Solution: 2018 Winter Midterm - 15
Author: Michiel SmidQuestion
You are given a fair die and roll it 12 times. Define the event
- A = "there are exactly two 6's in the sequence of 12 rolls".
(a)
$12^{2} \cdot \left. 5^{10} \middle/ 6^{12} \right.$
(b)
${12 \choose 2} \cdot \left. 5^{10} \middle/ 6^{12} \right.$
(c)
${12 \choose 2} \cdot \left. 5^{12} \middle/ 6^{12} \right.$
(d)
$12^{2} \cdot \left. 5^{12} \middle/ 6^{12} \right.$
Solution
We need to choose 2 rolls to be 6’s from the 12 rolls: $ \binom{12}{2} $
For each of the 2 rolls, there is a $ \frac{1}{6} $ chance of rolling a 6: $ {( \frac{1}{6})}^2 $
For each of the 10 remaining rolls, there is a $ \frac{5}{6} $ chance of not rolling a 6: $ {( \frac{5}{6})}^{10} $
Pr$ (A) = \binom{12}{2} \cdot {( \frac{1}{6})}^2 \cdot {( \frac{5}{6})}^{10} $
Pr$ (A) = \frac{ \binom{12}{2} \cdot 5^{10}}{6^{12}} $