Question

Suppose that 30% of mosquitoes possess a certain gene. A researcher wishes to obtain four mosquitoes that possess this gene. Find the probability that the researcher will obtain the fifth fly after sampling 15 flies from the general mosquito population?

Answer 1

Given,

The probability that mosquitoes possess cetain gene, p = 30% = 0.3 (probability of success)

Let X be the number of mosquitos sampled, and r be the number of mosquitos possess cetain gene.

Then X will be negative binomial variate, then the negative binomial probability is given by:

$\boldsymbol{NB(x;r,p)=\binom{x-1}{r-1}p^{r}(q)^{x-r}}$

where

x: The number of trials required to produce r successes in a negative binomial experiment.
r: The number of successes in the negative binomial experiment.
p: The probability of success on an individual trial.
q: The probability of failure on an individual trial.

In this context,

x= 15

r=5

p=0.3

q= 1 - 0.3 = 0.7

Now, the probability that the researcher will obtain the fifth fly after sampling 15 flies from the general mosquito population = NB(15;5,0.3)

$\boldsymbol{NB(15;5,0.3)=\binom{15-1}{5-1}0.3^{5}(0.7)^{15-5}}$

                                          $\boldsymbol{=0.0687}$

Answer: 0.0687