Question

Concrete blocks are tested and it is found that, on average, 7% fail to meet the required specification. For a batch of 9 blocks, determine the probabilities that (a) three blocks and (b) less than four blocks will fail to meet the specification. If the failure rate of the blocks in the previous problem rises to 15%, find the probabilities that (c) no blocks and (d) more than two blocks will fail to meet the specification in a batch of 9 blocks.

Answer 1

Ans:

Use binomial distribution with n=9 and p=0.07

P(x=r)=9C_r*0.07^r*(1-0.07)^9-r

a)P(x=3)=9C₃*0.07³*(1-0.07)⁶=0.0186

b)P(less than 4)=P(x<=3)

=P(x=0)+P(x=1)+P(x=2)+P(x=3)

=9C₀*0.07⁰*(1-0.07)⁹+9C₁*0.07¹*(1-0.07)⁸+9C₂*0.07²*(1-0.07)⁷+9C₃*0.07³*(1-0.07)⁶

=0.5204+0.3525+0.1061+0.0186

=0.9977

c)n=9 and p=0.15

P(x=0)=(1-0.15)^9=0.2316

d)

P(x>2)=1-P(x<=2)

=1-P(x=0)-P(x=1)-P(x=2)

=1-9C₀*0.15⁰*(1-0.15)⁹-9C₁*0.15¹*(1-0.15)⁸-9C₂*0.15²*(1-0.15)⁷

=1-0.2316-0.3679-0.2579

=0.1408