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

Answer: 0019

Let A: One ball drawn from the bowl B found to be blue.
$B_1:1R+4B$ from bowl $A$ to $B$.
$B_2:2R+3B$ from bowlA to $B$.
$B_3:3R+2B\text{ from bowl }A\text{ to }B.$
$B_4:4R+1B\text{ from bowl }A\text{ to }B$
$B_5:5R+0B\text{ from bowl }A\text{ to }B.$
$P\left(B_1\right)=\frac{{}^6{C}_1\cdot {}^4{C}_4}{{}^{10}{C}_5}=\frac{6}{{}^{10}{C}_5};P\left(B_2\right)=\frac{{}^6{C}_2\cdot {}^4{C}_3}{{}^{10}{C}_5}=\frac{60}{{}^{10}{C}_5};P\left(B_3\right)=\frac{{}^6{C}_3\cdot {}^4{C}_2}{{}^{10}{C}_5}=\frac{120}{{}^{10}{C}_5}$
$P\left(B_4\right)=\frac{{}^6{C}_4\cdot {}^4{C}_1}{{}^{10}{C}_5}=\frac{60}{{}^{10}{C}_5};P\left(B_5\right)=\frac{{}^6{C}_5\cdot {}^4{C}_0}{{}^{10}{C}_5}=\frac{6}{{}^{10}{C}_5}$
$$\begin{gathered} P\left(A/B_1\right)=\frac{4}{5};P\left(A/B_2\right)=\frac{3}{5};P\left(A/B_3\right)=\frac{2}{5};P\left(A/B_4\right)=\frac{1}{5};P\left(A/B_5\right)=0 \\ <br/>\therefore P\left(B_2/A\right)=\frac{P\left(B_2\right)\cdot P\left(A/B_2\right)}{{\sum }_{i=1}^{5}P\left(B_i\right)\cdot P\left(A/B_i\right)} \end{gathered}$$
$P\left(B_2/A\right)=\frac{36}{\frac{24}{5}+36+48+12}=\frac{3}{\frac{2}{5}+3+4+1}=\frac{3}{\frac{2+40}{5}}=\frac{15}{42}=\frac{5}{14}\equiv \frac{m}{n}$
Hence $(m+n)=19$.