1
$\begingroup$

Giving randomly 5 red balls and 5 blue balls to 5 kids, each kid get 2 balls. What is the expected value of number of kids got 2 different balls?

I solve it by:

Let $X_i$ be random variable that holds whether kid $i$ got different balls (i.e. red and blue), or same balls (i.e. red and red or blue and blue).

$X_i = 1 \iff$ got red and blue. $X_i = 0 \iff$ got blue and blue or red and red.

Get probabilities: $P(X_i = 0) = red\cdot red + blue\cdot blue = \frac{1}{2}\frac{1}{2} + \frac{1}{2}\frac{1}2{} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}.$

From here, $P(X_i = 1) = 1 - P(X_i = 0) = \frac{1}{2}$.

Find $E(X_i = 1) =\sum_1^5 X_i\cdot P(X_i = 1) = 5\cdot \frac{1}{2} = \frac{5}{2}$.

It's not the answer (got 0/9 on the exam sheet).

Why? What is the correct answer? Thanks in advance.

$\endgroup$
  • $\begingroup$ Note that given your initial conditions it is impossible to get RR BB for all of the kids. At least one kid will end up with two different colored balls. $\endgroup$ – adam May 13 '15 at 16:56
0
$\begingroup$

Your approach with indicator variables using linearity of expectation went exactly in the right direction. You just used the wrong probabilities. Give the first child a ball; no matter which colour it is, $5$ of the remaining $9$ balls are of the other colour, so the probability that the second ball you give them is of the other colour is $\frac59$. Thus the expected number of kids with different colours is $5\cdot\frac59=\frac{25}9$.

$\endgroup$
-1
$\begingroup$

I solved it with some brute force but I found it to be the easiest way. If there were a higher number of balls and kids It would be better to use some algebra.

You can have $5$, $3$ or $1$ kid(s) with different colored balls. You only need to count in how many ways each can happen.

For {{R,B},{R,B},{R,B},{R,B},{R,B}} there's only one possibility.

For {{R,B},{R,B},{R,B},{R,R},{B,B}} there's $5\times 4=20$ options (5 possible kids who have the red balls, then 4 possible kids who have the blue balls).

For {{R,B},{R,R},{R,R},{B,B},{B,B}} there's $5\times 6=30$ options (5 possible kids who have different balls and $4\choose 2$ to order the other 4).

All together you have $51$ possibilities. Now let's calculate the mean of kids with different boys:

$$\mu = 5\times\frac{1}{51}+3\times\frac{20}{51}+1\times\frac{30}{51}= \frac{5+3\times 20+30}{51}=\frac{95}{51}\\ \mu \approx 1.863 $$

Hope it helped you.

$\endgroup$
  • $\begingroup$ These $51$ outcomes are not equiprobable, so you can't calculate a probability as the fraction of favourable outcomes. There are $10!$ elementary outcomes (the permutations of the $10$ balls), so the denominator of the expected value of an integer can't contain a factor of $17$. $\endgroup$ – joriki Jun 19 '16 at 11:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.