Question

Hello - have I calculated the following correctly? Comments appreciated

Using a normal dice, rolled 5 times, the probability of rolling 2 sixes would be 1.61%

i.e 1/6 x 1/6 x 5/6 x 5/6 x 5/6 =125/7776

Using a normal dice, rolled 5 times, the probability of not rolling a six at all would be 40.19%

i.e 5/6 x 5/6 x 5/6 x 5/6 x 5/6 = 3125/7776

Answer 1

Answer on Question #43890-Math-Statistics and Probability

Using a normal dice, rolled 5 times, the probability of rolling 2 sixes would be 1.61%

i.e $1/6 \times 1/6 \times 5/6 \times 5/6 \times 5/6 = 125/7776$

Using a normal dice, rolled 5 times, the probability of not rolling a six at all would be 40.19%

i.e $5/6 \times 5/6 \times 5/6 \times 5/6 \times 5/6 = 3125/7776$

Solution

Let's use Bernoulli distribution.

Using a normal dice, rolled 5 times, the probability of rolling 2 sixes would be

$$P(2 \text{ sixes}) = \frac{5!}{(5 - 2)! \cdot 2!} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^{5 - 2} = \frac{5!}{3! \cdot 2!} \cdot \frac{5^3}{6^5} = 10 \cdot \frac{125}{7776} = 0.1607 = 16.07\%,$$

where $n! = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot n$, $\frac{1}{6}$ is the probability of rolling sixe rolled 1 time, $\frac{5}{6}$ is the probability of not-rolling sixe rolled 1 time.

Using a normal dice, rolled 5 times, the probability of not rolling a six at all would be

$$P(\text{not rolling a six}) = \frac{5!}{(5 - 0)! \cdot 0!} \left(\frac{1}{6}\right)^0 \left(\frac{5}{6}\right)^{5 - 0} = \frac{5!}{5! \cdot 0!} \cdot \left(\frac{5}{6}\right)^5 = 1 \cdot \frac{3125}{7776} = 0.4019 = 40.19\%.$$

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