Question

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
f(x)={ 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.

Answer 1

Answer on Question #39179 - Math - Statistics

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

$$f(x) = \begin{cases} 6 * (0.25 - (x - 1.5)^2) & \text{when } 1 \leq x \leq 2 \\ 0 & \text{otherwise} \end{cases}$$

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.

Solution: $T_1$ – the random variable of a lifetime (measured in multiples of 1000 hrs.) of i-th bulb, $i = 1, 2, 3$. Then, for $A$ = "lifetime of each bulb is longer than 1500",

$$P(A) = P\left((T_1 \geq 1.5) \land (T_2 \geq 1.5) \land (T_3 \geq 1.5)\right) = P(T_1 \geq 1.5) * P(T_2 \geq 1.5) * P(T_3 \geq 1.5) = P(T_1 \geq 1.5)^3.$$ $$P(T_1 \geq 1.5) = \int_{1.5}^{\infty} f(x) \, dx = \int_{1.5}^{2} 6 * (0.25 - (x - 1.5)^2) \, dx = \left(\frac{3}{2} x - 2(x - 1.5)^3\right) \Bigg|_{1.5}^{2} = (3 - 2 * 0.5^3) - \left(\frac{9}{4}\right) = 3 - \frac{1}{4} - \frac{9}{4} = 3 - \frac{5}{2} = \frac{1}{2}.$$

Then $P(A) = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$.

Answer: 1/8.