# Binomial distribution probability

A married couple decided to have $5$ children. Based on gene history, probability that any one of their children will need to wear eye glasses, independent of sex, is $60$%; probability that a child being a boy or a girl are equally $50$%. Let $X$ be the number of children that needs glasses and $Y$ be the number of boys in the family.

Probability distribution tables for $X$ and $Y$:

$$\begin{array}{} \begin{array}{c|c|c} \text{X} & \text{P(X)}\\ \hline \\0 & 0.01024 \\1 & 0.07680 \\2 & 0.23040 \\3 & 0.34560 \\4 & 0.25920 \\5 & 0.07776 \end{array} & \begin{array}{c|c|c} \text{Y} & \text{P(Y)}\\ \hline \\0 & 0.03125 \\1 & 0.15625 \\2 & 0.31250 \\3 & 0.31250 \\4 & 0.15625 \\5 & 0.03125 \end{array} \end{array}$$

What is P(X=E(X))?

Let W be the number of girls that wear glasses. What is P(W=E(W))?

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what do you calculate E(X) to be? – oks Feb 12 '13 at 21:39
E(X) = 5 * 0.6 = 3 – user60852 Feb 12 '13 at 21:45

First part: $X\sim B(5,0.6)$, so $E(X)=np=3$ and $P(X=3)=0.34560$
Since being a girl and requiring glasses are independent, $W\sim B(5,p)$ where $p=0.5\cdot 0.6=0.3$. Thus, $E(W)=1.5$, and $P(W=1.5)=0$ since $W$ is discrete.
Because $3$ is an integer, but $1.5$ is not (I presume you mean $P(X=E(X))$). – Daniel Littlewood Feb 12 '13 at 22:18
The probability of $3$ children needing glasses is worked out above ($0.34560$). The other probability is $0$ because it is impossible to have one and a half girls that wear glasses. – Daniel Littlewood Feb 12 '13 at 22:33