Urn problem with return

Question

In an urn there are 10 balls. Each ball can be white or black with the same probability. We will draw balls repeatedly with return. Find:

(a) The expected value and variance of black balls in the urn;

(b) The probability that there are only white balls in the urn, if we didn't draw any black ball in the four first draws;

(c) The probability that there are at least two black balls in the urn, if we drew one black ball in the first four draws.

My solution:

(a) There is not defined how many times we will draw balls from the urn and it seems to me as binomial distribution, so EX and varX should be:

$EX=np=\frac{n}{2}$

$varX=np(1-p)=\frac{n}{4}$

(b) Conditional probability

B ... I will not draw black ball = $(\frac{1}{2})^4$

$A \cap B = (\frac{1}{2})^n$

$P(A | B)=\frac{P(A \cap B)}{P(B)}=\frac{(\frac{1}{2})^n}{(\frac{1}{2})^4}$

(c)

B ... if we draw only one black ball in first four moves = $\binom{4}{1}(\frac{1}{2})(\frac{1}{2})^3$

A ... at least two black balls are in the urn = $\sum_{k=2}^{\infty}\binom{n}{k}(\frac{1}{2})^{k}(\frac{1}{2})^{n-k}$

and then I use conditional probability, right?

Someone told me that this example is on Bayes' theorem, but I cannot see how can I use it here. But also something is wrong in my solution ... could someone give me an advice?

Answer 1

(a) This is not about drawing; its about how many black balls are in the urn.

So, yes, this will be a binomial distribution. The number of trials is 10, the success rate is $1/2$ (there is equal and independent probability each of the ten balls is black).

$$X\sim\mathcal{Bin}(10,1/2)$$

(b) $\mathsf P(X=0\mid Y=0)$ where $X$ is the number of black balls in the urn (as above) and $Y$ is the number drawn in four draws with replacement.   $Y$ is binomial conditional on $X$; since in this case the rate of success is dependent on the number of black balls in the urn. $$Y\mid X~\sim~\mathcal{Bin}(4, x/{10})$$

Then $\mathsf P(Y=0\cap X=x) = (\frac{10-x}{10})^4\binom{10}{x}\frac{1}{2^{10}}$

So $\mathsf P(Y=0\cap X=0)= \tfrac 1{2^{10}}$

$\mathsf P(Y=0) =\tfrac 1{2^{10}10^4}\sum_{x=0}^{10} \binom{10}{x}(10-x)^4 =\frac{407}{4000}$

$\therefore\quad\ldots$

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(c) Similar to above, but this time: $\mathsf P(X\geq 2\mid Y=1) = \dfrac{\mathsf P(Y=1\cap X\geq 2)}{\mathsf P(Y=1)} \\ = \dfrac{\sum_{k=2}^{10} \binom{10}k k(10-k)^3}{\sum_{j=0}^{10} \binom{10}j j(10-j)^3}$