Jackson goes to the gym 0, 2, or 3 days per week, depending on work demands. The expected value of the number of days per week that Jackson goes to the gym is 2.05. The probability that he goes 0 days is 0.1, the probability that he goes 2 days is , and the probability that he goes 3 days is .

For a probability distribution the expected value is the summation of product of probabilities with their respective data values. Let x be the probability that Jackson goes gym for 2 days and y be the probability that he goes gym for 3 days. For the given case we have following values and their probabilities:0 : 0.12 : x 3 : ySo the expected value will be = 0(0.1) + 2(x) + 3(y)Expected value is given to be 2.05. So we can write the equation as:2x + 3y = 2.05 (Equation 1)Also for a probability distribution, the sum of probabilities must always equal to 1. So we can set up the second equation as:0.1 + x + y = 1x + y = 0.9 (Equation 2)From Equation 2 we can write the value of x to be x = 0.9 - y. Using this value in equation 1, we get:2(0.9 - y) + 3y = 2.051.8 - 2y + 3y = 2.051.8 + y = 2.05y = 0.25Using the value of y in equation 2 we get value of x to be 0.65Therefore we can conclude that:The probability that Jackson goes to gym for 2 days is 0.65 and the probability that he goes to gym for 3 days is 0.25