probability: A six-sided die is rolled 8 times (Multiple Problems - Locked)

#10 A piece of six sided die is rolled 8 times. Determine the probability that exactly five times the die will turn up 1 or 2.
 

One possible outcome is $$x_1, ~ (1 \text{ or } 2), ~ (1 \text{ or } 2), ~ x_2, ~ (1 \text{ or } 2), ~ x_3, ~ (1 \text{ or } 2), ~ (1 \text{ or } 2)$$

Where x₁, x₂, and x₃ are outcomes that neither 1 nor 2. The probability for the above outcome to happen is
$$ \left( \dfrac{4}{6} \right)^3 \left( \dfrac{2}{6} \right)^5 $$

The above outcome may happen or it can happen in different ways provided there are five (1 or 2)s and three non (1 or 2)s. The keyword here is "or", hence, the total number of ways we can roll a die 8 times so that (1 or 2) will appear exactly 5 times is
$$ \binom{8}{5} $$

The probability that exactly 5 times the die will turn up 1 or 2 is
$$ \binom{8}{5} \left( \dfrac{4}{6} \right)^3 \left( \dfrac{2}{6} \right)^5 $$
 

Note: $\displaystyle \binom{8}{5} = \dfrac{8!}{5!(8 - 5)!}$
 

Problem 14:
P₁(getting <20pins)= 50/1000
P₂(getting >20pins)= 950/1000

2 packets are chosen in random, favorable outcomes for one contains lesser than 20 pins and the other more than 20 pins is
(<20pins, >20pins) or (>20pins, <20pins),
Thus,
P₁*P₂ + P₂*P₁= 0.095