Probabilities?

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1 Answer

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