If we roll a fair dice 3 times, what is the probability that the sum of values of the three rolls is greater than 9?
I approached it like this:
Let X be the result of a single dice roll. Let Sn = X1 + X2 + X3.
We know that E(X) = 7/2 and that V(X) = 35/12
By the rules of expectation, we know that E(Sn) = 21/2 and that V(Sn) = 35/4.
Shouldn't the answer just be P(Z > (9-(21/2))/sqrt(35/4))? For some reason I keep getting a different answer than what the book has.
Best Answer
The problem is, the distribution of $S_n$ is not normal, nor closely approximated by a normal random variable, since $3$ rolls is a small number. Note that if the first two dice give a total of at most $3$ (which can happen in $3$ ways), then no matter what the third die shows, the total will be at most $9;$ this gives $3\cdot 6=18$ ways to get a total of at most $9$ if the first two dice total at most $3$. If the first two dice give a total of $4$ (which can happen in $3$ ways), then we must get at most $5$ on the third die; this gives $3\cdot5=15$ ways to get a total of at most $9$ if the first two dice give a total of $4.$ Similarly, there are (respectively) $4\cdot4=16,$ $5\cdot3=15,$ $6\cdot2=12,$ and $5\cdot1=5$ ways to get a total of at most $9$ if the first two dice give a total of (respectively) $5,6,7,$ and $8.$ There are no other totals on the first two dice that will allow us to get a total of at most $9$ on all three dice. Hence, adding them up, there are $81$ ways to get a total of at most $9$ on the three dice. Since there are $216$ ways to roll, it follows that the probability of a total of more than $9$ is $\frac58=0.625.$ (Why?)
I leave it to you to fill in the details, but let me know if you have any questions.