Probability problem with 3 dice

Question

I just wanted to make sure my reasoning was right. The question is: 3 dice with 6 sides are rolled. I must say whether the following statement is true or false: The probability that 3 times the same number will come up is less than the probability that at least a number less than or equal to 2 will come up. $P(A= "\text{3 times same number}")=6*(\frac{1}{6})^3=\frac{1}{36}$

$P(B="\text{No number less or equal to 2}")=1-(P(\text{at least a 1 will come up})+P(\text{at least a 2 will come up}))=1-(2-(P(\text{1 will not come up})+P(\text{2 will not come up}))=1-(2-(\frac{5}{6})^3)=\frac{17}{108}$

$P(B)>P(A)$ so the statement is true.

Is there something wrong? Is there a faster way to get to the result?

Answer 1

You have a typo in your last formula, it should be $1−(2−2(\frac{5}{6})^3)$. That being said, your mistake is that you consider the events "(1 will not come up)" and "(2 will not come up)" as independent, while in reality, they are not. If no 2 comes up, it will be more likely that one of the die is a "1". The correct probability for case B is therefore $1-P("\text{all 3 dice >2}") = 1-(\frac{2}{3})^3 = \frac{19}{27}$.

Answer 2

You could also work with the complementary event $$P(\text{“at least one number less or equal to two”})=1-P(\text{“all numbers are }>2)=1-\left(\frac46\right)^3.$$