Mathematics, 17.03.2020 05:24 rivman10

A factory received a shipment of 38 sprockets, and the vendor who sold the items knows there are 5 sprockets in the shipment that are defective. Before the receiving foreman accepts the delivery, he samples the shipment, and if too many of the sprockets in the sample are defective, he will refuse the shipment.
a) If a sample of 5 sockets is selected, find the probabilty that all in the sample are defective.
b) If a sample of 5 sprockets is selected, find the probability that none in the sample are defective.

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Answer from: bajnathhannavanna

Using the hypergeometric distribution, it is found that there is a

a) 0.000002 = 0.0002% probability that all in the sample are defective.

b) 0.4728 = 47.28% probability that none in the sample are defective.

The sprockets are chosen from the sample without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes. N is the size of the population. n is the size of the sample. k is the total number of desired outcomes.

In this problem:

Shipment of 38 sprockets, hence N = 38

.5 are defective, hence k = 5

.Sample of 5, hence n = 5

Item a:

This probability is P(X = 5), hence:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$P(X = 5) = h(5,38,5,5) = \frac{C_{5,5}C_{33,0}}{C_{38,5}} = 0.000002$

0.000002 = 0.0002% probability that all in the sample are defective.

Item b:

This probability is P(X = 0), hence:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,38,5,5) = \frac{C_{5,0}C_{33,5}}{C_{38,5}} = 0.4728$

0.4728 = 47.28% probability that none in the sample are defective.

A similar problem is given at link

Answer from: goosenar

a) 0.00019923%

b) 47.28%

Step-by-step explanation:

a) To find the probability of all sockets in the sample being defective, we can do the following:

The first socket will be in a group where 5 of the 38 sockets are defective, so the probability is 5/38

The second socket will be in a group where 4 of the 37 sockets are defective, as the first one picked is already defective, so the probability is 4/37

Expanding this, we have that the probability of having all 5 sockets defective is: (5/38)*(4/37)*(3/36)*(2/35)*(1/34) = 0.0000019923 = 0.00019923%

b) Following the same logic of (a), the first socket have a chance of 33/38 of not being defective, as we will pick it from a group where 33 of the 38 sockets are not defective. The second socket will have a chance of 32/37, and so on.

The probability will be (33/38)*(32/37)*(31/36)*(30/35)*(29/34) = 0.4728 = 47.28%

Answer from: Quest

cbnygjdyj

step-by-step explanation:

Answer from: Quest

$30.32

step-by-step explanation:

65.00*.12=2.80

65.00-2.80=57.20

57.20*1.06=60.632

60.632/2=30.316

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