Question

A bag X contains 4 white balls and 2 black balls, while another bag Y contains 3 white balls and 3 black balls. Two balls are drawn (without replacement) at random from one of the bags and were found to be one white and one black. Find the probability that balls were drawn from bag Y.

OR A and B throw a pair of dice alternatively, till one of them gets a total of 10 and wins the game, Find their respective probabilities of winning, if A starts first.

Answer 1

Let $E_1$ and $E_2$ be the events of drawing bag X and Y respectively.

Then $P\left(E_1\right)=P\left(E_2\right)=\frac{1}{2}$.

Let A be the event of drawing one white and one black ball from anyone of the bag without replacement.

Then, $P\left(A|E_1\right)=\frac{4}{6}\times \frac{2}{5}+\frac{2}{6}\times \frac{4}{5}=\frac{16}{30}$ and $P\left(A|E_2\right)=\frac{3}{6}\times \frac{3}{5}+\frac{3}{6}\times \frac{3}{5}=\frac{18}{30}$.

By Bayes' theorem, $P\left(E_2|A\right)=\frac{P\left(A|E_2\right)P\left(E_2\right)}{P\left(A|E_1\right)P\left(E_2\right)+P\left(A|E_2\right)P\left(E_2\right)}=\frac{\frac{18}{30}\times \frac{1}{2}}{\frac{16}{30}\times \frac{1}{2}+\frac{18}{30}\times \frac{1}{2}}=\frac{18}{34}or\frac{9}{17}$

OR

Let E : getting a total of 10. $\therefore E=(4,6),(5,5),(6,4)$

$\Rightarrow P\left(E\right)=\frac{3}{36}=\frac{1}{12},\Rightarrow P\left(E\right)=\frac{11}{12}$

If A starts the game then he may win in $1^{st},3^{rd},5^{th},.....$trials.

P(A wins) = $\frac{1}{12}+{\left(\begin{array}{c}\frac{11}{12}\end{array}\right)}^2\times \frac{1}{12}+\frac{1}{12}+{\left(\begin{array}{c}\frac{11}{12}\end{array}\right)}^4\times \frac{1}{12}+.....\Rightarrow P\left(Awins\right)=\frac{\frac{1}{12}}{1-\frac{121}{144}=\frac{12}{23}}$

And, P(B wins) = 1-P(A wins) = $1-\frac{12}{23}=\frac{11}{23}$.