Bag A contains 9 white and 8 black balls, while bag B contains 6 white and 4 black balls.

Question

Bag A contains 9 white and 8 black balls, while bag B contains 6 white and 4 black balls. One ball is randomly picked up from the bag B and mixed up with the balls in the bag A . Then a ball is randomly drawn from the bag A . If the probabi...

Accepted Answer

Bag A : 9 white, 8 black (Total $=17$ )

Bag B : 6 white, 4 black (Total $=10$ )

One ball is moved from Bag B to Bag A. There are two mutually exclusive events for this transfer:

Case 1 : The ball moved is white

The probability of choosing a white ball from Bag B is:

$P({W}_{B})=\frac{6}{10}$

After adding this white ball to Bag A, Bag A now contains 10 white and 8 black balls ( 18 total). The probability of then drawing a white ball from Bag A is :

$P({W}_{A}|{W}_{B})=\frac{10}{18}$

Case 2: The ball moved is black

The probability of choosing a black ball from Bag B is :

$P({B}_{B})=\frac{4}{10}$

After adding this black ball to Bag A, Bag A now contains 9 white and 9 black balls ( 18 total). The probability of then drawing a white ball from Bag A is :

$P({W}_{A}|{B}_{B})=\frac{9}{18}$

Using total probability theorem

$P({W}_{A})=P({W}_{B})\cdot P({W}_{A}|{W}_{B})+P({B}_{B})\cdot P({W}_{A}|{B}_{B})$

$\Rightarrow$$P({W}_{A})=(\frac{6}{10}	imes \frac{10}{18})+(\frac{4}{10}	imes \frac{9}{18})$

$P({W}_{A})=\frac{60}{180}+\frac{36}{180}=\frac{96}{180}=\frac{8}{15}$

It is given that $P({W}_{A})=\frac{p}{q}$.

$\begin{gathered} gcd(8,15)=1 \ \ p=8q=15 \end{gathered}$

$p+q=8+15=23$

Solutions · 1