0

An urn contains 15 Balls (5 white, 10 Black). Let's say we pick them one after the other without returning them. How many white balls are expected to have been drawn after 7 turns?

I can calculate it by hand with a tree model but is there a formula for this?

JMoravitz
  • 79,518
Marcus
  • 111
  • Why a down vote? – Marcus Aug 08 '16 at 06:39
  • 3
    Just do $7 \cdot 5 / (5 + 10) $ – AJJ Aug 08 '16 at 06:42
  • If you did not get the answer of $\frac{7}{3}$ then you must have made an error using your "tree-model." With enough caution, you should be able to get the correct answer using a brute force approach like that, but the point is that you really don't need to put in much effort by approaching the problem in a convenient manner. – JMoravitz Aug 08 '16 at 06:59
  • Ok, i found my fault. thx – Marcus Aug 08 '16 at 07:01

2 Answers2

0

An easier method: approach via indicator random variables.

Let $\chi_n = \begin{cases} 1&\text{if the}~n\text{'th ball is white}\\0&\text{otherwise}\end{cases}$

Let $X=\sum\limits_{n=1}^7\chi_n$. That is to say, $X$ counts the number of white balls in the first seven draws.

We wish to calculate $E[X]$

By the linearity of expectation:

$$E[X]=E\left[\sum\limits_{n=1}^7\chi_n\right]=\sum\limits_{n=1}^7E[\chi_n]$$

Now, recognize that $E[\chi_n]=E[\chi_1]$ for every $n\in\{1,2,\dots,7\}$. The probability that the first ball is white is $\frac{5}{15}$, hence $E[\chi_1]=\frac{5}{15}\cdot 1 + \frac{10}{15}\cdot 0 = \frac{5}{15}$

Why was $E[\chi_n]=E[\chi_1]$ for all $n$? By temporarily assigning labels to each ball, we see that there are $15!$ ways to arrange the balls in a line. The first seven balls in the line will be those seven which we will be pulling. Each of these $15!$ arrangements are equally likely to occur. In exactly $\frac{5}{15}$ of these arrangements (that is to say, in exactly $5\cdot 14!$ arrangements) the first ball is white. Similarly, in exactly $\frac{5}{15}$ of these arrangements the second ball is white, etc... In general, in precisely $\frac{5}{15}$ of the arrangements the $n$ 'th ball is white. Hence the conclusion.

We have then:

$$E[X]=\sum\limits_{n=1}^7E[\chi_n]=\sum\limits_{n=1}^7E[\chi_1]=7E[\chi_1]=7\cdot\frac{5}{15}=\frac{7}{3}\approx 2.3333\dots$$

JMoravitz
  • 79,518
  • The more important part of the answer is the "recognize that $E[\chi_n]=E[\chi_1]$" – Masacroso Aug 08 '16 at 07:15
  • @Masacroso added a paragraph about that. I had initially thought it an obvious statement and not needing explicit proof, but you are right that it might not be obvious to someone starting out. – JMoravitz Aug 08 '16 at 07:20
0

Hint: the linearity of expectations show us that

$$\Bbb E[W]=\sum_i\Bbb E[X_i]$$

where $W$ is the random variable that counts the total number of white balls, and each $X_i$ is the random variable that take value $1$ if the ball is white or zero if the ball is black for the $i$-nth pick.

The $X_i$ random variables are not independent, by example, if the first ball is white ($X_1=1$) the expectation of $X_2$ is different that if the first ball is black.

Then we use the total law of probability i.e.

$$\Pr[X_i=1]=\sum\Pr[X_i=1|X_{i-1}=a_{i-1},X_{i-2}=a_{i-2},...,X_1=a_1]\cdot\Pr[X_{i-1}=a_{i-1},X_{i-2}=a_{i-2},...,X_1=a_1]$$

where the sum ranges all combinations of $a_i$, where every $a_i\in\{0,1\}$, i.e. every ball can be white or black. This is just a bit of combinatorics. Check this answer that is similar to understand how to continue.

Community
  • 1
Masacroso
  • 30,417
  • The independence of the indicator random variables does not in any way affect our ability to manipulate them via linearity of expectation. It is good (albeit unnecessary) to recognize that they are in fact dependent, but what is most important is recognizing that they are identically distributed. Approaching with conditional probabilities is far too tedious for this problem. – JMoravitz Aug 08 '16 at 07:24
  • I dont understand @JMoravitz... the conditional probabilities are not tediuous at all, it is just the expression of the combinations behind the equality $E[\chi_n]=E[\chi_1]$. I dont know a formal way different than this to solve the problem. How you show, formally, that are equality distributed if is not doing this? – Masacroso Aug 08 '16 at 07:27
  • Have you read my answer? Which is easier to you? Just immediately stating that $Pr(X_3=1)=\frac{1}{3}$ by recognizing the symmetry in the problem? Or calculating $Pr(X_3=1)=\frac{3}{13}\cdot \frac{5\cdot4}{15\cdot14}+\frac{4}{13}\cdot\frac{5\cdot10}{15\cdot 14}+\frac{4}{13}\cdot\frac{10\cdot 5}{15\cdot14}+\frac{5}{13}\cdot\frac{10\cdot 9}{15\cdot 14}$? What about calculating $Pr(X_7)$? Will you really calculate all $2^6$ individual terms and then add? – JMoravitz Aug 08 '16 at 07:35
  • Of course I calculate all @JMoravitz because is an easy calculation, we group results, check the link in my answer. – Masacroso Aug 08 '16 at 07:41