is it ok to use the tables of the binomial distribution for the following question?

Question

Given at least one child in a family of 5 has blue eyes, and the probability any child has blue eyes = 1/4, what is the probability at least 2 children have blue eyes? (i don't see why they tell you at least one child has blue eyes if each event is independent anyway).

Answer 1

Note: they aren't asking just for the chance that at least 2 children have blue eyes. They are asking for the probability, given that you know at least one already does.

This is conditional probability.

P(B | A) = P(A and B) / P(A)

So the probability that you have at least 2, given you have at least 1 is the probability you have at least 2 divided by the probability you have at least 1.

For that part though, you would use binomal distribution.

First, what is the probability that at least one child has blue eyes?
This would be 1 minus the probability that nobody has blue eyes.
P ( blue >= 1 ) = 1 - (0.75)^5
P ( blue >= 1 ) = 0.762695312

Now the probability that at least *two* children have blue eyes:
P( blue = 2 ) + P ( blue = 3 ) + P ( blue = 4 ) + P ( blue = 5 ). Again you can figure the negative statement and subtract from 1. So this would be equivalent to:
1 - P ( blue = 0 ) - P ( blue = 1 ).

This would be:
P ( blue >= 2 ) = 1 - (0.75)^5 - (0.25)(0.75)^4 * C(5,1)
The C(5,1) means the ways to choose 1 person from 5, which is just 5.
P ( blue >= 2) = 1 - (0.75)^5 - (0.25)(0.75)^4 * 5
P ( blue >= 2) = 0.3671875

So putting it all together:
P ( blue >= 2 | blue >= 1 ) = P ( blue >= 2 ) / P ( blue >= 1 )
= 0.3671875 / 0.762695312
= 0.481434059
≈ 48.1%

Answer 2

yes