# Unknown Box problem

I have $2$ bags $A$ and $B$, both bags have $200$ hundred balls, in bag $A$ two of the balls are black and the rest are white, and in bag $B$ one of the balls is black and the rest are white.

a) If I am given a bag but I don't know if it is $A$ or $B$, what is the probability of picking a black ball.

b) I pull the first ball out of the bag and its white, what is the probability its from bag $A$.

c) What is the probability that that the second ball will be white also, given the first was white.

• These are all very very basic probability questions, are you sure you studied this material? – Avrham Aton Mar 31 '15 at 10:27
• Welcome to math.SE! It is helpful if you provide some context, what you have tried and where you are having difficulty. – user103828 Mar 31 '15 at 10:30
• Yeh I know they are elementary. I answered them with probability trees but I wanted to see how it would be answered with correct notation. – Tom Mar 31 '15 at 10:33
• Are there $202$ balls in bag $A$ and $201$ in bag $B$? Question a) can only be answered if the probability of receiving bag $A$ (or $B$) is known. – drhab Mar 31 '15 at 10:33
• There are 200 ba – Tom Mar 31 '15 at 10:34

I answered them with probability trees but I wanted to see how it would be answered with correct notation.

Let $A$ be the event of getting bag A, $B$ that of getting bag B, and $W$ the event of picking a white ball out of the bag.   Let $W^\complement$ be the complement event (the ball picked is black).   We'll use subscripts when more than one ball is picked.

Thus since bag A contains two back ball in 200 total, $\mathsf P(W^\complement \mid A) = \frac 1 {100}$ is the conditional probability of a single ball picked being a black ball when given bag A.   And so forth.

Note As others have mentioned, the prior probability of receiving bag A (or B) is not given, and deciding that there is no bias in the selection ($\mathsf P(A)=\tfrac 1 2$) is an assumption.   Can you assume this?

a) If I am given a bag but I don't know if it is A or B, what is the probability of picking a black ball.

Note: You did not mention how many balls you had picked.

Assuming you picked one ball, then by the Law of Total Probability:

$$\mathsf P(W^\complement) = \mathsf P(W^\complement\mid A)\mathsf P(A)+\mathsf P(W^\complement\mid B)\mathsf P(B)$$

If you had pick two balls from the bag and wish to know the probability at least one is white is:

\begin{align} \mathsf P(W_1^\complement\cup W_2^\complement) & = \mathsf P(W_1^\complement\cup W_2^\complement\mid A)\mathsf P(A)+\mathsf P(W_1^\complement\cup W_2^\complement\mid B)\mathsf P(B) \end{align}

b) I pull the first ball out of the bag and its white, what is the probability its from bag A.

Here we use the Bayes' Rule, and the Law of Total Probability.

$$\mathsf P(A\mid W) = \frac{\mathsf P(W\mid A)\mathsf P(A)}{\mathsf P(W\mid A)\mathsf P(A)+\mathsf P(W\mid B)\mathsf P(B)}$$

c) What is the probability that that the second ball will be white also, given the first was white.

The Law of Total Probability

$$\mathsf P(W_2\mid W_1) = \mathsf P(W_2\mid W_1\cap A)\mathsf P(A)+\mathsf P(W_2\mid W_1\cap B)\mathsf P(B)$$

So in our $2$ boxes, both labeled $A$ and $B$, we have that in box $A$ there are $200$ balls, where $2$ of the balls of blacks. Similarly, in box $B$ there are also $200$ balls, where only one of the balls is black.

Now, for part (a), first consider the probability of choosing one of the boxes. Since there are $2$ boxes, the probability of choosing either one is $\frac{1}{2}$.

Then, consider each dependent case, where the first case is the box we've chosen is box $A$. In this case, it follows that there are $2$ black balls present, so the probability is $\frac{2}{200} = \frac{1}{100}$.

Similarly, if we perform the same steps for our second case, where we have box $B$, we find that the probability is $\frac{1}{200}$

Now, adding up our probabilities, we find that our answer for part (a) is:

$$\frac{1}{2}*\frac{1}{100} + \frac{1}{2}*\frac{1}{200} = \frac{3}{400}$$

You can use the same process to answer parts (b) and (c) to your question.

• The OP is not choosing a bag (at random), but a bag is given unto him. So there is no ground for $\frac12$ as probability. The question can only be answered if things are preassumed. – drhab Mar 31 '15 at 10:49
• @drhab I understand. I assumed certain things when answering the question. – Varun Iyer Mar 31 '15 at 10:51

Let's denote the event that box $A$ is given unto you by $A$ and the event that box $B$ is given unto you by $B$. Note that $P\left(A\right)+P\left(B\right)=1$. Denote the event that the $i$-th ball taken out is white (or black) by $W_{i}$ (or $B_{i}$). Then the following equalities can be used to answer your question.

a)$$P\left(B_{1}\right)=P\left(B_{1}\mid A\right)P\left(A\right)+P\left(B_{1}\mid B\right)P\left(B\right)$$ b)$$P\left(A\mid W_{1}\right)\left[P\left(W_{1}\mid A\right)P\left(A\right)+P\left(W_{1}\mid B\right)P\left(B\right)\right]=$$$$P\left(A\mid W_{1}\right)P\left(W_{1}\right)=P\left(A\cap W_{1}\right)=P\left(W_{1}\mid A\right)P\left(A\right)$$ c)$$P\left(W_{2}\mid W_{1}\right)P\left(W_{1}\right)=P\left(W_{1}\cap W_{2}\right)=$$$$P\left(W_{1}\cap W_{2}\mid A\right)P\left(A\right)+P\left(W_{1}\cap W_{2}\mid B\right)P\left(B\right)$$