A fair coin is flipped 11 times. Find the probability that more than 3 of the flips turn up tails.

0.873308
0.910482
0.858837
0.808877
0.886719
0.899099

The problem asks for the probability that more than 3 flips turn up tails when a fair coin is flipped 11 times.

### Step-by-step solution:

1. **Model the problem**: Each flip of the coin has two possible outcomes (heads or tails), and we assume a fair coin, so the probability of tails (and heads) on any single flip is \( \frac{1}{2} \).

2. **Use the binomial distribution**:
   The number of tails in 11 flips follows a binomial distribution:
   \[
   X \sim \text{Binomial}(n=11, p=0.5)
   \]
   where \( n = 11 \) is the number of flips, and \( p = 0.5 \) is the probability of getting tails on each flip.

3. **Define the event**: We are interested in the probability that more than 3 flips turn up tails. This can be written as:
   \[
   P(X > 3)
   \]
   This is the complement of the event \( X \leq 3 \), so:
   \[
   P(X > 3) = 1 - P(X \leq 3)
   \]

4. **Calculate \( P(X \leq 3) \)**: 
   We need the cumulative probability of getting 0, 1, 2, or 3 tails, which is:
   \[
   P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
   \]
   Each term can be computed using the binomial probability formula:
   \[
   P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}
   \]
   Where \( \binom{n}{k} \) is the binomial coefficient.

5. **Calculate individual probabilities**:
   For \( X = 0 \):
   \[
   P(X = 0) = \binom{11}{0} (0.5)^0 (0.5)^{11} = 1 \cdot (0.5)^{11} = \frac{1}{2048}
   \]
   For \( X = 1 \):
   \[
   P(X = 1) = \binom{11}{1} (0.5)^1 (0.5)^{10} = 11 \cdot \frac{1}{2048} = \frac{11}{2048}
   \]
   For \( X = 2 \):
   \[
   P(X = 2) = \binom{11}{2} (0.5)^2 (0.5)^9 = 55 \cdot \frac{1}{2048} = \frac{55}{2048}
   \]
   For \( X = 3 \):
   \[
   P(X = 3) = \binom{11}{3} (0.5)^3 (0.5)^8 = 165 \cdot \frac{1}{2048} = \frac{165}{2048}
   \]

6. **Sum the probabilities**:
   \[
   P(X \leq 3) = \frac{1}{2048} + \frac{11}{2048} + \frac{55}{2048} + \frac{165}{2048} = \frac{232}{2048} \approx 0.11328
   \]

7. **Find the complement**:
   \[
   P(X > 3) = 1 - 0.11328 = 0.88672
   \]

The probability that more than 3 flips turn up tails is approximately **0.886719**.

Thus, the correct answer is **0.886719**.

---

Would you like further clarification? Here are 5 related questions to deepen your understanding:

1. How does the probability change if the number of flips is increased to 15?
2. What is the probability of getting exactly 5 tails in 11 flips?
3. How does the binomial distribution apply to real-world situations like coin flipping?
4. Can you compute the expected number of tails in 11 flips?
5. What is the variance of the number of tails in 11 flips?

**Tip**: The binomial distribution is very useful in situations where there are a fixed number of independent trials, each with two possible outcomes (success or failure).

A fair coin is flipped 11 times. Find the probability that more than 3 of the flips turn up tails.

Discover how to calculate the probability of getting more than 3 tails when flipping a fair coin 11 times. Learn the step-by-step process, including the binomial probability formula and complement rule, with a final solution of approximately 0.886719.

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