MHB Choosing a ball at random from a randomly selected box

[Sapphireluna]

I got this probability question as homework, but I got very confused as I didn't know if I should calculate the balls in the 2 boxes as whole or if I should calculate the probability of the first box then add the probability of the second. Any help is appreciated.
Thanks.Here is the question:
There are 2 boxes, box A contains 20 white balls and 20 black balls, box B contains 10 white balls and 5 black balls. A box is chosen at random then a ball is taken at random from the chosen box, what is the probability that the ball is white?

First I thought, it should be the probability of box A + the probability of box B;
20/40 + 10/15 = 7/6
but then I thought it doesn't really make sense to me as this tells that I will always get a white ball?!

So I thought, I could do the probability of the whole lot;
30/55 = 6/11
this answer seems more reasonable, but then I was confused, as there are 2 boxes, so that the chances of a white ball being taken out of each ball should be different?

 

[MarkFL]

Re: Help needed in this simple probability problem

Hello and welcome to MHB, Sapphireluna! (Wave)

I can see being tempted to approach the problem as if all the balls are in one big box:

$$P(\text{white})=\frac{30}{55}=\frac{6}{11}$$

However, we should approach it as follows:

$$P(\text{white})=P(\text{box 1 AND white})\,\text{OR}\,P(\text{box 2 AND white})=\frac{1}{2}\cdot\frac{20}{40}+\frac{1}{2}\cdot\frac{10}{15}=\frac{1}{4}+\frac{1}{3}=\frac{7}{12}$$

You were on the right track with your first approach, however you didn't account for the fact that we have only a 50% probability of choosing either box.

 

[Sapphireluna]

Re: Help needed in this simple probability problem

Hello and welcome to MHB, Sapphireluna! (Wave)

I can see being tempted to approach the problem as if all the balls are in one big box:

$$P(\text{white})=\frac{30}{55}=\frac{6}{11}$$

However, we should approach it as follows:

$$P(\text{white})=P(\text{box 1 AND white})\,\text{OR}\,P(\text{box 2 AND white})=\frac{1}{2}\cdot\frac{20}{40}+\frac{1}{2}\cdot\frac{10}{15}=\frac{1}{4}+\frac{1}{3}=\frac{7}{12}$$

You were on the right track with your first approach, however you didn't account for the fact that we have only a 50% probability of choosing either box.

Thank you very much!