One of two biased coins A and B is selected and flipped. Let A be the event that coin A is selected and B be the event that coin B is selected, with probabilities p(A) = 0.4 and p(B) = 0.6.

When coin A is flipped, the probability of heads is 0.3.
When coin B is flipped, the probability of heads is 0.9.Let H be the event that the selected coin comes up heads. Complete the values X, Y, and Z in Bayes’ Theorem to determine the probability coin B was chosen if the flip came up heads.

bayes therom

To solve this problem using **Bayes' Theorem**, let's define the relevant events and fill in the values for the formula.

### Problem Setup:

- \( A \): Event that coin A is selected.
- \( B \): Event that coin B is selected.
- \( H \): Event that the selected coin comes up heads.

Given probabilities:
- \( P(A) = 0.4 \) (probability of selecting coin A)
- \( P(B) = 0.6 \) (probability of selecting coin B)
- \( P(H|A) = 0.3 \) (probability of heads given coin A is selected)
- \( P(H|B) = 0.9 \) (probability of heads given coin B is selected)

### Goal:

We need to find the probability that coin B was chosen given that the result was heads, i.e., \( P(B|H) \). Bayes' Theorem can be applied as follows:

\[
P(B|H) = \frac{P(H|B) \cdot P(B)}{P(H)}
\]

Where:

- \( P(H|B) \): The probability of getting heads given coin B was selected (this is 0.9).
- \( P(B) \): The probability of selecting coin B (this is 0.6).
- \( P(H) \): The total probability of getting heads.

Now, \( P(H) \) (the total probability of getting heads) can be found using the **Law of Total Probability**:

\[
P(H) = P(H|A) \cdot P(A) + P(H|B) \cdot P(B)
\]

Substitute the known values:

\[
P(H) = (0.3 \cdot 0.4) + (0.9 \cdot 0.6)
\]
\[
P(H) = 0.12 + 0.54 = 0.66
\]

Now apply Bayes' Theorem:

\[
P(B|H) = \frac{0.9 \cdot 0.6}{0.66}
\]
\[
P(B|H) = \frac{0.54}{0.66} \approx 0.818
\]

### Conclusion:

The probability that coin B was chosen given that the result was heads is approximately 0.818.

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### In terms of **Bayes' Theorem**, the values are:

\[
P(B|H) = \frac{P(H|B) \cdot P(B)}{P(H)}
\]

- \( X = P(H|B) = 0.9 \)
- \( Y = P(B) = 0.6 \)
- \( Z = P(H) = 0.66 \)

Would you like to discuss this further or go into more details?

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Here are five related questions to deepen your understanding:

1. How would the result change if the probabilities of selecting coins A and B were equal?
2. What if the probability of heads for coin A was higher than for coin B?
3. How can Bayes' Theorem be used in other real-world applications?
4. What is the importance of the Law of Total Probability in this problem?
5. How would the probability of tails change the analysis?

**Tip**: When applying Bayes' Theorem, ensure you carefully define the events and compute the total probability for better clarity.

One of two biased coins A and B is selected and flipped. Let A be the event that coin A is selected and B be the event that coin B is selected, with probabilities p(A) = 0.4 and p(B) = 0.6. When coin A is flipped, the probability of heads is 0.3. When coin B is flipped, the probability of heads is 0.9. Let H be the event that the selected coin comes up heads. Complete the values X, Y, and Z in Bayes’ Theorem to determine the probability coin B was chosen if the flip came up heads.

Learn how to apply Bayes' Theorem to determine the probability that a biased coin B was selected given a heads outcome. This problem involves key probability concepts, including conditional probability and the Law of Total Probability. Perfect for undergraduate students studying probability and statistics.

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