Question

The probability of rolling a sum of 7 when rolling two dice simultaneously is 0.167. You decide to test that probability by rolling th dice 12 times. What is the probability that exactly 2 of the rolls is a sum of 7?

Answer 1

Answer:

  about 0.296

Explanation:

The probability is found from ...

... C(12, 2)×(1/6)^2×(5/6)^10 ≈ 66×0.0044863 ≈ 0.29609

_____

C(n, k) = n!/(k!(n-k)!)

C(12, 2) = (12·11)/(2·1) = 66 . . . . the number of ways 7s can show up in 12 rolls of the dice.

Answer 2

Answer:

There is a 29.61% probability that exactly 2 of the rolls is a sum of 7.

Explanation:

For each roll of the dice, there are only two possible outcomes. Either the sum is 7, or it is not. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

The probability of rolling a sum of 7 when rolling two dice simultaneously is 0.167. This means that $p = 0.167$

You decide to test that probability by rolling th dice 12 times. What is the probability that exactly 2 of the rolls is a sum of 7?

This is P(X = 2) when n = 12. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{12,2}.(0.167)^{2}.(0.833)^{10} = 0.2961$

There is a 29.61% probability that exactly 2 of the rolls is a sum of 7.