Two boxes contain respectively 3 red, 2 black balls and bred, 8 black balls. A fair coin is tossed. A ball is to be drawn from any of these boxes according to the toss of the coin. If the coin shows head a ball is taken from the first box and if the coin shows tail a ball is taken from box 2. Finally a red ball was chosen. Find the probability that coin was resulted in head.

This problem involves conditional probability, where we need to find the probability that the coin showed heads given that a red ball was drawn. We can solve this using **Bayes' Theorem**.

### Let’s define the events:
- \( H_1 \): The event that the coin shows heads (which means the ball is drawn from the first box).
- \( H_2 \): The event that the coin shows tails (which means the ball is drawn from the second box).
- \( R \): The event that a red ball is drawn.

We are asked to find \( P(H_1 \mid R) \), which is the probability that the coin showed heads given that a red ball was drawn. Using Bayes' theorem:

\[
P(H_1 \mid R) = \frac{P(R \mid H_1) \cdot P(H_1)}{P(R)}
\]

Where:
- \( P(H_1) = \frac{1}{2} \) is the probability that the coin shows heads (since it's a fair coin).
- \( P(H_2) = \frac{1}{2} \) is the probability that the coin shows tails.

Now, let's compute the probabilities:

1. **\( P(R \mid H_1) \)**: The probability of drawing a red ball from the first box (which contains 3 red and 2 black balls) is:

\[
P(R \mid H_1) = \frac{3}{3+2} = \frac{3}{5}
\]

2. **\( P(R \mid H_2) \)**: The probability of drawing a red ball from the second box (which contains no red balls and 8 black balls) is:

\[
P(R \mid H_2) = 0
\]

3. **\( P(R) \)**: The total probability of drawing a red ball, considering both possible outcomes (head or tail), is:

\[
P(R) = P(R \mid H_1) \cdot P(H_1) + P(R \mid H_2) \cdot P(H_2)
\]
\[
P(R) = \frac{3}{5} \cdot \frac{1}{2} + 0 \cdot \frac{1}{2} = \frac{3}{10}
\]

### Now apply Bayes' theorem:

\[
P(H_1 \mid R) = \frac{P(R \mid H_1) \cdot P(H_1)}{P(R)} = \frac{\frac{3}{5} \cdot \frac{1}{2}}{\frac{3}{10}} = 1
\]

Thus, the probability that the coin showed heads given that a red ball was drawn is **1**.

### Conclusion:
The probability that the coin showed heads given that a red ball was drawn is **1**. This makes sense because it's impossible to draw a red ball from the second box, meaning the red ball must have come from the first box, implying that the coin showed heads.

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Do you want details on any part of the solution or have any questions?

Here are 5 related questions:
1. What is Bayes' Theorem and how is it used in probability?
2. How can we calculate the total probability of an event happening?
3. What are conditional probabilities and how are they different from regular probabilities?
4. How would the solution change if the second box contained some red balls?
5. Can Bayes' Theorem be applied to more complex events involving multiple steps?

**Tip:** When solving probability problems with multiple conditions, always start by clearly defining the events and applying formulas like Bayes' Theorem to simplify the process.

Explore how to solve a probability problem involving two boxes of balls, a coin toss, and Bayes' Theorem. The solution details how to compute the probability that the coin showed heads given that a red ball was drawn. Suitable for students in Grades 10-12.

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