Question

Q1. The J.O. Supplies Company buys calculators from a Korean supplier. The probability of a defective calculator is 15%. If 16 calculators are selected at random, what is the probability that more than 5 of the calculators will be defective?

2. An important part of the customer service responsibilities of a cable company relates to the speed with which trouble in service can be repaired. Historically, the data show that the likelihood is 0.40 that troubles in a residential service can be repaired on the same day. For the first 7 troubles reported on a given day, what is the probability that more than 4 troubles will be repaired on the same day?

I need both question answer with step and methods for the solution of above two question

Answer 1

Q1. The J.O. Supplies Company buys calculators from a Korean supplier. The probability of a defective calculator is 15%. If 16 calculators are selected at random, what is the probability that more than 5 of the calculators will be defective?

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 = 0.15

N = number of trials = 16

R = desired success = more than 5

We know that sum of all the probabilities is = 1

So, P(more than 5) = 1 - (p(0) + p(1) + p(2) + p(3) + p(4) + p(5))

= 0.02354438115