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 probability, that the ball drawn is white, is $$\dfrac{p}{q},gcd(p,q)=1,$$ then $$p+q$$ is equal to

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

It is given that one ball is picked from Bag B and mixed with the balls in Bag A. Then one ball is picked from Bag A and we have to find the probability that the ball picked is white. 

Two cases will be formed. 

Case 1: White ball is picked from Bag B and mixed with the other balls in Bag A. 

Probability of picking a white ball from Bag B = $$\dfrac{6}{10}$$

Now, it is mixed with the balls in Bag A. 

Bag A: 10 white and 8 black balls. 

Probability of picking a white ball from Bag A = $$\dfrac{10}{18}$$

Probability of picking a white ball from Bag A given that a white ball is picked from Bag B = $$\dfrac{6}{10}\times\dfrac{10}{18}=\dfrac{1}{3}$$

Case 2: Black ball is picked from Bag B and mixed with the other balls in Bag A. 

Probability of picking a black ball from Bag B = $$\dfrac{4}{10}$$

Now, it is mixed with the balls in Bag A. 

Bag A: 9 white and 9 black balls. 

Probability of picking a white ball from Bag A = $$\dfrac{9}{18}$$

Probability of picking a white ball from Bag A given that a black ball is picked from Bag B = $$\dfrac{4}{10}\times\dfrac{9}{18}=\dfrac{1}{5}$$

So, the probability of picking a white ball from Bag A after picking a ball from Bag B = $$\dfrac{1}{3}+\dfrac{1}{5}=\dfrac{8}{15}$$

It is in the form of $$\dfrac{p}{q}$$ where the greatest common divisor of $$(p\ ,\ q)=1$$

So, $$\dfrac{p}{q}=\dfrac{8}{15}$$

$$p+q=8+15=23$$

Hence, the sum of $$p$$ and $$q$$ is 23. 

$$\therefore\ $$ The required answer is A.