Question

Research has shown the 30% of all houses ( p = 0.3) have faulty wiring. An inspection was done of 18 houses.

A. Find the probability that four (4) houses have faulty wiring.

B. Probability that at least 1 house has faulty wiring.

C. Find the mean and standard deviation.

Answer 1

Answer)

As there are fixed number of trials and probability of each and every trial is same and independent of each other

Here we need to use the binomial formula

P(r) = ncr*(p^r)*(1-p)^n-r

Ncr = n!/(r!*(n-r)!)

N! = N*n-1*n-2*n-3*n-4*n-5........till 1

For example 5! = 5*4*3*2*1

Special case is 0! = 1

P = probability of single trial = 0.3

N = number of trials = 18

R = desired success

A)

P(4) = 18c4*(0.3^4)*(1-0.3)^18-4 = 0.16810437084

B)

P(at least 1) = 1 - P(0) = 0.99837158641

C)

Mean = n*p = 18*0.3 = 5.4

S.d = √{n*p*(1-p)} = 1.94422220952