Figure 1: Components E1, . . . , E5 at operational at time t with probabilities
, e−t/3 and e
The system S consists of five independent elements Ei
, i = 1, . . . , 5, connected as in
Figure 1. Probability that the element Ei
is operational at the end of time interval [0, t] is
pi(t) = e
, t ≥ 0,
for λ1 = 1, λ2 = 2, λ3 = 1/2, λ4 = 1/3, and λ5 = 1.
(a) Find the probability that the system S will be operational at time t. Plot this
probability as a function of time t. What is this probability for t = 1/2.
(b) Find the probability that component E5 was operational at time t = 1/2, if the
system was operational at that time.
Hint: If you consider (b), it is conditional probability, more precisely, a posterior probability
of the hypothesis H1 : E5 operational at time t, given that system S is operational at t. Thus,
solve part (a) as a total probability with H1 and H2 = Hc
1 as hypotheses. Under the two
hypotheses the system simplifies as in Figure 2 and it is easy to find P(S|H1) and P(S|H2).
Then (b) is just a Bayes formula. The results for arbitrary t will be messy – do not simplify.
For plotting in part (a) take some reasonable interval for time t, say [0, 4].
Figure 2: Left: System under hypothesis H1 : E5 operational; Right: System under hypothesis H2 : E5 not operational.
2. Two Batches.
There are two batches of the same product. In one batch all products
are conforming. The other batch contains 20% non-conforming products. A batch is selected
at random and one randomly selected product from that batch is inspected. The inspected
product was found conforming and was returned back to its batch.
What is the probability that the second product, randomly selected from the same batch,
is found non-conforming?
Hint. This problem uses both Bayes’ rule and Total Probability. The two hypotheses
concern the type of batch.
For the first draw the hypotheses are equally likely (the batch is
selected at random), but for the second draw, the probabilities of hypotheses are updated
by the information on the result of the first draw via Bayes rule. Updated probabilities of
hypotheses are then used in the Total Probability Formula for the second draw.
In a machine learning classification procedure the items are classified as 1 or
0. Based on a training sample of size 120 in which there are 65 1’s and 55 0’s, the classifier
predicts 70 1’s and 50 0’s. Out of 70 items predicted by the classifier as 1, 52 are correctly
From the population of items where the proportion of 0-labels is 99% (and 1-labels 1%),
an item is selected at random. What is the probability that the item is of label 1, if the
classifier says it was.
Hint. Think about the following interpretation. If 1 is a specific disease present, 0 no
disease present, and the classifier is a medical test for the disease, then you are asked to
find a positive predictive value of a test for a subject coming from population where the
prevalence of the disease is 1%.