Question

A medical treatment is successful 30% of the time. Suppose that 9 patients try this
treatment.
a. What is the probability that at least 4 patients end with a successful
experience.
b. What is the probability that exactly 3 patients end with a successful
experience.
c. What is the probability that none of the patients end with a successful
experience.
d. What is the probability that at most 5 patients end with a successful
experience.
e. What is the mean number of patients who end with a successful experience.
f. What is the variance of number of patients who end with a successful
experience

Answer 1

This is a binomial distribution case with n=9, p=0.3, p=1-0.3=0.7

P(X=x) =${n\choose x} p^x q^{n-x}$

a) P(X $\geq$ 4)

$={\sum_{x=4}^{30}}0.3^x 0.7^{30-x}$

=0.2703

b) P(x=3)

$={9\choose 3}0.3^3 0.7^6$

=0.2668

c) P(X=0)

$={9\choose 0}0.3^0 0.7^9$

=0.0404

d) P(X$\le$ 5)

$=\sum_{x =0}^50.3^x 0.7^{9-x}$

=0.9747

e) E(X) =np

=9*0.3=2.7

$\approx 3 patients$

f) var(X) =npq

=9*0.3*0.7

=1.89

$\approx 2 patients$