Two unbiased dice are thrown 3 times. Find the probability that the sum nine would be obtained once.

To find the probability of obtaining a sum of nine exactly once when throwing two unbiased dice three times, we can break down the problem into a few steps.

Step 1: Calculate the Probability of Getting a Sum of Nine

First, we need to determine the probability of rolling a sum of nine with two dice. The possible combinations that yield a sum of nine are:

• (3, 6)
• (4, 5)
• (5, 4)
• (6, 3)

There are 4 successful outcomes. Since there are a total of 36 possible outcomes when rolling two dice (6 sides on the first die multiplied by 6 sides on the second die), the probability of rolling a sum of nine is:

P(sum of 9) = 4/36 = 1/9

Step 2: Determine the Probability of Not Getting a Sum of Nine

The probability of not rolling a sum of nine is:

P(not sum of 9) = 1 - P(sum of 9) = 1 - 1/9 = 8/9

Step 3: Use the Binomial Probability Formula

We can use the binomial probability formula to find the probability of getting a sum of nine exactly once in three rolls. The formula is:

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.
• p is the probability of success (1/9).
• q is the probability of failure (8/9).
• n is the number of trials (3).
• k is the number of successes (1).

Step 4: Calculate the Probability

Plugging in the values:

C(3, 1) = 3

P(X = 1) = 3 * (1/9)^1 * (8/9)^(3-1)

P(X = 1) = 3 * (1/9) * (8/9)^2

P(X = 1) = 3 * (1/9) * (64/81)

P(X = 1) = 3 * 64 / 729

P(X = 1) = 192 / 729

Final Result

The probability of obtaining a sum of nine exactly once when throwing two unbiased dice three times is:

192/729 or approximately 0.263.