SOLUTION: Urn A has 2 white and 13 red balls. Urn B has 14 white and 6 red balls. We flip a fair coin. If the outcome is heads, then a ball from urn A is selected, whereas if the outcome is

Question 1168481: Urn A has 2 white and 13 red balls. Urn B has 14 white and 6 red balls. We flip a fair coin. If the outcome is heads, then a ball from urn A is selected, whereas if the outcome is tails, then a ball from urn B is selected. Suppose that a white ball is selected. What is the probability that the coin landed heads?
Answer by CPhill(2264) (Show Source): You can put this solution on YOUR website!
Let's solve this problem using Bayes' Theorem.
**1. Define Events**
* **H:** Coin lands heads.
* **T:** Coin lands tails.
* **W:** A white ball is selected.
**2. Given Probabilities**
* P(H) = 1/2 (fair coin)
* P(T) = 1/2 (fair coin)
* P(W | H) = 2/15 (probability of white ball from urn A)
* P(W | T) = 14/20 = 7/10 (probability of white ball from urn B)
**3. Bayes' Theorem**
We want to find P(H | W), the probability that the coin landed heads given that a white ball was selected.
Bayes' Theorem states:
P(H | W) = [P(W | H) * P(H)] / P(W)
**4. Calculate P(W)**
We need to calculate P(W), the probability of selecting a white ball. We can use the law of total probability:
P(W) = P(W | H) * P(H) + P(W | T) * P(T)
P(W) = (2/15) * (1/2) + (7/10) * (1/2)
P(W) = 1/15 + 7/20
P(W) = (4 + 21) / 60
P(W) = 25/60 = 5/12
**5. Calculate P(H | W)**
Now, we can use Bayes' Theorem:
P(H | W) = [P(W | H) * P(H)] / P(W)
P(H | W) = [(2/15) * (1/2)] / (5/12)
P(H | W) = (1/15) / (5/12)
P(H | W) = (1/15) * (12/5)
P(H | W) = 12/75 = 4/25
**6. Final Answer**
The probability that the coin landed heads given that a white ball was selected is 4/25.

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