Answer to Question #202130 in Statistics and Probability for Aasia Tariq

Question

Q.No.1 i. A supplier shipped a lot of six parts to a company. The lot contained three defective parts. Suppose the customer decided to randomly select two parts and test them for defects. How large a sample space is the customer potentially working with? List the sample space. Using the sample space list, determine the probability that the customer will select a sample with exactly one defect.

ii. A bin contains six parts. Two of the parts are defective and four are acceptable. If three of the six parts are selected from the bin, how large is the sample space? Which counting rule did you use, and why? For this sample space, what is the probability that exactly one of the three sampled parts is defective?

iii.   A company places a seven-digit serial number on each part that is made. Each digit of the serial number can be any number from 0 through 9.Digits can be repeated in the serial number. How many different serial numbers are possible?

Answer 1

i)

Sample space size:

$N=C^2_7=21$

Let $a,b,c$ are defective, $A,B,C$ are non defective.

Sample space:

$(aa,ab,ac,bb,bc,cc,AA,AB,BB,AC,BC,CC,aA,aB,aC,bA,bB,bC,cA,cB,cC)$

The probability that the customer will select a sample with exactly one defect:

$p=9/21=3/7$

ii)

Sample space size:

Counting rule:

The number of ways to choose m items from total n without replacement:

$C^m_n=\frac{n!}{m!(n-m)!}$

So, the number of ways to choose 3 from 6:

$N_1=C^3_6=\frac{6!}{3!3!}=20$

For each of two defective parts we have the number of ways that equals the number of was to choose two non defective parts from total four.

So, the number of samples with one defective part is two times (two defective parts) by the number of ways to choose 2 from 4 (non defective parts):

$N_2=2C^2_{4}=2\cdot\frac{4!}{2!2!}=6$

The probability that exactly one of the three sampled parts is defective:

$p=N_2/N_1=6/20=3/10$

iii)

$N=10^7$