Final answer:
A PMF and the expected value E[X] for a discrete random variable X with values 1, 2, 3 and 4 have been constructed, and the expected value is found to be 2.5. The PMF conditioned on the even values has been described with probabilities adjusted to reflect only even outcomes.
Step-by-step explanation:
The question asks about constructing a probability mass function (PMF) and computing the expected value, E[X], for a discrete random variable X which is equally likely to take on values 1, 2, 3, and 4. To draw the PMF, we plot these values on the x-axis and assign each a probability of ¼, since they are equally likely. To find E[X], we apply the formula E[X] = Σ xP(x), which means we sum the products of each value of X and its probability. Calculating this gives us E[X] = (1)(1/4) + (2)(1/4) + (3)(1/4) + (4)(1/4) = 2.5.
For the PMF conditioned on 'the even' (assuming 'the even' refers to the event where X takes on even values), we only consider the even numbers 2 and 4. The probabilities now adjust to each being ½, since they are equally likely among the even numbers. The conditioned PMF will have only the values 2 and 4 on the x-axis with a probability of 0.5 each.