# A conditional probability of picking a colored ball from box

There are 10 balls in a box, each has same probability of being black or being white, i.e., $P[x_i=Black]=P[x_i=White]=0.5$ for $i=1 \to 10$. Every time a ball is picked at random, it is then returned back to the box.

Compute the following:

• The probability of only white balls in the box if no black ball appears in the first four picks.
• The probability that at least two black balls are in the box if we pick exactly one black ball in first four picks.
• The distribution of black balls in the box if we pick only white balls on the first ten picks.

I cannot understand how "putting a ball back in the box after ball is picked" express in a formula.

• The phrase "each of them has the same probability of being black or white" is not clear. Do you mean there are $5$ of each color? Presumably not, judging from the questions. Do you just mean there are $B$ black and $W$ white and the balls are indistinguishable (except for color)? – lulu Feb 8 '17 at 17:16
• I interpret the question to mean $P[x_i = Black] = P[x_i = White] = 0.5$, for $i = 1 \to 10$. So there is a prior probability all are white, or all are black, or various mixtures, as given by a binomial distribution. – David G. Stork Feb 8 '17 at 17:18
• @DavidG.Stork That certainly makes sense...maybe the OP can confirm? – lulu Feb 8 '17 at 17:19
• Yes, you are right. I edited the question. – Superian007 Feb 8 '17 at 17:20

To answer the first question (the others are handled in a similar way):

Let $P_n$ be the probability of drawing $WWWW$ if there are exactly $n$ white balls in the urn. It is an easy matter to compute that $$P_n=\left( \frac {n}{10}\right)^4$$

Now, what is the probability that we started with exactly $n$ white balls? From the binomial distribution we see that $$P(\#White = n)=\binom {10}n\frac 1{2^{10}}$$

It follows that the total probability of observing $WWWW$ is $$\sum_{n=0}^{10}\left( \frac {n}{10}\right)^4\times \binom {10}n\frac 1{2^{10}}=0.10175$$

As the contribution of the "all white" scenario to that sum is $\frac 1{2^{10}}$ we see that the desired answer is $$\frac 1{\sum_{n=0}^{10}\left( \frac {n}{10}\right)^4\times \binom {10}n}=\boxed{0.009597666}$$

Remark: as our prior, we believed that the probability we were in the all-white scenario was $\frac 1{2^{10}}=0.000976563$ so our estimate of that probability has gone up by, roughly, a factor of $10$.

• can u please tell me why my argument differs or gives a different answer? – Kiran Feb 8 '17 at 17:42
• @Kiran you appear to be looking at a version of the question without replacement, though the OP specifies that there is replacement here. – lulu Feb 8 '17 at 17:49
• ok, got it. deleting my answer. – Kiran Feb 8 '17 at 17:50