# Rolling a fair dice [closed]

If I roll a dice three times, how do you calculate the probability that at least 2 dice are less than four?

## closed as off-topic by Tim♦Apr 11 at 14:44

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Each toss of a die is independent of one another and uniformly distributed. Denote the number shown on each die $$X_i$$ for $$i=1,2,3$$. Then $$P(X_i \leq 3) = 3/6 = 1/2$$. Let this probability represent the probability of "success" (with respect to a Bernoulli trial). Then the probability of achieving two successes in three trials is found through the binomial distribution. Let $$Y_i = I(X_i \leq 3)$$. Then $$Y_1 + Y_2 + Y_3 \sim Binomial(3, p=1/2)$$. Thus, \begin{aligned} P(Y_1 + Y_2 + Y_3 \geq 2) &= \binom{3}{2}(1/2)^{2}(1/2)^{3-2} + (1/2)^3 \\ &= 1/2 \end{aligned}
$$1-\text{"all dices show 4 or more"} - \text{"exact two dices shows 4 or more"}$$ $$=1-\left(\frac{1}{2}\right)^{3}-\left(\begin{array}{c}3\\2\end{array}\right)\frac{1}{2}\left(\frac{1}{2}\right)^{2}=\frac{1}{2}$$