Binomial Distribution

Question

A fair die will be rolled 12 times. What is the probability that an even number is rolled more than 9 times? Round your answer to four decimal places

Answer 1

We've already established that the binomial distribution will be used for this example because this is a trial based scenario. Since we're using binomial, what's considered a success and what is the probability of it? We are asked to figure out the probability that an even number is rolled. We are rolling a dice, so values from 1-6 are the possibilities. Three of the numbers of the six are even (2, 4, 6), which makes the probability of success equal 50%, or P(Even) = 0.50. That is one component we need to solve this problem.

The other component would be the number of trials. It is mentioned that the die will be rolled 12 times. Therefore, the number of trials would be 12, or n = 12.

Now we have to figure out what we're looking for. In this case, it's the probability that an even number is rolled more than 9 times. We are essentially looking for P(X > 9), with X being binomially distributed. This can also be rewritten as P(X >= 10) since we are dealing with a discrete distribution. This can be further expanded by writing P(X = 10) + P(X = 11) + P(X = 12). So:

P(X > 9) = P(X >= 10) = P(X = 10) + P(X = 11) + P(X = 12)

The formula used for the binomial distribution pdf is P(X = x) = _nC_x * p^x * (1 - p)^n-x. Parameters include the number of successes (n) and probability of success (p). The combination formula expanded would be:

n!

——————

x! (n - x)!

To put it all together, we do:

P(X >= 10) = P(X = 10) + P(X = 11) + P(X = 12) = ₁₂C₁₀ * (0.50)¹⁰ * (0.50)² + ₁₂C₁₁ * (0.50)¹¹ * (0.50)¹ + ₁₂C₁₂ * (0.50)¹² * (0.50)⁰ = 0.016133 + 0.002930 + 0.0002441 = 0.0193.

This means P(X > 9) = 0.0193, or the probability that an even number is rolled more than 9 times is approximately 1.93%.