How do we solve the three variable systems
1 answer:
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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)
<em>{80C9 gives set of working transistor and 20C1 gives 20 defective transistor and 100C10 is combination of shipment of 10 transistors}</em>
P(X≥9)=
<em>(Use the permutation and combination calculator)</em>
P(X≥9)=(231900297200*20/17310309456440)
+(1646492110120/17310309456440)
P(X≥9)=0.267933+0.0951162
P(X≥9)=0.3630492