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How do you find the probability of a specific value when you only have the expected value and the function's variance?

For example, I'm asked to find

a) $P\{X = 4\}$ and

b) $P\{X>12\}$

If $E[X] = 7$ and $Var(X) = 2.1$

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You can't, unless you know the distribution of $X$; at best you can provide a bound for the probability. Are you given that $X$ is normally distributed? –  MJD Oct 14 '12 at 4:39
    
Or perhaps that it is uniformly distributed, or something like that? –  MJD Oct 14 '12 at 4:48
    
No, only that $X$ is a binomial random variable. –  Imray Oct 14 '12 at 5:35
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1 Answer 1

up vote 2 down vote accepted

You know that $X \sim Binomial(n,p)$. Hence

$$ P(X=4)=\binom{n}{4}p^{4}(1-p)^{n-4}\\ P(X \geq 13)=\sum_{k=13}^{n}\binom{n}{k}p^{k}(1-p)^{n-k} $$ The latte by the way does not exist in closed form, so you will need to approximate it, e.g. with Markov or Chernoff bounds.

OK, the most interesting part is how to find these paramters, $p$ and $n$. If the variable is Binomial, we know that $$ \mathbf{E}X=np\\ \mathbf{Var}X=np(1-p) $$ You have these tow values above, hence you need to solve the system of equations: $$ np=7\\ np(1-p)=2.1 $$ After you obtain the values for $n,p$ plug them in the equations above.

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I got n = 10, p = .7 Everything adds up. Thanks for the awesome explanation –  Imray Oct 15 '12 at 2:12
    
You are welcome –  Alex Oct 15 '12 at 2:12
    
wait- I got stuck. In the second binomial, k is bigger than n, I'm getting a math error when I do 10 choose 13. ?? –  Imray Oct 15 '12 at 2:15
    
you still online? Can you help me? –  Imray Oct 15 '12 at 2:28
    
If $n=10$ then clearly $P(X>12)=0$ –  Alex Oct 15 '12 at 2:42
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