Urn A has three red balls and two white balls and urn B has livro red balls and five white balls. A fair coin is tossed. If it land's heads up, a other wise ball is drawn from urn A, a bark is drawn from urw B. (a) what is the probability that a red ball

is drawn) (b) If a red ball is drawn, what is Probability that coin landed heads up!

Let’s break down the problem carefully. We have two urns and a fair coin toss. 

**Urn A** contains:
- 3 red balls
- 2 white balls

**Urn B** contains:
- 4 red balls (assuming "livro" is a typo for "four")
- 5 white balls

The problem states:
- If the coin lands **heads** (H), a ball is drawn from **Urn A**.
- If the coin lands **tails** (T), a ball is drawn from **Urn B**.

### Part (a): What is the probability that a red ball is drawn?

#### Step 1: Probability of heads and drawing a red ball from Urn A
The probability of landing heads is \( P(H) = \frac{1}{2} \).

If the coin lands heads, we are drawing from Urn A. The probability of drawing a red ball from Urn A is the ratio of red balls to total balls in Urn A:
\[
P(\text{Red} | H) = \frac{3}{3 + 2} = \frac{3}{5}
\]

Therefore, the combined probability of landing heads and drawing a red ball is:
\[
P(\text{Red and H}) = P(H) \times P(\text{Red} | H) = \frac{1}{2} \times \frac{3}{5} = \frac{3}{10}
\]

#### Step 2: Probability of tails and drawing a red ball from Urn B
The probability of landing tails is \( P(T) = \frac{1}{2} \).

If the coin lands tails, we are drawing from Urn B. The probability of drawing a red ball from Urn B is:
\[
P(\text{Red} | T) = \frac{4}{4 + 5} = \frac{4}{9}
\]

Thus, the combined probability of landing tails and drawing a red ball is:
\[
P(\text{Red and T}) = P(T) \times P(\text{Red} | T) = \frac{1}{2} \times \frac{4}{9} = \frac{4}{18} = \frac{2}{9}
\]

#### Step 3: Total probability of drawing a red ball
We now add the two cases (heads or tails) together:
\[
P(\text{Red}) = P(\text{Red and H}) + P(\text{Red and T}) = \frac{3}{10} + \frac{2}{9}
\]
To add these, we need a common denominator:
\[
P(\text{Red}) = \frac{27}{90} + \frac{20}{90} = \frac{47}{90}
\]

Thus, the probability that a red ball is drawn is:
\[
\boxed{\frac{47}{90}}
\]

### Part (b): If a red ball is drawn, what is the probability that the coin landed heads up?

We are asked to find the conditional probability \( P(H | \text{Red}) \), which is the probability that the coin landed heads given that a red ball was drawn.

Using **Bayes' theorem**:
\[
P(H | \text{Red}) = \frac{P(\text{Red and H})}{P(\text{Red})}
\]
We already know:
- \( P(\text{Red and H}) = \frac{3}{10} \)
- \( P(\text{Red}) = \frac{47}{90} \)

Substituting these values:
\[
P(H | \text{Red}) = \frac{\frac{3}{10}}{\frac{47}{90}} = \frac{3}{10} \times \frac{90}{47} = \frac{27}{47}
\]

Thus, the probability that the coin landed heads given that a red ball was drawn is:
\[
\boxed{\frac{27}{47}}
\]

---

Would you like a detailed breakdown of any part of the solution? Here are 5 related questions you can explore next:

1. What is the probability of drawing a white ball?
2. How would the probabilities change if Urn A had 4 red balls?
3. How would the result change if the coin was biased with a different probability of heads?
4. What is the probability that a white ball is drawn given the coin landed tails?
5. Can we generalize the method to calculate probabilities for \( n \) urns?

**Tip**: Always use Bayes' theorem when you need to reverse a conditional probability!

Urn A has three red balls and two white balls and urn B has livro red balls and five white balls. A fair coin is tossed. If it lands heads up, a ball is drawn from urn A; otherwise, a ball is drawn from urn B. (a) What is the probability that a red ball is drawn? (b) If a red ball is drawn, what is the probability that the coin landed heads up?

Explore the probability of drawing a red ball from two urns with a fair coin toss, using conditional probability and Bayes' Theorem. This detailed solution includes step-by-step calculations for both scenarios: when the coin lands heads or tails. Suitable for Grades 10-12 students learning probability.

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