Question

Some biology students were checking eye color in a large number of fruit flies. For the individual fly, suppose that the probability of white-eyes is 1/4 and the probability of red-eyes is 3/4. We may treat these observations as independent Bernoulli trials. what is the probability that at least 3 flies have to be checked for eye color to observe a white-eye fly?

Answer 1

Answer:

0.5625 = 56.25% probability that at least 3 flies have to be checked for eye color to observe a white-eye fly.

Step-by-step explanation:

A sequence of independent Bernoulli trials is the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

What is the probability that at least 3 flies have to be checked for eye color to observe a white-eye fly?

This is the probability of no white-eyes during the first two flies, that is, $P(X = 0)$ when $n = 2$

Probability of white-eyes is 1/4

This means that $p = \frac{1}{4} = 0.25$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2,0}.(0.25)^{0}.(0.75)^{2} = 0.5625$

0.5625 = 56.25% probability that at least 3 flies have to be checked for eye color to observe a white-eye fly.