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Find the probability of throwing a sum of 5 at least 8 times in 11 throws of a pair of fair dice.

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Answer:

See below

Step-by-step explanation:

To find the probability of throwing a sum of 5 at least 8 times in 11 throws of a pair of fair dice, we can use the concept of binomial probability.

First, let's determine the probability of getting a sum of 5 in a single throw. To get a sum of 5, we can have the following combinations: (1, 4), (2, 3), (3, 2), and (4, 1). Each combination has a probability of 1/36 (1 out of 36 possible outcomes) since there are 36 possible outcomes when rolling a pair of fair dice.

Now, we need to calculate the probability of getting a sum of 5 at least 8 times in 11 throws. To do this, we'll use the binomial probability formula:

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

Where:

- P(X = k) is the probability of getting exactly k successes,

- n is the total number of trials,

- k is the number of desired successes,

- p is the probability of success in a single trial,

- (n C k) is the binomial coefficient, calculated as n! / (k! * (n - k)!)

In our case, n = 11 (the total number of throws), k = 8 (the desired number of times to get a sum of 5), and p = 1/36 (the probability of getting a sum of 5 in a single throw).

Now, let's calculate the probability using the formula:

P(X >= 8) = P(X = 8) + P(X = 9) + ... + P(X = 11)

P(X = 8) = (11 C 8) * (1/36)^8 * (35/36)^(11 - 8)

P(X = 9) = (11 C 9) * (1/36)^9 * (35/36)^(11 - 9)

P(X = 10) = (11 C 10) * (1/36)^10 * (35/36)^(11 - 10)

P(X = 11) = (11 C 11) * (1/36)^11 * (35/36)^(11 - 11)

Let's calculate each probability:

P(X = 8) = (11 C 8) * (1/36)^8 * (35/36)^3

P(X = 9) = (11 C 9) * (1/36)^9 * (35/36)^2

P(X = 10) = (11 C 10) * (1/36)^10 * (35/36)^1

P(X = 11) = (11 C 11) * (1/36)^11 * (35/36)^0

After calculating each probability, we can add them up to find the final probability:

P(X >= 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11)

Keep in mind that the binomial coefficient (n C k) represents the number of ways to choose k successes from n trials, and it can be calculated as n! / (k! * (n - k)!).

I hope this explanation helps you understand how to find the probability of throwing a sum of 5 at least 8 times in 11 throws of a pair of fair dice using the binomial probability formula. If you have any further questions, feel free to ask.

User Andreas Zwerger
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