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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 $Y$:

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

Both $X$ and $Y$ are binomial distributions. What are the parameter values?

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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  • $\begingroup$ Houston, we lost zero. $\endgroup$ – Did Feb 12 '13 at 19:29
  • $\begingroup$ Zero is back, we happy. $\endgroup$ – Did Feb 13 '13 at 6:09
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The first table is wrong. Its probabilities should add up to 1. The second table is correct. X ~ Bin(5, 0.6). Y ~ Bin(5,0.5).

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  • $\begingroup$ I think when someone edited the data into the table format he deleted a 0. I meant P(X=5)=0.0776. Would that be a correct value? $\endgroup$ – user60852 Feb 12 '13 at 19:14
  • $\begingroup$ @user60852 Yes, but you also need to add P(X=0) and P(Y=0) as well. So in fact the second table is incomplete. Can you edit your question? $\endgroup$ – oks Feb 12 '13 at 19:20

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