Coin A is flipped 3 times and coin B is flipped 5 times. What is the probability that the number of heads obtained from flipping the two coins is the same?

We are given two coins, A and B, where:

- Coin A is flipped 3 times.
- Coin B is flipped 5 times.

We need to find the probability that the number of heads obtained from flipping Coin A is equal to the number of heads obtained from flipping Coin B.

### Step 1: Identifying possible outcomes
For Coin A:
- The number of heads \( X_A \) obtained from flipping Coin A can range from 0 to 3 (since it is flipped 3 times).

For Coin B:
- The number of heads \( X_B \) obtained from flipping Coin B can range from 0 to 5 (since it is flipped 5 times).

### Step 2: Calculating the probabilities for each outcome
Both coins are fair, so each flip is independent and has a probability of 0.5 for heads and 0.5 for tails. The number of heads follows a **binomial distribution**.

- For Coin A: \( X_A \sim \text{Binomial}(3, 0.5) \)
  \[
  P(X_A = k) = \binom{3}{k} \left(\frac{1}{2}\right)^3 = \binom{3}{k} \frac{1}{8}, \quad k = 0, 1, 2, 3
  \]
- For Coin B: \( X_B \sim \text{Binomial}(5, 0.5) \)
  \[
  P(X_B = k) = \binom{5}{k} \left(\frac{1}{2}\right)^5 = \binom{5}{k} \frac{1}{32}, \quad k = 0, 1, 2, 3, 4, 5
  \]

### Step 3: Matching probabilities
We need to find the probability that \( X_A = X_B \), meaning the number of heads for Coin A matches the number of heads for Coin B. We calculate the individual probabilities for each possible number of heads \( k \).

#### Probability of Coin A:
\[
\begin{aligned}
P(X_A = 0) &= \binom{3}{0} \frac{1}{8} = \frac{1}{8}, \\
P(X_A = 1) &= \binom{3}{1} \frac{1}{8} = \frac{3}{8}, \\
P(X_A = 2) &= \binom{3}{2} \frac{1}{8} = \frac{3}{8}, \\
P(X_A = 3) &= \binom{3}{3} \frac{1}{8} = \frac{1}{8}.
\end{aligned}
\]

#### Probability of Coin B:
\[
\begin{aligned}
P(X_B = 0) &= \binom{5}{0} \frac{1}{32} = \frac{1}{32}, \\
P(X_B = 1) &= \binom{5}{1} \frac{1}{32} = \frac{5}{32}, \\
P(X_B = 2) &= \binom{5}{2} \frac{1}{32} = \frac{10}{32}, \\
P(X_B = 3) &= \binom{5}{3} \frac{1}{32} = \frac{10}{32}, \\
P(X_B = 4) &= \binom{5}{4} \frac{1}{32} = \frac{5}{32}, \\
P(X_B = 5) &= \binom{5}{5} \frac{1}{32} = \frac{1}{32}.
\end{aligned}
\]

### Step 4: Calculating the total probability
Now, we sum the probabilities that \( X_A = X_B \) for each value of \( k \):
\[
P(X_A = X_B) = P(X_A = 0)P(X_B = 0) + P(X_A = 1)P(X_B = 1) + P(X_A = 2)P(X_B = 2) + P(X_A = 3)P(X_B = 3)
\]
Substitute the values:

\[
\begin{aligned}
P(X_A = X_B) &= \left(\frac{1}{8} \times \frac{1}{32}\right) + \left(\frac{3}{8} \times \frac{5}{32}\right) + \left(\frac{3}{8} \times \frac{10}{32}\right) + \left(\frac{1}{8} \times \frac{10}{32}\right) \\
&= \frac{1}{256} + \frac{15}{256} + \frac{30}{256} + \frac{10}{256} \\
&= \frac{56}{256} = \frac{7}{32}.
\end{aligned}
\]

### Final Answer:
The probability that the number of heads obtained from flipping the two coins is the same is \( \frac{7}{32} \).

Would you like further details on this, or do you have any other questions?

Here are 5 related questions to expand your knowledge:
1. How is a binomial distribution defined, and when is it used?
2. What are the general properties of binomial coefficients?
3. How would the answer change if Coin A were flipped 4 times instead of 3?
4. What is the probability that Coin A produces more heads than Coin B?
5. How does the probability change if the coins are biased instead of fair?

**Tip:** Understanding binomial distribution is crucial in probability when dealing with independent binary outcomes like coin flips!

Explore the probability of obtaining the same number of heads when flipping Coin A three times and Coin B five times. This problem involves key concepts such as binomial distribution and combinatorics. A step-by-step solution using binomial probability formulas and combination calculations. Suitable for Grades 10-12.

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