{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

CEE 202 Homework 6 Solution

# Better to find the probability that 2 or fewer pipes

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: = number of pipes failing a) You could do a very long sum of probability of 3 or more pipes will fail, but this would take a long time. Better to find the probability that 2 or fewer pipes will fail. Then find out the probability of at least 3 pipes will fail by deducting the probability of 2 or fewer will fail from 1. Here are the calculations. Let ‘p’ be the probability of failure, ‘q’ be the probability of success which is (1 ­p). Then p=0.001, and q =0.999 No. trials(n) 4000 p failure 0.001 q success 0.999 ȹ n ȹ n(x)= ȹ ȹ # failures(x) px q(n-x) P(x)=n(x)*px*q(n-x) ȹ x Ⱥ 0 1 2 1 4000 7998000 1 0.018279 0.001 0.018297 0.000001 0.018316 0.018279 0.073189 0.146488 P(0 ­2) = 0.0183 + 0.0732 + 0.1465 =0.2380 P(3 or more) = 1 – P(0 ­2) = 1 ­0.2380 = 0.7620 b) The expected value of the binomial distribution is n p = 4000 x 0.001 = 4...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online