Let X be a discrete random variable with the following PMF:PX (x) =0.1 for x = 0.20.2 for x = 0.40.2 for x = 0.50.3 for x = 0.80.2 for x = 10 otherwise(a) Find RX , the range of the random variable X.(b) Find P (X ≤ 0.5).(c) Find P (0.25 < X < 0.75).(d) Find P (X = 0.2 | X < 0.6).

(a) The range of the random variable X, RX, is the set of all possible values that X can take. From the given probability mass function (PMF), we can see that X can take the values 0.2, 0.4, 0.5, 0.8, and 1. So, RX = {0.2, 0.4, 0.5, 0.8, 1}.

(b) P(X ≤ 0.5) is the probability that X takes on a value less than or equal to 0.5. From the PMF, we can see that the values less than or equal to 0.5 are 0.2, 0.4, and 0.5. So, P(X ≤ 0.5) = P(X = 0.2) + P(X = 0.4) + P(X = 0.5) = 0.1 + 0.2 + 0.2 = 0.5.

(c) P(0.25 < X < 0.75) is the probability that X takes on a value between 0.25 and 0.75 (exclusive). From the PMF, we can see that the only value in this range is 0.5. So, P(0.25 < X < 0.75) = P(X = 0.5) = 0.2.

(d) P(X = 0.2 | X < 0.6) is the conditional probability that X equals 0.2 given that X is less than 0.6. From the PMF, we can see that the values less than 0.6 are 0.2, 0.4, and 0.5. So, P(X < 0.6) = P(X = 0.2) + P(X = 0.4) + P(X = 0.5) = 0.1 + 0.2 + 0.2 = 0.5. Therefore, P(X = 0.2 | X < 0.6) = P(X = 0.2) / P(X < 0.6) = 0.1 / 0.5 = 0.2.