Question

1) Suppose an antibiotic has been shown to be 70% effective against a common bacteria.
a) If the antibiotic is given to five unrelated individuals with the bacteria, what is the probability that it will be effective in exactly three?
b) What is the probability that the antibiotic is effective in exactly three patients out of five (when the probability of effectiveness in a single patient is 70%).
c) What is the probability that the antibiotic will be effective in none of the five?

Answer 1

1. Suppose an antibiotic has been shown to be 70% effective against a common bacteria.

a) If the antibiotic is given to five unrelated individuals with the bacteria, what is the probability that it will be effective in exactly three?

b) What is the probability that the antibiotic is effective in exactly three patients out of five (when the probability of effectiveness in a single patient is 70%).

c) What is the probability that the antibiotic will be effective in none of the five?

Use this formula: $P_n(k) = C_n^k p^k q^{n-k}$, where $q=1-p$.

a) We are using five unrelated individuals, so $n=5$.

We should find the probability that it will be effective in exactly three, so $k=3$, probability of success is 70%, or $p = \frac{70}{100} = 0.7$,

Probability of failure is $q=1-p=1-0.7=0.3$

$P_5(3) = C_5^3 0.7^3 0.3^2 = \frac{5!}{3!2!} 0.7^3 0.3^2 = 0.3087$

b) (the same with task (a), but instead of individuals of bacteria using the patients)

$n=5, k=3, p=0.7$, so $P_5(3) = C_5^3 0.7^3 0.3^2 = \frac{5!}{3!2!} 0.7^3 0.3^2 = 0.3087$

c) We should find the probability that it will be effective in none ($k=0$) of the five ($n=5$) probability of success is 70%, or $p = \frac{70}{100} = 0.7$,

Probability of failure is $q=1-p=1-0.7=0.3$

$n=5, k=0, p=0.7$, so $P_5(0) = C_5^0 0.7^0 0.3^5 = \frac{5!}{0!5!} 0.7^0 0.3^5 = 0.00243$