A pharmacist receives a shipment of 22 bottles of a drug and has 4 of the bottles tested. If 6 of the 22 bottles are contaminated, what is the probability that more than 1 of the tested bottles is contaminated? Express your answer as a fraction or a decimal number rounded to four decimal places.

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14-10-2019
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A pharmacist receives a shipment of 22 bottles of a drug and has 4 of the bottles tested. If 6 of the 22 bottles are contaminated, what is the probability that more than 1 of the tested bottles is contaminated? Express your answer as a fraction or a decimal number rounded to four decimal places.

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Jenirelle Jenirelle · Answer 1 · 2019-10-21T07:02:27+02:00

Explanation:

A hypergeometric distribution as a function of statistical distribution where members from two groups are analysed, without replacing those members. Unlike binomial distributions, in which the probability remains the same for each trial, in hypergeometric distributions, these members are not replaced- each trial's probability affects the probability within next trial.

This is used in the process of random sampling and as a means of quality control in statistics; for example selecting members of a sports team from a mixed population of boys and girls.

Hypergeometric Formula..,

For a population consisting of N items, k are successes. And a random sample from that population consisting of n items, x of which are successes.

∴ hypergeometric probability is:

h(x; N, n, k) = [ kCx ] [ N-kCn-x ] / [ NCn ]

this distribution always has...

its mean= n * k / N .
its variance is n * k * ( N - k ) * ( N - n ) / [ N2 * ( N - 1 ) ] .

Thus….

Total number of bottles = 22

Number of contaminated bottles = 6

Number of tested bottles = 4

X= a random variable representing the number of contaminated bottles selected for the test.

N = 22

K = 6

n = 4

The probability mass function, f(x), is given by...

P(X=x)) = [ kCx] [ N-kCn-x ] / [ NCn ] x= 0,1,2,3,...,n

= [tex]\frac{[6 C 1] [22-6 C 4-1]}{[22 C 4]}[/tex]

OR

[tex]\frac{C\left \ {{k=6} \atop {x=1}} \right. C\left \ {{22-6} \atop {4-1}} \right. }{C \left \ {22} \atop {4}} \right. }[/tex]

[tex]Cxk= C14= 4!(1!(4 - 1)!)^{-1}= 24(6)^{-1}= 4\\Cn-xN-k=C\left \ {{22-6} \atop {4-1}} \right. } =16!(4!(16-4)!)^{-1}\\(CnN)^{-1}=(C \left \ {{y=22} \atop {x=6}} \right. ) ^{-1}=[22!(6!(22-6)!)^{-1}]^{-1}[/tex]

f(x)= p= P(X=1)=0.45933014354067

P(that more than 1 of the tested bottles is contaminated);

P(X>1)=1−P(X≤1)=1−(P(X=0)+P(X=2))

P(X>1)=0.291866028708134

Probability that one is contaminated: 0.4593(4 dec places)

Probability that more than one is contaminated: 0.2919 (4 dec places)

Learn more about calculating probability at https://brainly.com/question/4021035

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