This is a mixture binomial distribution question. I know how to get the $\mu$ and $\sigma^2$ of the mixture, but I am not sure how to use it to get the probably of specific number.

States that $X_1=B(2,0.52), X_2=B(3,0.41), X_3=B(4,0.38)$ are binomial distributions with 43%, 36%, 21% users respectively. Find the probability that occurrence is more than 2. The question don't state whether the variables are independence of each other. I will assume this is a mixture distribution question.

Binomial distribution: https://en.wikipedia.org/wiki/Binomial_distribution

Reference for mixture distribution: https://en.wikipedia.org/wiki/Mixture_distribution

My solution:
$X_1 = B(2, 0.52), E(X_1)=1.04, E(X_1^2)=1.5808, Var(X_1)=0.4992$

$X_2 = B(3, 0.41), E(X_2)=1.23, E(X_2^2)=2.2386, Var(X_2)=0.7257$

$X_3 = B(4, 0.38), E(X_3)=1.52, E(X_3^2)=3.2528, Var(X_3)=0.9424$

I see S as a mixture distribution of $X_1, X_2, X_3$

$P(S=0) = 0.43P(X_1=0) + 0.36P(X_2=0) + 0.21P(X_3=0)$

$E(S) = 0.43E(X_1) + 0.36E(X_2) + 0.21E(X_3)=1.2092$

$E(S^2)=0.43E(X_1^2) + 0.36E(X_2^2) + 0.21E(X_3^2)=2.168728$


How do I get $P(S<=2)$? Do I do $P(\frac{S-1.2092}{\sqrt{0.706563}}<=\frac{2-1.2092}{\sqrt{0.706563}})$? Is normal distribution method of getting the probability correct?

Or should I do $P(S<=2) = 1-P(S>2)$ where $P(S>2)=0.43P(X_1>2)+0.36P(X_2>2)+0.21P(X_3>2)$


2 Answers 2


First $-$ your calculation of $\text{VAR}(S)$ is wrong $-$ the coefficients have to be squared when $S$ is a linear combination (assuming of course that $X_1,X_2,X_3$ are independent random variables). Second $-$ the normal approximation does not seem to be accurate with such small numbers as $2,3,4$. I would use "brute force" calculation $-$ $X_1$ can be 0,1,2 ; $X_2$ can be $0,1,2,3$ ; $X_3$ can be $0,1,2,3,4$. There are 60 possible results - see which ones result in $S\leq 2$ and add up their probabilities (each probability in the sum will be a product of three binomial probabilities). Looking at the numbers, it will be easier to calculate $P(S>2)$ and then subtract it from $1$. There are less combinations which will result in $S>2$.

  • $\begingroup$ I got $Var(S) = E(S^2)-(E(S))^2$, what do you mean by squaring the coefficients? $\endgroup$
    – shawnngtq
    Sep 10, 2017 at 1:48

Can you find the probability that $P(X_i ≤ 2)$ for each $i$? If so, it's straightforward to use the law of total probability to compute $P(S ≤ 2)$, similarly to how you propose to find $P(S > 2)$. It should be clear that the answer you get this way is exact. The normal approximation you propose doesn't make a lot of sense because $S$ is not defined as a sum of random variables.

  • $\begingroup$ I'm a little confused by your answer. As a mixture, the possible values taken by S will be all the possible values taken by the components (the $X_i$). So S will take values in $\{0,1,2,3,4\}$ ... i.e. always integer. Are you sure you understood the question? $\endgroup$
    – Glen_b
    Sep 10, 2017 at 8:08
  • $\begingroup$ @Glen_b The question has been edited. It originally defined $S$ as $S = 0.43X_1 + 0.36X_2 + 0.21X_3$, rather than a conventional mixture distribution. $\endgroup$ Sep 10, 2017 at 14:17
  • $\begingroup$ Ah. I see. However, every version of the question opens with "This is a mixture binomial distribution question". The first version (the one up at the time you answered) does contain the line $S = 0.43X_1 + 0.36X_2 + 0.21X_3$ but in the context of that opening statement, one would have to be cautious about interpreting that at face value (since it would not be a mixture interpreted the way you did). $\endgroup$
    – Glen_b
    Sep 10, 2017 at 16:41
  • $\begingroup$ @Glen_b I figured that the equation would be likelier to be copied from the textbook exactly, and hence would be more trustworthy, than the words. Looks like I guessed wrong. I'll edit my answer. $\endgroup$ Sep 10, 2017 at 17:09

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