \frac{6! \cdot 12!}{18!} \\ = & \frac{3003}{18564} \\ = & \frac{11}{68} \end{array}$

Step 2

Probability for one substandard pieces out of 6 pieces,

$\begin{array}{rcl} P (X = 1) & = & \frac{{}^{14}C_{5} \cdot {}^{4}C_{1}}{{}^{18}C_{6}} \\ = & \frac{2002 \cdot 4}{18564} \\ = & \frac{22}{51} \end{array}$

Probability for two substandard pieces out of 6 pieces,

$\begin{array}{rcl} P (X = 2) & = & \frac{{}^{14}C_{4} \cdot {}^{4}C_{2}}{{}^{18}C_{6}} \\ = & \frac{1001 \cdot 6}{18564} \\ = & \frac{11}{34} \end{array}$

Step 3

Probability for three substandard pieces out of 6 pieces,

$\begin{array}{rcl} P (X = 3) & = & \frac{{}^{14}C_{3} \cdot {}^{4}C_{3}}{{}^{18}C_{6}} \\ = & \frac{364 \cdot 4}{18564} \\ = & \frac{4}{51} \end{array}$

Probability for four substandard pieces out of 6 pieces,

$\begin{array}{rcl} P (X = 4) & = & \begin{array}{l} \frac{{}^{14}C_{2} \cdot {}^{4}C_{4}}{{}^{18}C_{6}} \end{array} \\ = & \frac{91 \cdot 1}{18564} \\ = & \frac{1}{204} \end{array}$

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