One bag-A contains $3$ red and $5$ white balls. Second bag-B contains $4$ red and $8$ white balls.
If a blind man is presented with these two bags and allowed to pick a single ball from any of the above(only 1 ball, not 1 ball from each) then what is the probability of picking white ball by him?
I tried it and came to an answer $31/48$, but I am not sure if I am right.
Following is my answer, please correct me if I am wrong:-
- For bag-A, probability is $\dfrac{5}{8}$ for white
- For bag-B, probability is $\dfrac{2}{3}$ for white
But chance of getting either bag is $50\%$, so adding them
Total probability = $\left(\dfrac{5}{8}\times \dfrac{1}{2}\right) +\left(\dfrac{2}{3} \times \dfrac{1}{2}\right) = \dfrac{31}{48}$
My question is..
Am i right? And if yes, is my method right?
P.S.- This is my first question on this site so pardon me if I am making any mistakes, sorry.