If the random variable X assumes the values 0 and 1 only and is such that P(X = 0) = 3P(X =

1. It seems like your question got cut off. However, I can start explaining with the information given.

A random variable X can take on the values of 0 and 1. The probability of X being 0 is three times the probability of X being 1. This can be written as P(X = 0) = 3P(X = 1).

2. The sum of the probabilities of all possible outcomes of a random variable should equal 1. This is a fundamental rule in probability theory. So, we can set up the following equation: P(X = 0) + P(X = 1) = 1.

3. Substituting P(X = 0) = 3P(X = 1) into the equation, we get: 3P(X = 1) + P(X = 1) = 1.

4. Simplifying the left side of the equation, we get: 4P(X = 1) = 1.

5. Solving for P(X = 1), we get: P(X = 1) = 1/4 = 0.25.

6. Substituting P(X = 1) = 0.25 back into the equation P(X = 0) = 3P(X = 1), we get: P(X = 0) = 3 * 0.25 = 0.75.

So, the probability that X = 0 is 0.75 and the probability that X = 1 is 0.25.