Question

MO U USE Social media? According to the US Dept of Health and Human Services, 29% of 21-25 year olds say they have driven while under the influence. If four 21-25 vear olds are selected at random, a) what is the probability that all three say they have driven under the influence b) What is the probability that at least one has NOT driven under the influence? c) What is the probability that none of the three have driven under the influence?

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 = 0.29

N = number of trials = 3

R = desired success

A)

P(3) = 3c3*(0.29^3)*(1-0.29)^3-3 = 0.024389

B)

P(at least 1) = 1 - p(0) = 0.642089

C)

P(0) = 0.357911