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An urn contains four white balls and six black. Three balls are drawn with replacement. Let $x$ be the number of white balls. Calcaulate $E (x)$, $VAR(x)$ and $\sigma x$.

I don't know how to calculate $E(x) =\sum\limits_{i=1}^{n} X_i P(X_i)$.

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    Hint: Number and name the balls as $W_1, W_2, W_3, W_4$ and $B_1, B_2, \ldots, B_6$, and make a list of all possible outcomes of three draws with replacement. Then figure out the corresponding value of $X$ for each outcome to deduce the probability mass function of $X$ and proceed. (There are other ways of doing this problem if you know about the binomial distribution that I won't get into).2012-05-25
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    I Know that take a white ball is: $\frac{4}{10}$ and black is: $\frac{6}{10}$.2012-05-25

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Let me give you some hints: let $X$ be a number of white balls

  1. Use combinatorics to find $p_0 = P(X = 0)$, $p_1 = P(X = 1)$, $p_2 = P(X = 2)$, $p_3 = P(X = 3)$. For example, the probability $p_0$ that there are no white balls at all is $b^3$ where $b = \frac{6}{10}$ is a probability to draw a black ball.

  2. For $E(X)$ you have $E(X) = \sum\limits_{k=0}^3 k\cdot p_k = p_1+2p_2+3p_3$.

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    So, how can I calcule th ecombinatories of P(X = 0) ?2012-05-25
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    I Know that take a white ball is: $\frac{4}{10}$ and black is: $\frac{6}{10}$.2012-05-25
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    @mastergoo: edited2012-05-25
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    Soh I'll make a table corresponding.2012-05-25
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    Xi = 3W = $P(w,w,w)$ = $0,4*0.4*0.4 = 0.064$2012-05-25
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    Xi = 2W = $P(w,w,p)$ = $0,4*0.4*0.6 = 0.096$2012-05-25
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    Xi = 1W = $P(w,p,p)$ = $0,4*0.6*0.6 = 0.144$2012-05-25
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    Xi = 0W = $P(p,p,p)$ = $0,6*0.6*0.6 = 0.216$2012-05-25
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    @mastergoo: For $2$ white, $1$ black there is $wwb$, $wbw$, $bww$. Each has probability $(0.4)^2(0.6)$, so probability of $2$ white $1$ black is $3(0.4)^2(0.6)$. Same issue with $1$ white $2$ black.2012-05-25
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    @mastergoo: two whites is exactly one black, isn't it?2012-05-25
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    E(X = 3) = (0.064 * 3) + (0.096 * 2) + (0.144) = 0.528, this is correctly?2012-05-25
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    @mastergoo: what is $E(X = 3)$? there is only $E(X)$2012-05-25
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    Ok The answer of A) is: (3 * 0.064) + (2*0.0288) + (0.432) + 0 = 1,2 ?2012-05-25
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    @mastergoo: seems to be right2012-05-25
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    OK, Thanks. Can you help-me to solve VAR(X) now?2012-05-25
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    @Mastergoo: sure, just use the formula: $Var(X) = E[(X - E(X))^2] = \sum\limits_{k=0}^3 (k-1.2)^2\cdot p_k$2012-05-25
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    OK, what is $K$ and what is $Pk$ ?2012-05-25
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    @mastergo it is an index: just like $i$ in the formula in your question. You can use $i$ instead of $k$2012-05-25
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    Oh I understood, sorry the inconvenience.2012-05-25
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    @mastergoo: no problem, you're welcome2012-05-25
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Because you use replacement the probability of white will be 4/10 =2/5=0.40 each time you draw. Each draw is independent of previous ones so for a given sequnce of white and black you can get the probability by multiplication. So for example the sequence W B W has probability that is the following product: (2/5)x(3/5)x(2/5). Since this os the set up the number of w balls drwn in three draws has a binomial distribution with p=2/5 and n=3. With all those hints you should be able to solve this.