# An urn contains 5 white and 10 black balls. Fair die is rolled and that number of balls is randomly chosen

An urn contains 5 white and 10 black balls. A fair die is rolled and that number of balls is randomly chosen from the urn. What is the probability that all of the balls selected are white? What is the conditional probability that the die landed on 3 if all the balls selected are white?
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insonsipthinye

1.Events:
A- ll of the choose balls are white
${E}_{i}$ - result of the die rill is i, $i=1,2,3,4,5,6$
Probabilities:
Since the die is fair:
$P\left({E}_{i}\right)=\frac{1}{6}f\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}i=\left\{1,2,3,4,5,6\right\}$
If the die rolls i we choose a combination of i balls, among black and five white balls, therefore $P\left(A\mid {E}_{1}\right)=\frac{\left(\begin{array}{c}5\\ 1\end{array}\right)}{\left(\begin{array}{c}15\\ 1\end{array}\right)}=\frac{5}{15}=\frac{1}{3}$
$P\left(A\mid {E}_{2}\right)=\frac{\left(\begin{array}{c}5\\ 2\end{array}\right)}{\left(\begin{array}{c}15\\ 2\end{array}\right)}=\frac{10}{105}=\frac{2}{21}$
$P\left(A\mid {E}_{3}\right)=\frac{\left(\begin{array}{c}5\\ 3\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}=\frac{10}{455}=\frac{2}{91}$
$P\left(A\mid {E}_{4}\right)=\frac{\left(\begin{array}{c}5\\ 4\end{array}\right)}{\left(\begin{array}{c}15\\ 4\end{array}\right)}=\frac{1}{273}$
$P\left(A\mid {E}_{5}\right)=\frac{\left(\begin{array}{c}5\\ 5\end{array}\right)}{\left(\begin{array}{c}15\\ 5\end{array}\right)}=\frac{1}{3003}$
$P\left(A\mid {E}_{6}\right)=\frac{\left(\begin{array}{c}5\\ 6\end{array}\right)}{\left(\begin{array}{c}15\\ 6\end{array}\right)}=0$
Calculate: $P\left(A\right),P\left({E}_{3}\mid A\right)$
2.${E}_{1},{E}_{2},{E}_{3},{E}_{4},{E}_{5},{E}_{6}$ are competing hypothesis, that is, mutually exclusive events, union of which is the whole outcome space, so conditioning on the roll of the die: $P\left(A\right)=\sum _{i=1}^{6}P\left(A\mid {E}_{i}\right)P\left({E}_{i}\right)$
Substituting $P\left({E}_{i}\right),P\left(A\mid {E}_{i}\right)$ yields: $P\left(A\right)=\frac{1}{6}\left(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}\right)=\frac{5}{66}$
$P\left({E}_{3}\mid A\right)$ can be calculated from the definition if we note the identity $P\left({}_{}$

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Jeffrey Jordon

A-event that all balls drawn are white

${D}_{i}$ - outcome of roll of the die is $i,i=1,2,\dots ,6$

$P\left(A\right)=\sum _{i=1}^{6}P\left(A|{D}_{i}\right)P\left({D}_{i}\right)$

$=\frac{1}{6}\left(P\left(A|{D}_{i}\right)+\dots +P\left(A|{D}_{6}\right)\right)$

$=\frac{1}{6}\left(\frac{5}{15}+\frac{{5}_{{C}_{2}}}{{15}_{{C}_{2}}}+\frac{{5}_{{C}_{3}}}{{15}_{{C}_{3}}}+\frac{{5}_{{C}_{4}}}{{15}_{{C}_{4}}}+\frac{{5}_{{C}_{5}}}{{15}_{{C}_{5}}}+0\right)=\frac{5}{66}$

$P\left({D}_{3}|A\right)=\frac{P\left(A|{D}_{3}\right)P\left({D}_{3}\right)}{P\left(A\right)}$

$=\frac{\frac{{5}_{{C}_{3}}}{{15}_{{C}_{3}}}}{\sum _{i=1}^{5}\frac{{5}_{{C}_{i}}}{{15}_{{C}_{i}}}}=\frac{22}{455}$