What is the probability that 9 rolls of a fair die will show two fives?
To calculate the probability of obtaining two fives in nine rolls of a fair die, we need to determine the total number of possible outcomes and the number of favorable outcomes.
First, let's consider the total number of possible outcomes for nine rolls of a fair die. Since each roll has six possible outcomes (numbers 1 to 6), the total number of possible outcomes for nine rolls can be calculated as 6^9, which is 1,000,000, as each roll is an independent event.
Next, we need to find the number of favorable outcomes, i.e., the number of ways two fives can be drawn in nine rolls.
To calculate this, we can use the concept of combinations. The formula to calculate combinations is given by:
C(n, r) = n! / (r!(n-r)!)
In this case, we want to find the number of ways we can choose 2 positions out of 9 for the fives to appear, while the remaining positions can be anything other than fives.
Using the combination formula, we have:
C(9, 2) = 9! / (2!(9-2)!) = 36
This means there are 36 ways to choose two positions out of nine for the fives to appear.
Now, we can calculate the number of ways the remaining seven rolls can result in any number other than five. Since each of the seven rolls has five possible outcomes (numbers 1 to 4 and 6), the total number of outcomes for the remaining rolls is 5^7, which is 78125.
Lastly, we multiply the number of favorable outcomes by the number of possible outcomes for the remaining rolls:
36 * 78125 = 2,812,500
Therefore, the probability of obtaining two fives in nine rolls of a fair die is 2,812,500 / 1,000,000, which simplifies to 2.8125 or 2.81% (rounded to two decimal places).
So, the probability is approximately 2.81%.