What's the probability of getting at least two 4s from rolling three dice? I think that the equation would be $\frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{1}{2}$, because each dice can roll a 4, and if I sum them up, I'll get 4 on each dice.
(The dice is fair, in case you were wondering)
 A: Required probability of at least two 4's,
is the probability of getting all three 4's, i.e. $\frac{1}{6^3}=\frac{1}{216}$, added to the probability of getting exactly two 4's, which is:$\frac{15}{216}$ 

(There are fifteen ways to get 2 fours: $441,414,144,442,424,244..$ and so on until $446,464,644$; another way to get the solution is using permutations and combinations: $\frac{(^1C_1)(^1C_1)(^5C_1)+(^1C_1)(^5C_1)(^1C_1)+(^5C_1)(^1C_1)(^1C_1)}{(^6C_1)^3}$)

Therefore your answer is simply: $\frac{1+15}{216}=\frac{2}{27}$
A: First thing you have to take cases with 2 time 4's out of 3 rolls.
Probability = $\frac 16 \times \frac 16 \times \frac 56 + \frac 16 \times \frac 56 \times \frac 16 + \frac 56 \times \frac 16 \times \frac 16$
=$\frac 5{216} + \frac 5{216} + \frac 5{216}$
=$\frac {15}{216}$
Second here at least is used. So may be there is 4's on all dice. So you have to take cases of 3 time 4's.
Probability = $\frac 16 \times \frac 16 \times \frac 16$
=$\frac 1{216}$
Total probability = $\frac {15}{216} + \frac 1{216}$
A: Total number of possibilities are $6^3$ since there are three dices.
Now, there is one way of getting all $4$ in all dices $\geq$ at least two 4's and $15=3{5 \choose 1}$ ways of getting two fours $\geq$ at least two 4's. So...
$$\text{Probability of event} = \frac{16}{216}=\frac {2}{27}$$
