Question

Suppose that 4 children are planned in a family. Let the random variable X stand for the number of boys. Discuss the probability distribution of X.

• Find the mean and variance of the random variable X.

Answer 1

the couple can have no boys at all (x = 0), 1 boy (x = 1), 2 boys (x=2), 3 boys (x=3) or 4 boys (x=4)

so the distribution is as follows:

X = {0, 1, 2, 3,4}

The mean: $\mu=\frac{0+1+2+3+4}5=2$

The variance: $\sigma^2=\frac{\sum {(X_i-\mu)^2}}{n}=\frac{4+1+1+4}5=2$

The sample space: {gggg}, {gggb}, {ggbg}, {gbgg}, {bggg}, {ggbb}, {gbgb}, {bggb}, {bgbg}, {gbbg}, {bbgg}, {gbbb}, {bgbb}, {bbgb}, {bbbg}, {bbbb}.

So, the probability that couple will have 0 boys:

$P(x=0)=P( {\{gggg}\})=1/16$

the probability that couple will have 1 boy:

$P(x=1)=P(\{gggb\}, \{ggbg\}, \{gbgg\}, \{bggg\})=1/4$

the probability that couple will have 2 boys:

$P(x=2)=P( \{ggbb\},\{gbgb\}, \{bggb\}, \{bgbg\}, \{gbbg\}, \{bbgg\})=6/16=3/8$

the probability that couple will have 3 boys:

$P(x=3)=P(\{gbbb\}, \{bgbb\}, \{bbgb\}, \{bbbg\})=1/4$

the probability that couple will have 4 boys:

$P(x=4)=P(\{bbbb\})=1/16$