What is the probability of throwing two dice and getting the sum of the fallen numbers greater than 3? [closed]

Do you know how to calculate this probability? The total number of possible outcome is $12$. Numbers that is greater than $3$ is $4,5,6$. For 2 dices that would be $6/12$ or $1/2$.

Is this a correct answer?

closed as off-topic by Xander Henderson, Namaste, Jendrik Stelzner, Gibbs, Deepesh MeenaSep 12 '18 at 7:23

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• As alluded to in a comment of mine on another of your posts, you should be careful when using ratios of number of "good" outcomes divided by total number of outcomes only in situations where the outcomes are equally likely to occur. The eleven possible sums of two fair six sided dice are not equally likely to occur. (The sum of $2$ only occurs with probability $\frac{1}{36}$ while the sum of $7$ occurs with probability $\frac{6}{36}=\frac{1}{6}$). Your sample space should be the 36 equally likely possible ordered pairs of numbers on the dice. – JMoravitz Sep 11 '18 at 19:07

Its easier to figure out the probability of getting sum less than $4$, which is only three cases out of $36$: $(1,1); (1,2); (2,1)$. Thus the probability of getting sum greater than three is $\large{1-\frac{3}{36}=\frac{11}{12}}$
• Why do you use $1$ to minus the probability of getting the sum less than 4? – James Warthington Sep 11 '18 at 21:10
• @JamesWarthington: we have two mutually exclusive cases: $A$) sum is less than four and $B$) the sum is grater than three. These two cases cover all possible outcomes, thus $P(A \ or \ B)=1=P(A)+P(B)$. You need to think about the number of outcomes when we throw two dice. There are 36 possible outcomes, 3 of them cover case $A$ and the remaining $36-3=33$ cases cover case $B$. So the probability of $B$ is a ratio of favorable outcomes to the number of all outcomes, $33/36=11/12$ – Vasya Sep 12 '18 at 2:31