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

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

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) = (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 = (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.