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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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If this is homework, please add the homework tag. 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). –  Dilip Sarwate May 25 '12 at 13:11
    
I Know that take a white ball is: $\frac{4}{10}$ and black is: $\frac{6}{10}$. –  mastergoo May 25 '12 at 13:54

2 Answers 2

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) ? –  mastergoo May 25 '12 at 13:47
    
I Know that take a white ball is: $\frac{4}{10}$ and black is: $\frac{6}{10}$. –  mastergoo May 25 '12 at 13:49
    
@mastergoo: edited –  Ilya May 25 '12 at 13:54
    
Soh I'll make a table corresponding. –  mastergoo May 25 '12 at 14:00
    
Xi = 3W = $P(w,w,w)$ = $0,4*0.4*0.4 = 0.064$ –  mastergoo May 25 '12 at 14:02

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.

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