Question

Let Z be a discrete random variable, where Z can take the values {1,2,3} with strictly positive probabilities. Let f(z) be the probability mass function of Z.f(1)+f(2)<1.

A. True
B. False?

Answer 1

The statement 'f(1) + f(2) < 1' is true because each probability in a discrete probability distribution must be between 0 and 1, inclusive, and the sum of all probabilities must equal 1. To accommodate the strictly positive probability for f(3), f(1) and f(2) must together sum to less than 1.

Step-by-step explanation:

The question asks whether the following statement is true or false: 'f(1) + f(2) < 1', where f(z) is the probability mass function (PMF) of a discrete random variable Z that can take the values {1,2,3}.

A discrete random variable is defined by a set of values it can take, each with its own probability. For this random variable, there are only three possible outcomes: 1, 2, and 3. The essential characteristics of a discrete probability distribution function for a such random variable are that each probability must be between 0 and 1, inclusive, and that the sum of all probabilities must equal 1.

Based on this, the statement 'f(1) + f(2) < 1' is True. This is because the probability of the third value, f(3), must be strictly positive, as given by the condition of the question. Thus, if f(1) and f(2) were to sum up to 1 or more together, this would leave no probability mass for f(3), which would violate the requirements of a discrete probability distribution. Therefore, f(1) + f(2) must indeed total less than 1 to allow for a strictly positive probability for f(3).