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yawa3891 [41]
3 years ago
14

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________
Mathematics
1 answer:
Andrej [43]3 years ago
3 0

Answer:

1 time unit

Step-by-step 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) = \frac{1}{3}*0+\frac{2}{3} *\frac{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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