Final answer:
The probability of exactly two out of five 6-sided dice showing a 1 or 2 is found using the binomial probability formula and is 80/243.
Step-by-step explanation:
The question involves calculating the probability that exactly two out of five 6-sided dice show a 1 or a 2.
Each die has a 1/3 chance of showing 1 or 2, and 2/3 chance of showing another number.
The probability of exactly two dice showing 1 or 2 can be calculated using the binomial probability formula:
P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))
where:
- C(n, k) = Number of combinations of n items taken k at a time
- n = Total number of trials (in this case, 5 dice)
- k = Number of successful trials (in this case, 2 dice showing 1 or 2)
- p = Probability of a single successful trial (1 or 2 showing on a die, which is 1/3)
- 1-p = Probability of a single unsuccessful trial
Using this formula, we find:
P(X=2) = C(5, 2) * (1/3)^2 * (2/3)^3
C(5, 2) is the number of ways to pick 2 successful dice out of 5, which is 10. Therefore:
P(X=2) = 10 * (1/9) * (8/27) = 10 * (8/243) = 80/243
The probability of exactly two dice showing a 1 or 2, when five 6-sided dice are rolled is 80/243.