Suppose that on average, 8% of circuit boards are defective. What is the probabi

Question

Suppose that on average, 8% of circuit boards are defective. What is the probability of finding exactly 2 defective boards in a batch of 10 randomly chosen circuit boards? What is the probability of finding at most 2 defective boards in a batch of 10 randomly chosen circuit boards? What is the probability of finding more than 1 but less than 4 defective boards in a batch of 10 randomly chosen circuit boards? What is the probability that we have to test at least 3 circuit boards to find the first defective one?

Explanation / Answer

Binomial distribution used

P=0.08

N=10

P(X=x) = (nCx) px (1-p)n-x

a).

P( x=2) = 0.1478

b). P( x 2)= P( x=0)+P( x=1) +P( x=2)

= 0.4344+0.3777+0.1478

P=0.9599

c). P( more than1 but less than 4)

= P( x=2) +P( x=3) =0.1478+ 0.0343

= 0.1821

d).

geometric distribution used

for 3 board , P=P * Qx – 1   =0.08*0.923-1 =0.0677

for 2 board , P=0.0736

for 1 board , P=0.08

The required P =0.0677+0.0736+0.08 = 0.2213

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