Question

Probabilities?

Answer 1

See below:

Explanation:

a

Gender is a 50/50 thing, and so the probability of the oldest child being a girl is `1/2`

b

With no further information about the family, the same gender probability holds here. `1/2`

c

There is a `1/2` probability for each gender choice, and so:

`(1/2)^5=1/32`

d

This is a binomial probability - we're being asked to look at all the different ways we can have gender choices in 5 children - from 5 boys to 5 girls and every mix in between.

That general form looks like:

`sum_(k=0)^(n)C_(n,k)(p)^k(~p)^(n-k)=1`

What we're interested in is where 5 children, `n=5`, we have 3 girls, `k=3`. The probability of a child being a girl is `1/2`, and so `p=1/2` and therefore not a girl is also `1/2`, and so `~p=1/2`:

`C_(5,3)(1/2)^3(1/2)^2=10(1/8)(1/4)=10/32=5/16`

e

To see what the probability of having at least one girl is, we can take the full set of possibilities (5 girls to 5 boys and everything in between) and subtract out the probability of having 5 boys (all the other possibilities will have at least 1 girl).

We can use the binomial probability again with `n=5, k=0, p=1/2, ~p=1/2`. The probability of having all five boys is:

`C_(5,0)(1/2)^0(1/2)^5=1(1)(1/32)=1/32`

And so the probability of having at least one girl is:

`1-1/32=31/32`