A public interest group hires students to solicit donations by telephone. After

Question

A public interest group hires students to solicit donations by telephone. After a brief training period students make calls to potential donors and are paid on a commission basis. Experience indicates that early on, these students tend to have only modest success and that 70% of them give up their jobs in their first two weeks of employment. The group hires 6 students, which can be viewed as a random sample. What is the probability that at least 2 of the 6 will give up in the first two weeks? What is the probability that at least 2 of the 6 will not give up in the first two weeks?

Explanation / Answer

Binomial Distribution

PMF of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
Where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial

a.
( X < 2) = P(X=1) + P(X=0)
= ( 6 1 ) * 0.7^1 * ( 1- 0.7 ) ^5 + ( 6 0 ) * 0.7^0 * ( 1- 0.7 ) ^6
= 0.0109

P( X > = 2 ) = 1 - P( X < 2) = 0.9891

b.
P( X < 2) = P(X=1) + P(X=0)
= ( 6 1 ) * 0.3^1 * ( 1- 0.3 ) ^5 + ( 6 0 ) * 0.3^0 * ( 1- 0.3 ) ^6
= 0.4202

P( X > = 2 ) = 1 - P( X < 2) = 0.5798

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