Question

A couple intends to have two children, and suppose that approximately 52% of births are male and 48% are female.

Answer 1

a) Probability of both being males is 27%

b) Probability of both being females is 23%

c) Probability of having exactly one male and one female is 50%

Step-by-step explanation:

a)

The probability that the birth is a male can be written as

$p(m) = 0.52$ (which corresponds to 52%)

While the probability that the birth is a female can be written as

$p(f) = 0.48$ (which corresponds to 48%)

Here we want to calculate the probability that over  2 births, both are male. Since the two births are two independent events (the probability of the 2nd to be a male  does not depend on the fact that the 1st one is a male), then the probability of both being males is given by the product of the individual probabilities:

$p(mm)=p(m)\cdot p(m)$

And substituting, we find

$p(mm)=0.52\cdot 0.52 = 0.27$

So, 27%.

b)

In this case, we want to find the probability that both children are female, so the probability

$p(ff)$

As in the previous case, the probability of the 2nd child to be a female is independent from whether the 1st one is a male or a female: therefore, we can apply the rule for independent events, and this means that the probability that both children are females is the product of the individual probability of a child being a female:

$p(ff)=p(f)\cdot p(f)$

And substituting

$p(f)=0.48$

We find:

$p(ff)=0.48\cdot 0.48=0.23$

Which means 23%.

c)

In this case, we want to find the probability they have exactly one male and exactly one female child. This is given by the sum of two probabilities:

- The probability that 1st child is a male and 2nd child is a female, namely $p(mf)$

- The probability that 1st child is a female and 2nd child is a male, namely $p(fm)$

So, this probability is

$p(mf Ufm)=p(mf)+p(fm)$

We have:

$p(mf)=p(m)\cdot p(f)=0.52\cdot 0.48=0.25$

$p(fm)=p(f)\cdot p(m)=0.48\cdot 0.52=0.25$

Therefore, this probability is

$p(mfUfm)=0.25+0.25=0.50$

So, 50%.

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