I roll 12 4-sided dice. What is the probability that I roll at least one 2? What is the probability that I roll exactly three 2s?
So i take it, on any give roll, the probability of rolling a 2 is .25 (25%); the probability of rolling something else must be .75
The probability of rolling at least one 2 is: one minus the probability of never rolling a two. And, with 12 chances, that probability is .75^12
To calculate the probability of rolling exactly three 2s, first count the numbe of ways can happen. That is, calculate 12-choose-3. This will be (10*11*12)/(1*2*3) = 220. So, the probability of rolling exactly three 2s will be 220*(.75^9)*(.25^3) = 25.8%
To calculate the probability that you roll at least one 2, you need to find the probability of never rolling a 2 and subtract it from 1. Since the probability of rolling something other than a 2 on any given roll is 0.75, the probability of never rolling a 2 in 12 rolls is (0.75)^12. Therefore, the probability of rolling at least one 2 is 1 - (0.75)^12.
To calculate the probability of rolling exactly three 2s, you first need to find the number of ways that can happen. This can be calculated using the combination formula, also known as "n choose r". In this case, it would be 12 choose 3, which is (12!)/(3!*(12-3)!), which simplifies to (10*11*12)/(1*2*3) = 220.
Once you have the number of ways that you can roll exactly three 2s, you multiply it by the individual probabilities. The probability of rolling a 2 is 0.25, so the probability of rolling exactly three 2s would be 220 * (0.75)^9 * (0.25)^3.
Therefore, the probability of rolling at least one 2 is 1 - (0.75)^12, and the probability of rolling exactly three 2s is 220 * (0.75)^9 * (0.25)^3, which is approximately 25.8%.