Question

1. Suppose warranty records show the probability that a new car needs a warranty repair in the first 90 days is .05. If a sample of twenty (20) cars is selected, what is the probability that:
   a) none of the cars will need a warranty repair?
   b) at least one needs a warranty repair?
   c) more than three need a warranty repair?
   d) no more than five need a warranty repair?

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 = 0.05

N = number of trials = 20

R = desired success

A)

None

P(0) = 20c0*(0.05^0)*(1-0.05)^20-0 = 0.35848592241

B)

At least 1 = 1 - p(0)

As the sum of all the probabilities is = 1

At least 1 = 0.64151407759

C)

More than 3

= 1 - (p(0) + p(1) + p(2) + p(3) )

= 0.01590152602

D)

No more than 5

P(0) + p(1) + p(2) + p(3) + p(4) + p(5)

= 0.99967070568