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A lightbulb is installed. With probability 1/3 , it burns out immediately when it is first installed. With probability 2/3 , it burns out after an amount of time that is uniformly distributed on [0,3] . The expected value of the time until the lightbulb burns out is________

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User BScutt
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7 votes

Answer:

1 time unit

Explanation:

There is a 1 in 3 chance that the time until the bulb burns out is zero, and a 2 in 3 chance that the time is the average between 0 and 3 (since it is uniformly distributed). Therefore, the expected value for the time until it burns is:


E(x) = (1)/(3)*0+(2)/(3) *(0+3)/(2)\\ E(X) = 1

The expected value of the time until the lightbulb burns out is 1 time unit.

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User Jamie Hartnoll
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