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
pr(burnoutduringfirst three hrs)=2/3, avg time is 1.5hrs
Pr(burnoutduringon)=2/3 * t/3 where t is time burning.
now, during the burn time of the bulbs burning, the average time is 1.5 hrs burning
expected life=2/3*1.5=1hr
Well, that lightbulb sure likes to keep things interesting! Let's do some calculations, shall we?
First, let's consider the immediate burnout scenario with probability 1/3. In this case, the time until the lightbulb burns out is 0.
Now, for the scenario where it burns out after an amount of time uniformly distributed on [0,3], we can find the expected value using the formula for the geometric mean of a uniform distribution, which is (a + b)/2.
The average time until the lightbulb burns out in this scenario is (0 + 3)/2 = 1.5.
Since these are mutually exclusive scenarios, we can calculate the overall expected value by taking the weighted average of the two cases, based on their probabilities.
Overall, the expected value of the time until the lightbulb burns out would be:
(1/3) * 0 + (2/3) * 1.5 = 1.
So, the expected value of the time until the lightbulb burns out is 1. Keep an eye on that lightbulb, who knows when it'll decide to bring some brightness into your life or leave you in the dark!
To find the expected value of the time until the lightbulb burns out, we need to calculate the expected value of two scenarios: when the lightbulb burns out immediately and when it burns out after an amount of time uniformly distributed on [0,3].
Scenario 1: The lightbulb burns out immediately (probability = 1/3)
In this case, the time until the lightbulb burns out is 0. Therefore, the expected value for this scenario is E1 = 0 * (1/3) = 0.
Scenario 2: The lightbulb burns out after an amount of time uniformly distributed on [0,3] (probability = 2/3)
The time until the lightbulb burns out in this scenario follows a uniform distribution on the interval [0,3]. The expected value for a uniform distribution is calculated as the average of the minimum and maximum value. In this case, the minimum value is 0 and the maximum value is 3, so the expected value for this scenario is E2 = (0 + 3) / 2 = 1.5.
Now, we can calculate the overall expected value by taking the weighted average of the two scenarios:
Expected value = (probability_scenario1 * expected_value_scenario1) + (probability_scenario2 * expected_value_scenario2)
= (1/3 * 0) + (2/3 * 1.5)
= 0 + 1
= 1.
Therefore, the expected value of the time until the lightbulb burns out is 1.
To find the expected value of the time until the lightbulb burns out, we need to compute the weighted average of the possible burnout times. Let's break it down into two cases.
Case 1: The lightbulb burns out immediately after installation (with probability 1/3):
In this case, the time until the lightbulb burns out is 0. Therefore, the contribution from this case to the expected value is (1/3) * 0 = 0.
Case 2: The lightbulb burns out after an amount of time uniformly distributed on [0,3] (with probability 2/3):
In this case, the time until the lightbulb burns out is the average of the possible burnout times, which is (0 + 3) / 2 = 1.5. Therefore, the contribution from this case to the expected value is (2/3) * 1.5 = 1.
Now, to find the overall expected value, we sum up the contributions from each case:
Expected value = Contribution from Case 1 + Contribution from Case 2 = 0 + 1 = 1.
Therefore, the expected value of the time until the lightbulb burns out is 1.