Three sixes in ten die rolls

Question

A fair die is rolled $10$ times. What is the probability of obtaining exactly $3$ sixes? (Binomial)

Solution

Each roll is an independent trial with success = “a $6$ comes up”, probability $p=1/6$. So $X\sim B(10;\,1/6)$: $P(X=3) = \binom{10}{3}\left(\frac16\right)^3\left(\frac56\right)^7 = 120\cdot\frac{1}{216}\cdot\frac{78\,125}{279\,936} \approx \boxed{0{,}155}.$ About $15{,}5\%$.

Links

Topics: Probability distributions
Concepts: Binomial distribution
Methods: Binomial
Skills: Probability calculation
Exercise type: Probability problem