Given $X_1,X_2,...,X_{20}$ are independent variables of the same distribution such that $X_n$ only equal to 2 or 3 and $P(X_n=3)=0.2$. Find the probability of their sum equals to 46.
My guess was to try convolution but it needs a lot of time to calculate. Can you guys advice some theory(laws, distributions and so on) to move a little bit forward?
Thank you.
Since the random variable attains only the values 2 or 3 you can see $X_n = 2+Y_n$ where $P(Y_n=1) = 0.2$ and $P(Y_n=0) = 0.8$. Then your problem is to evaluate $P\left( \sum_{n=1}^{20} 2+Y_n = 46 \right) = P\left( \sum_{n=1}^{20} Y_n = 6 \right)$. Now you must notice that since all $X_n$'s are independent then all $Y_n$ are independent and since $Y_n$ has a Bernoulli distribution their sum has a Binomial distribution, therefore:
$$ \sum_{n=1}^{20} Y_n \sim Bin(20,0.2)$$
And thus $P\left( \sum_{n=1}^{20} Y_n = 6 \right) = {20 \choose 6} (0.2)^6(0.8)^{14}$.