The random variable X takes on the values 1, 2 (or) 3 with probabilities $$\frac{{2 + 5{\text{P}}}}{5},\frac{{1 + 3{\text{P}}}}{5}$$ and $$\frac{{1.5 + 2{\text{P}}}}{5}$$ respectively the values of P and E(X) are respectively. A. 0.05, 1.87 B. 1.90, 5.87 C. 0.05, 1.10 D. 0.25, 1.40

The correct answer is: D. 0.25, 1.40

The expected value of a random variable is the sum of the products of each possible value and its probability. In this case, the possible values of X are 1, 2, and 3, and the probabilities are $\frac{{2 + 5{\text{P}}}}{5},\frac{{1 + 3{\text{P}}}}{5}$ and $\frac{{1.5 + 2{\text{P}}}}{5}$, respectively. Therefore, the expected value is:

$$E(X) = \frac{{2 + 5{\text{P}}}}{5}(1) + \frac{{1 + 3{\text{P}}}}{5}(2) + \frac{{1.5 + 2{\text{P}}}}{5}(3) = 1.40$$

To solve for P, we can use the fact that the sum of the probabilities of all possible values must be 1. In this case, the possible values are 1, 2,