The question is asking for the probability that a battery chosen at random is of type A, given that it lasted for 5 years. This is a conditional probability problem that can be solved using Bayes' theorem.

Bayes' theorem states that P(A|B) = P(B|A) * P(A) / P(B), where:

- P(A|B) is the probability of event A given event B is true
- P(B|A) is the probability of event B given event A is true
- P(A) and P(B) are the probabilities of events A and B

In this case, event A is that the battery is of type A, and event B is that the battery lasted for 5 years.

First, we need to calculate the individual probabilities:

- P(A) is the probability that a randomly chosen battery is of type A. There are 100 type A batteries and 280 type B batteries, so P(A) = 100 / (100 + 280) = 0.263
- P(B) is the probability that a randomly chosen battery lasts for 5 years. Since the battery life is exponentially distributed, this is calculated as 1 - e^(-5/average life). For type A batteries, the average life is 10 years, so P(B|A) = 1 - e^(-5/10) = 0.393. For type B batteries, the average life is 14 years, so P(B|B) = 1 - e^(-5/14) = 0.307. The overall probability P(B) is then a weighted average of these, so P(B) = 0.263 * 0.393 + (1 - 0.263) * 0.307 = 0.334

Then, we can substitute these values into Bayes' theorem to find P(A|B):

P(A|B) = P(B|A) * P(A) / P(B) = 0.393 * 0.263 / 0.334 = 0.308

So, the probability that the battery is of type A given that it lasted for 5 years is approximately 0.31, or 31%, when rounded to two decimal places.

Question

The question is asking for the probability that a battery chosen at random is of type A, given that it lasted for 5 years. This is a conditional probability problem that can be solved using Bayes' theorem.

Bayes' theorem states that P(A|B) = P(B|A) * P(A) / P(B), where:

- P(A|B) is the probability of event A given event B is true
- P(B|A) is the probability of event B given event A is true
- P(A) and P(B) are the probabilities of events A and B

In this case, event A is that the battery is of type A, and event B is that the battery lasted for 5 years.

First, we need to calculate the individual probabilities:

- P(A) is the probability that a randomly chosen battery is of type A. There are 100 type A batteries and 280 type B batteries, so P(A) = 100 / (100 + 280) = 0.263
- P(B) is the probability that a randomly chosen battery lasts for 5 years. Since the battery life is exponentially distributed, this is calculated as 1 - e^(-5/average life). For type A batteries, the average life is 10 years, so P(B|A) = 1 - e^(-5/10) = 0.393. For type B batteries, the average life is 14 years, so P(B|B) = 1 - e^(-5/14) = 0.307. The overall probability P(B) is then a weighted average of these, so P(B) = 0.263 * 0.393 + (1 - 0.263) * 0.307 = 0.334

Then, we can substitute these values into Bayes' theorem to find P(A|B):

P(A|B) = P(B|A) * P(A) / P(B) = 0.393 * 0.263 / 0.334 = 0.308

So, the probability that the battery is of type A given that it lasted for 5 years is approximately 0.31, or 31%, when rounded to two decimal places.

Knowee AI · Accepted Answer

The question is asking for the probability that a battery chosen at random is of type A, given that it lasted for 5 years. This is a conditional probability problem that can be solved using Bayes' theorem.

Bayes' theorem states that P(A|B) = P(B|A) * P(A) / P(B), where:

- P(A|B) is the probability of event A given event B is true
- P(B|A) is the probability of event B given event A is true
- P(A) and P(B) are the probabilities of events A and B

In this case, event A is that the battery is of type A, and event B is that the battery lasted for 5 years.

First, we need to calculate the individual probabilities:

- P(A) is the probability that a randomly chosen battery is of type A. There are 100 type A batteries and 280 type B batteries, so P(A) = 100 / (100 + 280) = 0.263
- P(B) is the probability that a randomly chosen battery lasts for 5 years. Since the battery life is exponentially distributed, this is calculated as 1 - e^(-5/average life). For type A batteries, the average life is 10 years, so P(B|A) = 1 - e^(-5/10) = 0.393. For type B batteries, the average life is 14 years, so P(B|B) = 1 - e^(-5/14) = 0.307. The overall probability P(B) is then a weighted average of these, so P(B) = 0.263 * 0.393 + (1 - 0.263) * 0.307 = 0.334

Then, we can substitute these values into Bayes' theorem to find P(A|B):

P(A|B) = P(B|A) * P(A) / P(B) = 0.393 * 0.263 / 0.334 = 0.308

So, the probability that the battery is of type A given that it lasted for 5 years is approximately 0.31, or 31%, when rounded to two decimal places.

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