What is the probability of throwing 2 dice 10 times and obtaining a sum of 6 4 times?

To find the probability of throwing 2 dice 10 times and obtaining a sum of 6 exactly 4 times, we need to use the concept of binomial probability.

The probability of obtaining a sum of 6 when rolling 2 dice is given by the number of favorable outcomes divided by the total number of possible outcomes.

The favorable outcomes occur when the sum of the two dice is 6. We can list these outcomes as follows:

(1, 5), (2, 4), (3, 3), (4, 2), (5, 1).

Thus, there are 5 favorable outcomes.

The total number of possible outcomes when rolling 2 dice is 36 (since each die has 6 possible outcomes, and there are 6x6=36 possible outcomes in total).

Now, we can use the formula for binomial probability:

P(x=k) = (n C k) * p^k * (1-p)^(n-k)

Where:
P(x=k) is the probability of obtaining exactly k successes,
n is the total number of trials,
k is the number of desired successes,
p is the probability of success in each trial.

In this case, n = 10 (we throw the dice 10 times), k = 4 (we want to obtain a sum of 6 exactly 4 times), and p = 5/36 (the probability of obtaining a sum of 6).

Now, we can plug in the values into the formula:

P(x=4) = (10 C 4) * (5/36)^4 * (1 - 5/36)^(10 - 4)

Calculating this expression:

P(x=4) = (10 C 4) * (5/36)^4 * (31/36)^6

P(x=4) = (10! / (4!(10-4)!)) * (5/36)^4 * (31/36)^6

P(x=4) = (10! / (4!6!)) * (5/36)^4 * (31/36)^6

Simplifying the factorials:

P(x=4) = (10! / (4!6!)) * (5^4 / 36^4) * (31^6 / 36^6)

P(x=4) = (10 * 9 * 8 * 7) / (4 * 3 * 2 * 1) * (5^4 / 36^4) * (31^6 / 36^6)

P(x=4) = (10 * 9 * 8 * 7) / (4 * 3 * 2 * 1) * (5^4 * 31^6) / (36^4 * 36^6)

Calculating this expression:

P(x=4) ≈ 0.03186

Answer: The probability of throwing 2 dice 10 times and obtaining a sum of 6 exactly 4 times is approximately 0.03186.