Question

1. within a warehouse there are lightbulbs that are classified into three groups, 0 to 1 year old, 1 to 2 years old, and 2 to 3 years old. 50% of the bulbs are in the first group, while 35% are in the second group and 15% are in the last group. the probability that a bulb in the first group will burn out in the next week is.05, the probability that a bulb in the second group will burn out in the next week is.35, and the probability that a bulb in the third group will burn out in the next week is.65.

(a) summarize this information in a tree diagram.
(b) after the week passes a randomly selected bulb has been found to have burned out. for each of the three groups determine the probability that the bulb was in that group.

Explanation:

Step1: Calculate the probability of a bulb burning out

Let \(A_1\) be the event that a bulb is in the first group (\(0 - 1\) year old), \(A_2\) be the event that a bulb is in the second group (\(1 - 2\) years old), \(A_3\) be the event that a bulb is in the third group (\(2 - 3\) years old), and \(B\) be the event that a bulb burns out.

We know that \(P(A_1)=0.5\), \(P(A_2) = 0.35\), \(P(A_3)=0.15\), \(P(B|A_1)=0.05\), \(P(B|A_2)=0.35\), \(P(B|A_3)=0.65\)

By the law of total - probability, \(P(B)=\sum_{i = 1}^{3}P(A_i)P(B|A_i)\)

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Step2: Use Bayes' theorem to find the conditional probabilities

By Bayes' theorem \(P(A_i|B)=\frac{P(A_i)P(B|A_i)}{P(B)}\)

For \(i = 1\):

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For \(i = 2\):

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For \(i = 3\):

$$
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Answer:

The probability that the burned - out bulb was in the first group (\(0 - 1\) year old) is \(\frac{5}{49}\approx0.102\), in the second group (\(1 - 2\) years old) is \(0.5\), and in the third group (\(2 - 3\) years old) is \(\frac{39}{98}\approx0.398\)