Suppose that you flip a coin 11 times. What is the probability that you achieve at least 4 tails?

this is a binomial probability

... p(h) = p(t) = .5

(h + t)^11 = h^11 + 11 h^10 t + 55 h^9 t^2 + 165 h^8 t^3 ...

p(at least 4 tails) is 1 minus the sum of the 1st four terms of the expansion
... 1 - [.5^11 * (1 + 11 + 55 + 165)] = .887

To find the probability of getting at least 4 tails when flipping a coin 11 times, we can use the binomial probability formula:

P(X ≥ k) = 1 - P(X < k)

where P(X ≥ k) is the probability of achieving at least k successes, P(X < k) is the cumulative probability of achieving less than k successes, and X follows a binomial distribution.

In this case, we want to find the probability of achieving at least 4 tails, so k = 4.

Now, let's calculate the probability of achieving less than 4 tails first.

The probability of getting exactly k tails in n coin flips can be calculated using the binomial probability formula:

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

Where (n C k) represents combinations, p is the probability of getting tails in a single coin flip and (1-p) is the probability of getting heads in a single coin flip.

The number of combinations (n C k) is calculated as:
(n C k) = n! / (k! * (n-k)!)

In this case, n = 11 (number of coin flips) and p = 0.5 (probability of getting tails in a single coin flip).

Let's calculate the probability of getting less than 4 tails:

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

P(X < 4) = [(11 C 0) * (0.5^0) * (1-0.5)^(11-0)] + [(11 C 1) * (0.5^1) * (1-0.5)^(11-1)] + [(11 C 2) * (0.5^2) * (1-0.5)^(11-2)] + [(11 C 3) * (0.5^3) * (1-0.5)^(11-3)]

Using the formula, we can calculate the value of each term.

(11 C 0) = 1
(11 C 1) = 11
(11 C 2) = 55
(11 C 3) = 165

P(X < 4) = [(1 * 1 * 0.5^0 * 0.5^11)] + [(11 * 1 * 0.5^1 * 0.5^10)] + [(55 * 1 * 0.5^2 * 0.5^9)] + [(165 * 1 * 0.5^3 * 0.5^8)]
= [1 * 1 * 0.5^11] + [11 * 1 * 0.5^10] + [55 * 1 * 0.5^9] + [165 * 1 * 0.5^8]
= 0.00048828125 + 0.00439453125 + 0.017578125 + 0.04296875
= 0.0654296875

Now, we can calculate the probability of achieving at least 4 tails:

P(X ≥ 4) = 1 - P(X < 4)
= 1 - 0.0654296875
≈ 0.9345703125

Therefore, the probability of achieving at least 4 tails when flipping a coin 11 times is approximately 0.9346 or 93.46%.

To find the probability of achieving at least 4 tails when flipping a coin 11 times, we need to calculate the individual probabilities of getting 4 tails, 5 tails, 6 tails, 7 tails, 8 tails, 9 tails, 10 tails, and 11 tails, and then add them up.

The probability of getting tails on a single coin flip is 1/2, assuming a fair coin. The probability of getting heads is also 1/2.

For any specific number of tails, we can use the concept of binomial probability. The formula for the binomial probability of achieving exactly k successes (in this case, tails) in n trials (in this case, coin flips) is:

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

Where:
- C(n, k) is the binomial coefficient, equal to n! / (k! * (n - k)!),
- p is the probability of success (getting tails in this case),
- n is the number of trials (coin flips), and
- k is the number of successes (number of tails).

Given that we want to find the probability of getting at least 4 tails, we need to sum up the probabilities of getting 4, 5, 6, 7, 8, 9, 10, and 11 tails. This can be done as follows:

P(X >= 4) = P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11)

Now we can calculate each individual probability and sum them up:

P(X = 4) = C(11, 4) * (1/2)^4 * (1 - 1/2)^(11 - 4)
P(X = 5) = C(11, 5) * (1/2)^5 * (1 - 1/2)^(11 - 5)
P(X = 6) = C(11, 6) * (1/2)^6 * (1 - 1/2)^(11 - 6)
P(X = 7) = C(11, 7) * (1/2)^7 * (1 - 1/2)^(11 - 7)
P(X = 8) = C(11, 8) * (1/2)^8 * (1 - 1/2)^(11 - 8)
P(X = 9) = C(11, 9) * (1/2)^9 * (1 - 1/2)^(11 - 9)
P(X = 10) = C(11, 10) * (1/2)^10 * (1 - 1/2)^(11 - 10)
P(X = 11) = C(11, 11) * (1/2)^11 * (1 - 1/2)^(11 - 11)

Finally, we add up all these probabilities to get the total probability of achieving at least 4 tails.

I'll calculate the probabilities and give you the final answer.

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