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What are the odds that I will get at most one 2 in rolling three

dice simultaneously?
Use the classical probability AND binomial
theorem!!!!

User Tgmath
by
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1 Answer

6 votes

Answer:

Explanation:

To calculate the odds of getting at most one 2 in rolling three dice simultaneously, we can use both classical probability and the binomial theorem.

Classical Probability Approach:

In classical probability, we determine the number of favorable outcomes and divide it by the total number of possible outcomes. Let's analyze the possible outcomes:

Total number of outcomes when rolling three dice simultaneously: 6^3 = 216 (each die has 6 possible outcomes: 1, 2, 3, 4, 5, 6)

Number of favorable outcomes:

To get at most one 2, we need to consider two cases:

No 2's: We can have any number except 2 on each of the three dice. For each die, there are 5 possibilities (1, 3, 4, 5, 6). So, the number of outcomes with no 2's is 5^3 = 125.

Exactly one 2: We can have a 2 on any one of the three dice, and any number except 2 on the other two dice. So, the number of outcomes with exactly one 2 is 3 * 5^2 = 75.

Total number of favorable outcomes: 125 + 75 = 200

Probability (odds) of getting at most one 2: favorable outcomes / total outcomes = 200 / 216 ≈ 0.9259 or 92.59%

Binomial Theorem Approach:

We can also use the binomial theorem to calculate the probability of getting at most one 2.

The binomial probability formula states that for a given number of trials (n) and the probability of success in a single trial (p), the probability of getting exactly k successes is given by:

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

Where C(n, k) represents the number of ways to choose k successes from n trials, and p represents the probability of success in a single trial.

In our case, n = 3 (number of dice), k = 0 or 1 (at most one 2), and p = 1/6 (the probability of getting a 2 on a single die).

Using the binomial formula, we can calculate the probability of getting at most one 2:

P(X ≤ 1) = P(X = 0) + P(X = 1)

P(X = 0) = C(3, 0) * (1/6)^0 * (5/6)^(3 - 0) = 1 * 1 * (5/6)^3 = (5/6)^3

P(X = 1) = C(3, 1) * (1/6)^1 * (5/6)^(3 - 1) = 3 * (1/6) * (5/6)^2 = 3 * (5/6)^2

P(X ≤ 1) = (5/6)^3 + 3 * (5/6)^2

Evaluating this expression will give you the probability of getting at most one 2 using the binomial theorem.

Note: The classical probability approach and the binomial theorem approach should yield the same result, as they are mathematically equivalent methods of calculating probabilities.

User Thomas Li
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