Let's denote \( T \) as the time until the lightbulb burns out. We can break down the expected value \( E[T] \) based on the two scenarios provided:
- Burns out immediately: With probability \( \frac{1}{3} \), the lightbulb burns out immediately, which contributes \( 0 \) to the expected time.
- Burns out after some time: With probability \( \frac{2}{3} \), the lightbulb burns out after a time uniformly distributed on the interval \( [0, 3] \).
To find the expected value of the time until the bulb burns out in the second case, we need to calculate the expected value of a uniform distribution on \( [0, 3] \).
The expected value \( E[X] \) of a uniformly distributed random variable \( X \) on the interval \( [a, b] \) is given by:
\[ E[X] = \frac{a + b}{2} \]
In our case \( a = 0 \) and \( b = 3 \), so:
\[ E[X] = \frac{0 + 3}{2} = \frac{3}{2} \]
Now we can compute the overall expected value \( E[T] \):
\[ E[T] = P(\text{immediate burn out}) \cdot E[T | \text{immediate}] + P(\text{later burn out}) \cdot E[T | \text{later}] \]
Substituting in the probabilities and expected values we have:
\[ E[T] = \frac{1}{3} \cdot 0 + \frac{2}{3} \cdot \frac{3}{2} \]
Calculating this gives:
\[ E[T] = 0 + \frac{2}{3} \cdot \frac{3}{2} = \frac{2 \cdot 3}{3 \cdot 2} = 1 \]
Therefore, the expected value of the time until the lightbulb burns out is:
\[ \boxed{1} \]
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