person randomly chooses a battery from a store which has 120 batteries of type A and 360 batteries of type B. Battery life of type A and type B batteries are exponentially distributed with average life of 12.0 years and 18.0 years, respectively. If the chosen battery lasts for 5 years, what is the probability that the battery is of type A? Enter your answer correct to two decimals accuracy.


To solve this problem, we need to find the probability that the battery is of type A, given that it lasted for 5 years. We can use Bayes' theorem for this purpose.

Let:
- \( A \): the event that the chosen battery is of type A.
- \( B \): the event that the chosen battery is of type B.
- \( T = 5 \): the event that the battery lasts for 5 years.

### Step 1: Find the Priors
The probability of choosing a type A battery (\( P(A) \)) and a type B battery (\( P(B) \)) are based on their proportions in the store:
\[
P(A) = \frac{120}{120 + 360} = \frac{120}{480} = 0.25
\]
\[
P(B) = \frac{360}{120 + 360} = \frac{360}{480} = 0.75
\]

### Step 2: Find the Conditional Probabilities \( P(T = 5 | A) \) and \( P(T = 5 | B) \)

Since the battery life is exponentially distributed, we can use the probability density function for an exponential distribution:
\[
P(T = t | \text{Type}) = \frac{1}{\theta} e^{-t / \theta}
\]
where \( \theta \) is the mean life of the battery for each type.

- For type A batteries, the mean life \( \theta_A = 12 \) years.
  \[
  P(T = 5 | A) = \frac{1}{12} e^{-5 / 12}
  \]

- For type B batteries, the mean life \( \theta_B = 18 \) years.
  \[
  P(T = 5 | B) = \frac{1}{18} e^{-5 / 18}
  \]

Calculating these values:

1. **For \( P(T = 5 | A) \):**
   \[
   P(T = 5 | A) = \frac{1}{12} e^{-5 / 12} \approx \frac{1}{12} \times 0.6703 \approx 0.0559
   \]

2. **For \( P(T = 5 | B) \):**
   \[
   P(T = 5 | B) = \frac{1}{18} e^{-5 / 18} \approx \frac{1}{18} \times 0.7594 \approx 0.0422
   \]

### Step 3: Use Bayes' Theorem to Find \( P(A | T = 5) \)

Using Bayes' theorem:
\[
P(A | T = 5) = \frac{P(T = 5 | A) \cdot P(A)}{P(T = 5 | A) \cdot P(A) + P(T = 5 | B) \cdot P(B)}
\]

Substitute the values we found:
\[
P(A | T = 5) = \frac{0.0559 \times 0.25}{(0.0559 \times 0.25) + (0.0422 \times 0.75)}
\]

Calculate the numerator and denominator:

1. **Numerator:** \( 0.0559 \times 0.25 = 0.013975 \)
2. **Denominator:** \( (0.0559 \times 0.25) + (0.0422 \times 0.75) = 0.013975 + 0.03165 = 0.045625 \)

So:
\[
P(A | T = 5) = \frac{0.013975}{0.045625} \approx 0.31
\]

### Final Answer
The probability that the battery is of type A, given that it lasted for 5 years, is approximately:
\[
\boxed{0.31}
\]

Would you like more details on any step, or have any questions?

### Additional Related Questions

1. How would the answer change if the battery life of type A were different?
2. How can we generalize this calculation for any given battery life \( t \)?
3. What is the probability that a chosen battery is of type B if it lasts for 10 years?
4. How would the result differ if there were an equal number of type A and type B batteries?
5. What is the expected lifespan of a randomly chosen battery from this store?

### Tip
Remember that Bayes' theorem is especially useful for updating probabilities based on new evidence, making it powerful in cases where you have prior information.

A person randomly chooses a battery from a store which has 120 batteries of type A and 360 batteries of type B. Battery life of type A and type B batteries are exponentially distributed with average life of 12.0 years and 18.0 years, respectively. If the chosen battery lasts for 5 years, what is the probability that the battery is of type A? Enter your answer correct to two decimals accuracy.

Solve for the probability that a randomly chosen battery from a store's stock of type A and type B batteries is of type A, given it lasted for 5 years. This problem uses Bayes' theorem with exponential distribution to calculate conditional probability based on prior and likelihood values. Suitable for undergraduate students studying probability and statistics.

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