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In an urn there are 5 balls with 1 written on them, 2 balls with 2 on them and 3 balls with 5 on them. 5 balls are extrated without replacement.
Let the random variable $X$ represent the sum of the number exctracted.
Determine:

  1. $P(X=6)$
  2. $E[X]$
  3. $Var(X)$

    Since the only combinations whose sum is 6 are the ones consisting of 1 ball with 2 and 4 balls with 1 then $$P(X=6)=\frac{\binom{5}{4}\binom{2}{1}}{\binom{10}{5}}=\frac{5}{126}$$
    Now, I don't know how to evaluate $E[X]$.
    I could define $X_i$ as the result of the i-th extraction, $i=1,2,3,4,5$ and $X=\sum_{i=1}^5 X_i$ but $X_i$ are not indipendent and it wouldn't make the calculation easier.
Alex_28
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1 Answers1

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Expected value is linear. It doesn't matter in the slightest whether or not the $X_i$ are independent or not... you still have $E[X]=E[X_1+X_2+\dots+X_5]=E[X_1]+E[X_2]+\dots+E[X_5]$

Recall further that $E[X_i]=E[X_1]$ for all $i$.

$~$

$E[X]=5\cdot E[X_1] = 5\cdot \frac{5\cdot 1 + 2\cdot 2 + 3\cdot 5}{10}=12$

For $\text{Var}[X]$ recall this is $E[X^2]-E[X]^2$

Recall that $X^2 = (X_1+X_2+\dots+X_5)(X_1+X_2+\dots+X_5)$ leaving you with five terms of the form $X_i^2$ and twenty terms of the form $X_iX_j$ with $i\neq j$

$~$

$E[X^2] = 5 E[X_1^2] + 20 E[X_1X_2]$

$~$

$E[X^2] = 5\cdot \frac{5\cdot 1 + 2\cdot 4 + 3\cdot 25}{10} + 20\cdot \frac{\binom{5}{2}\cdot 1 + 5\cdot 2\cdot 2 + 5\cdot 3 \cdot 5 + \binom{2}{2}\cdot 4 + 2\cdot 3\cdot 10 + \binom{3}{2}\cdot 25}{\binom{10}{2}} = 152+\frac{4}{9}$

$~$

$\text{Var}[X] = E[X^2]-E[X]^2 = 152+\frac{4}{9} - 12^2 = 8+\frac{4}{9}$

JMoravitz
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  • I know that expected value is linear, but I don't understand why $E[X_i]=E[X_1]$ since the result of an extraction influences the result of subsequent extractions. – Alex_28 Mar 12 '24 at 16:33
  • @Alex_28 https://math.stackexchange.com/questions/1287393/if-you-draw-two-cards-what-is-the-probability-that-the-second-card-is-a-queen – JMoravitz Mar 12 '24 at 16:33
  • Thank you. Is there any other way i could define $X$ that would make expectation and variance easier to evaluate? Does $X$ follow some known distribution? – Alex_28 Mar 12 '24 at 16:43
  • @Alex_28 No, $X$ is not going to follow some well-known distribution. The alluded to current approach is the suggested approach and is already relatively easy to evaluate. Much better than having to break all the way into cases of how all five balls behave. Calculating expectation here can be done in one's head. For variance, you just have to look at the different products of two values... $1,2,5,4,10,25$. – JMoravitz Mar 12 '24 at 16:48
  • $E[X^2]=\sum_{i=1}^5E[X_i]+2\sum_{i=1}^4\sum_{j>i}^5E[X_iX_j]$, so to calculate $E[X_iX_j]$ i need the joint probability of $X_i$ and $X_j$ and, again, they are not indipendent. Am I missing something? – Alex_28 Mar 12 '24 at 17:37
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    $X_1X_2$ will take value $1$ when both balls were $1$, will take value $2$ when one ball is a $1$ and the other a $2$... will take value $4$ when both balls were $2$, and so on... You already showed you could perform calculations when talking about five balls, four of which being $1$'s and one being a $2$... now we are talking about just two balls. Why could you do it then and not now? How is this such a different problem in your mind? – JMoravitz Mar 12 '24 at 17:40