To find the probability of rolling exactly one 3 when a fair 6-sided die is rolled three times, we can use the binomial probability formula. The formula is given by:
P(X = k) = C(n, k) * (p^k) * (q^(n-k))
Where:
- C(n, k) is the number of combinations of n items taken k at a time.
- n is the number of trials (in this case, 3 rolls).
- k is the number of successful outcomes (rolling a 3, which is 1).
- p is the probability of success on a single trial (probability of rolling a 3, which is 1/6).
- q is the probability of failure (not rolling a 3, which is 5/6).
In this scenario:
- n = 3
- k = 1
- p = 1/6
- q = 5/6
Now, we calculate C(3, 1):
C(3, 1) = 3
Putting it all into the formula:
P(X = 1) = C(3, 1) * (1/6)^1 * (5/6)^(3-1)
P(X = 1) = 3 * (1/6)^1 * (5/6)^2
P(X = 1) = 3 * (1/6) * (25/36)
P(X = 1) = 3 * (25/216)
P(X = 1) = 75/216
This simplifies to:
P(X = 1) = 25/72
Therefore, the probability that exactly one 3 is rolled when a fair 6-sided die is rolled three times is 25/72.