We roll three fair six-sided dice.
(a) What is the probability that at least two of the dice land on a number greater than $4$?
(b) What is the probability that we roll a sum of at least $15$?
(c) Now we roll three fair dice $n$ times. How large need $n$ be in order to guarantee a better than $50\%$ chance of rolling a sum of at least $15$, at least once?
Of course, the size of the sample is $6^3=216$.
For (a), my thinking is that 2 options (5 and 6) on the first die, 2 options on the 2nd die, 6 options on the 3rd die, multiplied by 3 orders of the dices resultings; So $2\cdot2\cdot6\cdot3 = 72$. Which gets me $\frac{72}{216}$. However I'm not sure if I'm correct.
For (b) and (c), I am not sure how to even approach the question.
Best Answer
PART A
Separate into 2 cases, 2 dice and 3 dice greater than 4.
2 dice greater than 4 $=3\cdot(2\cdot2\cdot4)=48$. The three represent the possible choices of the 2 dice from 3 dice.
3 dice greater than 4 $=2\cdot2\cdot2=8$.
$$P(A)={48+8\over216}=\frac7{27}$$