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

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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, $$ \begin{aligned} P(Y_1 + Y_2 + Y_3 \geq 2) &= \binom{3}{2}(1/2)^{2}(1/2)^{3-2} + (1/2)^3 \\ &= 1/2 \end{aligned} $$

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  • $\begingroup$ It's a good explanation, but usually "less than 4" is interpreted as being equal to 1, 2, or 3. $\endgroup$ – whuber Apr 11 '19 at 14:45
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    $\begingroup$ Thank you for pointing it out. I overlooked that. Made the edit. $\endgroup$ – Mr. Wayne Apr 11 '19 at 14:49
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$1-\text{"all dices show 4 or more"} - \text{"exact two dices shows 4 or more"}$ $=1-\left(\frac{1}{2}\right)^{3}-\left(\begin{array}{c}3\\2\end{array}\right)\frac{1}{2}\left(\frac{1}{2}\right)^{2}=\frac{1}{2}$

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