## Friday, September 27, 2013

Consider an urn $A$, with 15 blue balls, and 10 red balls, and an urn $B$, with 10 blue balls, and 15 red balls. We select randomly one urn (with probability 50% for each urn).
We draw a ball, which turns out to be blue, and we put it back in the urn, Now, we draw a (second) ball. What is the probability that this (second) ball is blue?

Solution after the break!

The question is here to compute
$\mathbb{P}(B_2=\text{blue}\vert B_1=\text{blue})$
and according to Bayes formula, it is
$\frac{\mathbb{P}(B_2=\text{blue}\cap B_1=\text{blue})}{\mathbb{P}( B_1=\text{blue})}$
Now, to compute those two probabilities, we have to condition on the urn,
$\mathbb{P}({\color{red}\bullet})=\mathbb{P}({\color{red}\bullet}\vert A)\cdot\mathbb{P}( A)+\mathbb{P}({\color{red}\bullet}\vert B)\cdot\mathbb{P}( B)$
Given the urn, since we replace the ball,
$\mathbb{P}(B_2=\text{blue}\cap B_1=\text{blue}\vert\text{urn})=\mathbb{P}(B_2=\text{blue}\vert\text{urn})\cdot\mathbb{P}(B_1=\text{blue}\vert\text{urn})$
i.e.
$\mathbb{P}(B_2=\text{blue}\cap B_1=\text{blue}\vert\text{urn})=\mathbb{P}(B_1=\text{blue}\vert\text{urn})^2$
So if we substitute numerical probabilities to get a blue ball in the previous formula, we get
$\frac{\displaystyle{\frac{1}{2}\left(\frac{15}{25}\right)^2+\frac{1}{2}\left(\frac{10}{25}\right)^2}}{\displaystyle{\frac{1}{2}\left(\frac{15}{25}\right)+\frac{1}{2}\left(\frac{10}{25}\right)}}$
which not the same as
$\frac{1}{2}\left(\frac{15}{25}\right)+\frac{1}{2}\left(\frac{10}{25}\right)$
Here, we get
`0.52 `

By : http://freakonometrics.hypotheses.org/8718

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