Quiz 3 Solution

1. You purchase a certain product. The manual states that the lifetime $T$ of the product, defined as the amount of time (in years) that product works properly until it fails, satisfies

$Pr (T \geq t) = \exp (- \frac{t^{3}}{30})$ , for all $t \geq 0$ .

You use the product for two years without any problems. What is the probability that it fails some time in the third year?

Solution: Let $A = [2, \infty)$ be the event that the lifetime exceeds two years. You are told that $A$ has occurred, i.e., $A$ is given. Let $B = [2, 3)$ be the event that the product fails some time in the third year. We must compute $Pr (B ∣ A)$ . We have

$Pr (B ∣ A) = \frac{Pr (B \cap A)}{Pr (A)}$ $= \frac{Pr ([2, \infty) \cap [2, 3))}{Pr ([2, \infty))}$ $= \frac{Pr ([2, 3))}{Pr ([2, \infty))}$

$= \frac{Pr ([2, \infty)) - Pr ([3, \infty))}{Pr ([2, \infty))}$

$= \frac{\exp (- 8 / 30) - \exp (- 27 / 30)}{\exp (- 8 / 30)}$

$= 1 - \exp (- 19 / 30) \approx 0.46918$

2. A computer manufacturer buys 60% of its memory chips from vendor A, 30% of its memory chips from vendor B, and 10% of its memory chips from vendor C. Memory chips from vendor A function correctly with probability 0.99, memory chips from vendor B function correctly with probability 0.96, and memory chips from vendor C function correctly with probability 0.93.

A memory chip is selected for testing uniformly at random from all chips purchased. What is the probability that it functions correctly?

Solution: Let $A$ be the event that the chip was obtained from vendor A, let $B$ be the event that the chip was obtained from vendor B, and let $C$ be the event that the chip was obtained from vendor C. Let $W$ be the event that the chip is working, i.e., functioning correctly.

We are told: $Pr (A) = 0.6$ , $Pr (B) = 0.3$ , and $Pr (C) = 0.1$ . Furthermore, $Pr (W ∣ A) = 0.99$ , $Pr (W ∣ B) = 0.96$ and $Pr (W ∣ C) = 0.93$ . Since $A$ , $B$ and $C$ are pairwise disjoint, but their union is the entire sample space, we apply the total probability theorem to obtain

$Pr (W) = Pr (W ∣ A) Pr (A) + Pr (W ∣ B) Pr (B) + Pr (W ∣ C) Pr (C)$

$= 0.99 \cdot 0.6 + 0.96 \cdot 0.3 + 0.93 \cdot 0.1 = 0.975$ .

Another way to get to this answer is as follows. Suppose we buy 1000 chips in total, 600 from vendor A, 300 from B and 100 from C. Furthermore, suppose that the probabilities give exact proportions, so 594 of the chips from vendor A are working (i.e., 99%), 288 from B are working and 93 from C are working. The total number of working chips is thus $594 + 288 + 93 = 975$ , and the probability of selecting one of these is 975/1000 = 0.975.

3. The memory chip selected in the previous question is found not to be functioning correctly. What is the probability that it was supplied by vendor B?

Solution: We are interested in computing $Pr (B ∣ W^{c})$ . Applying Bayes' Theorem we get

$Pr (B ∣ W^{c}) = \frac{Pr (W^{c} ∣ B) Pr (B)}{Pr (W^{c} ∣ A) Pr (A) + Pr (W^{c} ∣ B) Pr (B) + Pr (W^{c} ∣ C) Pr (C)}$

$= \frac{(0.04) (0.3)}{(0.01) (0.6) + (0.04) (0.3) + (0.07) (0.1)}$

$= 0.48$ .

Another way to see this parallels the second approach in the previous question. Out of 1000 chips purchased, 25 are not working (6 from vendor A, 12 from B, and 7 from C). The probability that the bad chip came from vendor B is $12 / 25 = 0.48$ .