What is the probability of getting at least two 1's on 5 dice

Question

If I want to calculate the probability of getting at least one "1" by rolling 5 fair six-sided dice, I know I have to compute the probability of not getting it, hence 5/6. Since the events are independent, the probability is

$$(\frac{5}{6})^5$$

or 0.4, the complementary of which is 0.6.

Now, how can I calculate the probability of getting at least two 1's, since the events are dependent?

Answer 1

Split it into disjoint events, and then add up their probabilities:

• The probability of getting exactly $\color\red2$ ones is $\binom{5}{\color\red2}\cdot\left(\frac16\right)^{\color\red2}\cdot\left(1-\frac16\right)^{5-\color\red2}$
• The probability of getting exactly $\color\red3$ ones is $\binom{5}{\color\red3}\cdot\left(\frac16\right)^{\color\red3}\cdot\left(1-\frac16\right)^{5-\color\red3}$
• The probability of getting exactly $\color\red4$ ones is $\binom{5}{\color\red4}\cdot\left(\frac16\right)^{\color\red4}\cdot\left(1-\frac16\right)^{5-\color\red4}$
• The probability of getting exactly $\color\red5$ ones is $\binom{5}{\color\red5}\cdot\left(\frac16\right)^{\color\red5}\cdot\left(1-\frac16\right)^{5-\color\red5}$

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Hence the probability of getting at least $\color\green2$ ones on $\color\purple5$ dice is:

$$\sum\limits_{n=\color\green2}^{\color\purple5}\binom{\color\purple5}{n}\cdot\left(\frac16\right)^{n}\cdot\left(1-\frac16\right)^{\color\purple5-n}$$

Answer 2

The probability of getting zero 1's is $(\frac{5}{6})^5=\frac{3125}{7776}$

There are 5 ways of getting one 1, so the probability is $5\times\frac{1}{6}\times(\frac{5}{6})^4$

You want the probability of neither of these events:

$$1-(\frac{5}{6})^5-(5\times\frac{1}{6}\times(\frac{5}{6})^4)=\frac{763}{3888}$$