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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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    No, only that $X$ is a binomial random variable.2012-10-14

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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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    $n$ is the number of experiments, $X$ is the number of times of getting some 'success', e.g. 6 or '3 or more' or whatever. Obviously in 10 trials it is impossible to get $X$ 12 times, so P(X>12) is certainly 0. Check the values for EX and VarX2012-10-15