Question

Bowl A has 6 red balls and 4 blue balls. 5 balls are drawn simultaneously and kept in a bowl $B$ which was empty. Now a ball is randomly taken out from the bowl B and found to be blue. If the probability that 2 red and 3 blue balls were transferred from bowl $A$ to $B$ is $(m / n)$ where $m$ and $n$ are relatively prime, find $( m + n )$.

Answer 1

Let A: One ball drawn from the bowl B found to be blue.
$B_1: 1 R+4 B$ from bowl $A$ to $B$.
$B _2: 2 R +3 B$ from bowlA to $B$.
$B _3: 3 R +2 B \text { from bowl } A \text { to } B . $
$B _4: 4 R +1 B \text { from bowl } A \text { to } B $
$B _5: 5 R +0 B \text { from bowl } A \text { to } B .$
$P \left( B _1\right)=\frac{{ }^6 C _1 \cdot{ }^4 C _4}{{ }^{10} C _5}=\frac{6}{{ }^{10} C _5} ; P \left( B _2\right)=\frac{{ }^6 C _2 \cdot{ }^4 C _3}{{ }^{10} C _5}=\frac{60}{{ }^{10} C _5} ; P \left( B _3\right)=\frac{{ }^6 C _3 \cdot{ }^4 C _2}{{ }^{10} C _5}=\frac{120}{{ }^{10} C _5} $
$P \left( B _4\right)=\frac{{ }^6 C _4 \cdot{ }^4 C _1}{{ }^{10} C _5}=\frac{60}{{ }^{10} C _5} ; P \left( B _5\right)=\frac{{ }^6 C _5 \cdot{ }^4 C _0}{{ }^{10} C _5}=\frac{6}{{ }^{10} C _5} $
$P \left( A / B _1\right)=\frac{4}{5} ; P \left( A / B _2\right)=\frac{3}{5} ; P \left( A / B _3\right)=\frac{2}{5} ; P \left( A / B _4\right)=\frac{1}{5} ; P \left( A / B _5\right)=0 \\
\therefore P\left(B_2 / A\right)=\frac{P\left(B_2\right) \cdot P\left(A / B_2\right)}{\sum_{i=1}^5 P\left(B_i\right) \cdot P\left(A / B_i\right)} $
$P\left(B_2 / A\right)=\frac{36}{\frac{24}{5}+36+48+12}=\frac{3}{\frac{2}{5}+3+4+1}=\frac{3}{\frac{2+40}{5}}=\frac{15}{42}=\frac{5}{14} \equiv \frac{m}{n} $
Hence $( m + n )=19$.