Question

A coin is weighted so that there is a 58.2% chance of it landing on heads when flipped. The coin is flipped 12 times. Find the probability that at most 8 of the flips resulted in "heads".

Answer 1

The probability that at most 8 of the flips resulted in "heads" when a coin is flipped 12 times with a 58.2% chance of landing on heads can be calculated using the binomial probability formula. The probability is approximately 1.394884.

Step-by-step explanation:

To find the probability that at most 8 of the flips resulted in "heads", we need to find the sum of the probabilities of getting 0, 1, 2, 3, 4, 5, 6, 7, and 8 heads. The probability of getting k heads in n flips is given by the binomial probability formula: P(k) = C(n, k) * p^k * (1-p)^(n-k), where C(n, k) is the number of ways to choose k objects from a set of n objects, p is the probability of heads, and (1-p) is the probability of tails.

Given that there is a 58.2% chance of the coin landing on heads, p = 0.582. Using the formula, we can calculate the probabilities for each value of k and sum them up:

1. P(0) = C(12, 0) * 0.582^0 * (1-0.582)^(12-0) = 0.001446
2. P(1) = C(12, 1) * 0.582^1 * (1-0.582)^(12-1) = 0.014526
3. P(2) = C(12, 2) * 0.582^2 * (1-0.582)^(12-2) = 0.069660
4. P(3) = C(12, 3) * 0.582^3 * (1-0.582)^(12-3) = 0.194125
5. P(4) = C(12, 4) * 0.582^4 * (1-0.582)^(12-4) = 0.332086
6. P(5) = C(12, 5) * 0.582^5 * (1-0.582)^(12-5) = 0.356530
7. P(6) = C(12, 6) * 0.582^6 * (1-0.582)^(12-6) = 0.259938
8. P(7) = C(12, 7) * 0.582^7 * (1-0.582)^(12-7) = 0.126271
9. P(8) = C(12, 8) * 0.582^8 * (1-0.582)^(12-8) = 0.040322

The probability of getting at most 8 heads is the sum of these probabilities: P(at most 8) = P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6) + P(7) + P(8) = 0.001446 + 0.014526 + 0.069660 + 0.194125 + 0.332086 + 0.356530 + 0.259938 + 0.126271 + 0.040322 = 1.394884