Question

Suppose 88% of all students taking a beginning programming course fail to get their first program to run on the first submission. Consider a sample of n = 6 students. Use a binomial distribution to find probabilities below, when we define success as "first program fails on the first submission". (a) Probability that at least 5 students fail on their first submission: Incorrect: Your answer is incorrect. (3 decimal places) (b) Probability that fewer than 5 students fail on their first submission: (3 decimal places) (c) Probability that exactly 4 students fail on their first submission (3 decimal places) (d) Continuing to use this binomial model, what is the mean number of students whose programs will fail on their first submission? (2 decimal places)

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 = 0.88 (88%)

N = number of trials = 6

R = desired success

A)

At least 5

P(5) + p(6)

= 0.84437106688

= 0.844

B)

P(fewer than 5) = 1 - p(atleast 5)

As sum of all the.probabilities is = 1

P(fewer than 5) = 1 - 0.844 = 0.156

C)

P(4) = 6c4*(0.88^4)*(1-0.88)^6-4 = 0.12953419776 = 0.13

D)

Mean is given by n*p

N = 6

P = 0.88

Mean = 5.28