The lifetime X of a bulb is a random variable with the probability density function:

f(x)=6[0.25-(x-1.5)^2] when 1<=x<=2
0 otherwise
X is measured in multiples of 1000 hrs. What is the probability that none of the three
bulbs in a traffic signal have to be replaced in the first 1500 hrs of their operation.

To find the probability that none of the three bulbs in a traffic signal have to be replaced in the first 1500 hours of their operation, we need to calculate the cumulative probability up to 1500 hours for each bulb and then multiply these probabilities together, assuming the bulbs are independent.

First, let's find the cumulative probability function for each individual bulb.

The cumulative probability function is found by integrating the probability density function (PDF) from the lower bound to the variable of interest. Since the PDF is defined piecewise, we need to integrate each segment separately.

For the interval 1 <= x <= 2, the cumulative distribution function (CDF) can be found as follows:

F(x) = ∫[0.25 - (t-1.5)^2] * 6 dt, for 1 <= t <= x

Integrating this function gives:

F(x) = 6 * ∫[0.25 - (t-1.5)^2] dt, for 1 <= t <= x

= 6 * [0.25t - (1/3)(t-1.5)^3] evaluated from 1 to x

Now, we can find the probability that a single bulb lasts at least 1500 hours:

P(X >= 1500) = 1 - F(1.5)

Substituting x = 1.5 into the cumulative distribution function:

P(X >= 1500) = 1 - F(1.5) = 1 - [6 * (0.25 * 1.5 - (1/3)(1.5-1.5)^3 - (0.25 * 1 - (1/3)(1-1.5)^3))]

Next, we multiply the probability for a single bulb by itself three times since the bulbs are independent:

P(No bulb replacement in 1500 hrs) = P(X >= 1500)^3

Finally, substitute the calculated value of P(X >= 1500) into the above equation to find the answer.