The lifespan of a particular brand of CFL light bulb is uniformly distributed between 19 and 39 months. Assume lifespans are independent from light bulb to light bulb. In a package that contains two of these CFL light bulbs, find the probability that the first bulb lasts for more than 28.3 months or the second bulb lasts for less than 28.2 months.
Question
The lifespan of a particular brand of CFL light bulb is uniformly distributed between 19 and 39 months. Assume lifespans are independent from light bulb to light bulb.
In a package that contains two of these CFL light bulbs, find the probability that the first bulb lasts for more than 28.3 months or the second bulb lasts for less than 28.2 months.
Solution 1
Let X be the lifespan (in months) of the first CFL light bulb in the package, and let Y be the lifespan (in months) of the second CFL light bulb in the package. We know that X and Y are both uniformly distributed between 19 and 39 months.
To find the probability that the first bulb lasts for more than 28.3 months or the second bulb lasts for less than 28.2 months, we can use the probability of the union of two events:
P(X > 28.3 or Y < 28.2) = P(X > 28.3) + P(Y < 28.2) - P(X > 28.3 and Y < 28.2)
Since X and Y are independent, we have:
P(X > 28.3 and Y < 28.2) = P(X > 28.3) * P(Y < 28.2)
Using the uniform distribution, we can find:
P(X > 28.3) = (39 - 28.3) / (39 - 19) = 0.47 P(Y < 28.2) = (28.2 - 19) / (39 - 19) = 0.26
Therefore:
P(X > 28.3 or Y < 28.2) = 0.47 + 0.26 - (0.47 * 0.26) = 0.6782
So the probability that the first bulb lasts for more than 28.3 months or the second bulb lasts for less than 28.2 months is approximately 0.6782 or 67.82%.
Solution 2
Let X1 and X2 be the lifespans of the first and second CFL light bulbs, respectively. According to the problem, X1 and X2 are uniformly distributed between 19 and 39 months, so the probability density function of X1 and X2 is:
f(x) = 1/20, 19 ≤ x ≤ 39
We need to find the probability that either X1 > 28.3 or X2 < 28.2. We can use the probability addition rule:
P(X1 > 28.3 or X2 < 28.2) = P(X1 > 28.3) + P(X2 < 28.2) - P(X1 > 28.3 and X2 < 28.2)
To find P(X1 > 28.3), we can use the cumulative distribution function of X1:
F(x) = (x - 19) / 20, 19 ≤ x ≤ 39
So, P(X1 > 28.3) = 1 - F(28.3) = 1 - (28.3 - 19) / 20 = 0.285
To find P(X2 < 28.2), we can use the same cumulative distribution function:
P(X2 < 28.2) = F(28.2) = (28.2 - 19) / 20 = 0.46
To find P(X1 > 28.3 and X2 < 28.2), we need to use the joint probability density function of X1 and X2, which is:
f(x1, x2) = f(x1) * f(x2) = 1/400, 19 ≤ x1 ≤ 39, 19 ≤ x2 ≤ 39
So, P(X1 > 28.3 and X2 < 28.2) = ∫∫[x1 > 28.3 and x2 < 28.2] f(x1, x2) dx1 dx2
= ∫19^28.2 ∫28.3^39 f(x1, x2) dx1 dx2
= ∫19^28.2 ∫28.3^39 1/400 dx1 dx2
= (28.2 - 19) * (39 - 28.3) / 400
= 0.027
Therefore, P(X1 > 28.3 or X2 < 28.2) = 0.285 + 0.46 - 0.027 = 0.718
So, the probability that the first CFL light bulb lasts for more than 28.3 months or the second CFL light bulb lasts for less than 28.2 months is 0.718.
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