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A contractor purchases a shipment of 100 transistors. It is his policy to test 10 of these transistors and to keep the shipment only if at least 9 of the 10 are in working condition. If the shipment contains 20 defective transistors, what is the probability it will be kept?

Respuesta :

Answer:

Hence the probability of the at least 9 of 10 in working condition is 0.3630492

Step-by-step explanation:

Given:

total transistors=100

defective=20

To Find:

P(X≥9)=P(X=9)+P(X=10)

Solution:

There  are 20 defective and 80 working transistors.

Probability of at least 9 of 10 should be working out 80 working transistors

is given by,

P(X≥9)=P(X=9)+P(X=10)

{80C9 gives set of working transistor and 20C1 gives 20 defective transistor and 100C10 is combination of shipment of 10 transistors}

P(X≥9)=[tex]{80C9*20C(10-9)}/(100C10)+{80C10*20(10-10)}/(100C10)[/tex]

(Use the permutation and combination calculator)

P(X≥9)=(231900297200*20/17310309456440)

+(1646492110120/17310309456440)

P(X≥9)=0.267933+0.0951162

P(X≥9)=0.3630492

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