Hey guys, Im having trouble with this question..I'd appreciate the help.

A study of the numbers of male and female children in families in a certain population is being carried out.

a) A sample model is that each child in any family is equally likely to be male or female, and that the sex of each child is independent of the sex of any previous children in the family. Using this model calculate the probability that, in a randomly chosen family of 4 children, there are 2 males and 2 females.

b) An alternative model is the first child in any family is equally likely to be male or female, but that, for any subsequent children, the probability that tthey will be of the same sex as the previous child is 3/5.

Usin tis model, calculate the probability that, in a randomly chosen family of 4 children,

i) all four will be the same sex

ii) no two consecutive children will be of the same sex,

iii) there will be 2 males and 2 females

Thanks.

**a)**
Since the probabilities are always 0.5, you can use the binomial expansion,

choosing the term with 2 of both.

Alternatively, 2 girls and 2 boys can be born in the following sequences

BBGG

BGBG

BGGB

GGBB

GBGB

GBBG

Hence the probability is \(\displaystyle 6(0.5)^4\)

**b)**
This one is not binomial due to the varying probabilities

**(i)** The sequences are

BBBB

GGGG

For the 4 boys, the probability is

\(\displaystyle \frac{1}{2}\ \frac{3}{5}\ \frac{3}{5}\ \frac{3}{5}=\frac{1}{2}\left(\frac{3}{5}\right)^3\)

same for the 4 girls, hence the solution is

\(\displaystyle \left(\frac{3}{5}\right)^3\)

**(ii)** The sequences are

BGBG

GBGB

\(\displaystyle \frac{1}{2}\ \frac{2}{5}\ \frac{2}{5}\ \frac{2}{5}\) for both sequences, giving

\(\displaystyle \left(\frac{2}{5}\right)^3\)

**(iii)** The sequences are

GGBB

GBGB

GBBG

BGBG

BBGG

BGGB

The probability is

\(\displaystyle \frac{1}{2}\ \frac{3}{5}\ \frac{2}{5}\ \frac{3}{5}\)

\(\displaystyle +\frac{1}{2}\ \frac{2}{5}\ \frac{2}{5}\ \frac{2}{5}\)

\(\displaystyle +\frac{1}{2}\ \frac{2}{5}\ \frac{3}{5}\ \frac{2}{5}\)

\(\displaystyle +\frac{1}{2}\ \frac{2}{5}\ \frac{2}{5}\ \frac{2}{5}\)

\(\displaystyle +\frac{1}{2}\ \frac{3}{5}\ \frac{2}{5}\ \frac{3}{5}\)

\(\displaystyle +\frac{1}{2}\ \frac{2}{5}\ \frac{3}{5}\ \frac{2}{5}\)

\(\displaystyle =\frac{1}{2}\left(2\frac{2}{5}\left(\frac{3}{5}\right)^2+2\frac{3}{5}\left(\frac{2}{5}\right)^2+2\left(\frac{2}{5}\right)^3\right)\)

\(\displaystyle =\frac{2}{5}\left(\frac{3}{5}\right)^2+\frac{3}{5}\left(\frac{2}{5}\right)^2+\left(\frac{2}{5}\right)^3\)