# Sampling Balls with Replacement: Independence

Question: A sample size of 10 is taken with replacement from an urn that contains 100 balls, which are numbered 1, 2, ..., 100. (At each draw, each ball has the same probability of being selected).

There are 3 parts to the question, and I've included my work below. However, I'm not sure if independence applies in part ii and iii.

i) P(ball 1 is in the sample) = 1 - P(ball 1 is not in the sample) = $$1 - (\frac{99}{100})^{10}$$

ii) P(neither ball 1 nor ball 2 are in the sample) = P($(1 ∪ 2)^c)$ = 1 - P(1 ∪ 2) = 1 - [P(1) + P(2) - P(1 n 2)]

I think that P(1) = P(2), but I'm not sure if I can apply independence here and assume that P(1 n 2) = P(1) * P(2).

Since we are sampling with replacement, does this mean that we can assume the event: ball 1 is in the sample, and event: ball 2 is in the sample are independent?

iii) Explain how you could calculate (with formulas) P(ball 1 is in the sample | ball 2 is in the sample).

If the two events are independent, then the probability would equal P(ball 1 is in the sample), but I'm confused as to whether I can assume independence.

Any help would be appreciated!

Thanks!

• If sampling is with replacement, events on two draws should be independent. Jan 13, 2017 at 23:40
• On (ii) the two events are not independent. Better to use $\frac{98}{100}$ and repeat what you did with (i) Jan 13, 2017 at 23:45

Let $B_i$ be the event "ball $i$ is in the sample". i) There are $100^{10}$ possible samples (if order is considered), and $99^10$ are not favorable to the event $B_1$, so probability is $${\rm P}(B_1)=1-\frac{99^{10}}{100^{10}}$$ ii) Now there are $98^{10}$ samples favorable to the event $\overline{B_1}\cap\overline{B_2}$, so probability is $${\rm P}(\overline{B_1}\cap\overline{B_2})=\frac{98^{10}}{100^{10}}$$ iii) The conditional probability is $${\rm P}(B_1|B_2)=\frac{{\rm P}(B_1\cap B_2)}{{\rm P}(B_2)}$$ We have \begin{align}{\rm P}(B_1\cap B_2)&=1-{\rm P}(\overline{B_1}\cup\overline{B_2}) \\ &=1-({\rm P}(\overline{B_1})+{\rm P}(\overline{B_2})-{\rm P}(\overline{B_1}\cap\overline{B_2})) \\ &=1-(\frac{99^{10}}{100^{10}}+\frac{99^{10}}{100^{10}}-\frac{98^{10}}{100^{10}}) \end{align} and of course ${\rm P}(B_2)={\rm P}(B_1)=1-\frac{99^{10}}{100^{10}}$ so $${\rm P}(B_1|B_2)=\frac{100^{10}-2.99^{10}+98^{10}}{100^{10}-99^{10}}$$ so $B_1$ and $B_2$ are not independent :-)

You cannot assume independence.   Clearly any knowledge about the count of balls#2 among the ten draws affects the (conditional)probability that ball#1 occurs among the same ten draws.

However, let us test this.

The probability that neither of the two balls occurs among the ten draws is: $$\mathsf P\left((A\cup B)^\complement\right)~=~\left(\frac{98}{100}\right)^{10}$$

The probability that both of the two balls occurs among the ten draws is thusly: \begin{align}\mathsf P(A\cap B) ~& =~ \mathsf P(A)+\mathsf P(B)-\mathsf P(A\cup B) \\[1ex] &=~ 2\left(1-\left(\frac {99}{100}\right)^{10}\right) - \left(1-\left(\frac{98}{100}\right)^{10}\right)\\[1ex] & =~\dfrac{100^{10}-2\cdotp 99^{10}+98^{10}}{100^{10}}\end{align}

Where as$$\mathsf P(A)\mathsf P(B) = \dfrac{{100^{10}}-{2}\cdotp{99^{10}}+{(99^{20}/100^{10})}}{100^{10}}$$

Therefore they are not independent.   Because $99^2\neq 9800$.

$\blacksquare$