The test results for 20 samples of polycarbonate plastic from a supplier showed that 15 samples had high resistance whereas five samples had low resistance. four samples were selected randomly and without replacement.

a. what is the probability that exactly one sample shows high resistance?

b. what is the probability that at least one sample shows high resistance?

c. what is the probability that three or more parts in the sample show high resistance?

d. in addition to the 5 samples having low resistance, six samples had high scratch. what is the probability that exactly one sample has low resistance and exactly one sample has high scratch

Question

Mathematics

Isabella Heath

30 March, 01:04

The test results for 20 samples of polycarbonate plastic from a supplier showed that 15 samples had high resistance whereas five samples had low resistance. four samples were selected randomly and without replacement.

a. what is the probability that exactly one sample shows high resistance?

b. what is the probability that at least one sample shows high resistance?

c. what is the probability that three or more parts in the sample show high resistance?

d. in addition to the 5 samples having low resistance, six samples had high scratch. what is the probability that exactly one sample has low resistance and exactly one sample has high scratch

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Answers (1)

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Braelyn Bell · Answer 1

This is NOT a binomial distribution because the population mean of defects is not known.

Use hypergeometric distribution instead, since the number of defects in the sample (batch) is known, and samples are taken without replacement.

Let

D=number of defective (low resistance) samples in batch=5

d=number of defective samples in draw

N=number of normal (high resistance) samples in batch=15

n=number of defective samples in draw.

Then by the hypergeometric distribution

P (D, d, N, n) = C (D, d) * C (N, n) / C (D+N, d+n)

Since four samples are selected from the twenty, we have d+n=4.

(a) n=1 = > d=4-1=3

P (n=1) = C (15,1) * C (5,3) / C (20,4) = 15*10/4845 = 10/323 = 0.03096

(b) n>=1 = > d<4-1=3

P (n>=1)

=1-P (n=0)

=1-C (15,0) * C (5,4) / C (20,4)

= 1-1*5/4845

= 1-1/969

= 1 - 0.00103

= 0.99897

(c) n>=3

P (n>=3)

=P (n=3) + P (n=4)

= C (15,3) C (5,1) / C (20,4) + C (15,4) C (5,0) / C (20,4)

=455*5/4845+1365*1/4845

=455/969+1365/4845

=0.46956+0.28173

=0.75129

(d) For this part, we need to assume independence of low resistance and scratches.

We proceed separate to find probabilities of 1 sample with low resistance/scratch.

P (d=1) [ one part shows low resistance]

=C (15,3) C (5,1) / C (20,4)

=455*5/4845

=455/969

P (s=1) [ one part shows scratches ]

=C (14,3) C (6,1) / C (20,4)

=364*6/4845

=728/1615

Since the two effects are independent, the probability of both happening is the product of the individual probabilities

=455/969*728/1615

=66248/312987

=0.21166