Suppose a biased coin comes up heads 40% of the time and tails 60% of the time. What is the probability of getting between 2 and 4 heads in 10 flips? 

We can model this problem using the **binomial distribution**, as we are dealing with a fixed number of independent trials (coin flips), where each trial has two possible outcomes (heads or tails). The probability of getting \( k \) heads in \( n \) flips is given by the binomial probability formula:

\[
P(k \text{ heads}) = \binom{n}{k} p^k (1-p)^{n-k}
\]

Where:
- \( \binom{n}{k} \) is the binomial coefficient, representing the number of ways to choose \( k \) heads out of \( n \) flips: \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \).
- \( p \) is the probability of heads (0.40 in this case).
- \( n \) is the number of flips (10 in this case).
- \( k \) is the number of heads we want.

You are asked to find the probability of getting between 2 and 4 heads (inclusive), so we need to calculate the probability for \( k = 2, 3, 4 \) and then sum them up.

### Step 1: Calculate for \( k = 2 \)
\[
P(2 \text{ heads}) = \binom{10}{2} (0.4)^2 (0.6)^8 = \frac{10!}{2!(10-2)!} (0.4)^2 (0.6)^8
\]
\[
= 45 \cdot (0.16) \cdot (0.0168) = 45 \cdot 0.002688 = 0.12096
\]

### Step 2: Calculate for \( k = 3 \)
\[
P(3 \text{ heads}) = \binom{10}{3} (0.4)^3 (0.6)^7 = \frac{10!}{3!(10-3)!} (0.4)^3 (0.6)^7
\]
\[
= 120 \cdot (0.064) \cdot (0.028) = 120 \cdot 0.001792 = 0.21446
\]

### Step 3: Calculate for \( k = 4 \)
\[
P(4 \text{ heads}) = \binom{10}{4} (0.4)^4 (0.6)^6 = \frac{10!}{4!(10-4)!} (0.4)^4 (0.6)^6
\]
\[
= 210 \cdot (0.0256) \cdot (0.0467) = 210 \cdot 0.001196 = 0.25084
\]

### Step 4: Total Probability

Now, we sum the probabilities for 2, 3, and 4 heads:
\[
P(2 \leq \text{heads} \leq 4) = P(2 \text{ heads}) + P(3 \text{ heads}) + P(4 \text{ heads})
\]
\[
P(2 \leq \text{heads} \leq 4) = 0.12096 + 0.21446 + 0.25084 = 0.58626
\]

Thus, the probability of getting between 2 and 4 heads in 10 flips is approximately **0.586** or **58.6%**.

Would you like further details on any step or have any questions?

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Here are 5 related questions to extend this topic:
1. What is the probability of getting exactly 5 heads in 10 flips?
2. How would the probability change if the coin were fair (50% heads and 50% tails)?
3. What is the expected number of heads in 10 flips of this biased coin?
4. How would you calculate the variance and standard deviation for the number of heads in 10 flips?
5. What is the probability of getting at least 7 heads in 10 flips?

**Tip**: For binomial probabilities, using a scientific calculator or software tools can speed up the process, especially for larger values of \(n\) and \(k\).

Suppose a biased coin comes up heads 40% of the time and tails 60% of the time. What is the probability of getting between 2 and 4 heads in 10 flips?

Explore the probability of getting between 2 and 4 heads in 10 flips of a biased coin with a 40% chance of heads and 60% chance of tails. This solution uses the binomial distribution and covers key concepts in probability and statistics. Suitable for students in Grades 9-12.

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