Assume that we randomly pick a battery from a box which contains 10 batteries of type A and 40 batteries of type B. The lifetime of both batteries is exponentially distributed with λ = 0.04 per hour for type A battery and λ = 0.06 for type B. If the randomly selected battery lasts longer than 20 hours, what is the probability that it is a type B battery? (Choose the closest answer) 0.300.670.730.87
Question
Assume that we randomly pick a battery from a box which contains 10 batteries of type A and 40 batteries of type B. The lifetime of both batteries is exponentially distributed with λ = 0.04 per hour for type A battery and λ = 0.06 for type B. If the randomly selected battery lasts longer than 20 hours, what is the probability that it is a type B battery? (Choose the closest answer) 0.300.670.730.87
Solution
To solve this problem, we need to use Bayes' theorem, which is a fundamental theorem in probability theory and statistics that describes how to update the probabilities of hypotheses when given evidence.
First, we need to calculate the prior probabilities of picking a type A or type B battery:
P(A) = 10 / (10 + 40) = 0.2 P(B) = 40 / (10 + 40) = 0.8
Next, we need to calculate the likelihood of a battery lasting longer than 20 hours given that it is a type A or type B battery. This is given by the exponential distribution:
P(T > 20 | A) = e^(-0.04 * 20) = 0.135 P(T > 20 | B) = e^(-0.06 * 20) = 0.082
Now we can apply Bayes' theorem to calculate the posterior probability that the battery is a type B battery given that it lasted longer than 20 hours:
P(B | T > 20) = P(T > 20 | B) * P(B) / [P(T > 20 | A) * P(A) + P(T > 20 | B) * P(B)] = 0.082 * 0.8 / [0.135 * 0.2 + 0.082 * 0.8] = 0.066 / [0.027 + 0.066] = 0.066 / 0.093 = 0.71
So, the closest answer is 0.73.
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