The discrete random variable X has probability distribution given by | x | -1 | 0 | 1 | 2 | 3 | |-----|-------|------|------|-----|-----| | P(X = x) | 1/5 | a | 1/10 | a | 1/5 | where a is a constant - Edexcel - A-Level Maths Statistics - Question 3 - 2010 - Paper 2

To calculate the expected value E(X):
E(X) = $ext{sum of } (x_i imes P(X = x_i))$:

E(X) = (-1) imes rac{1}{5} + (0) imes rac{1}{4} + (1) imes rac{1}{10} + (2) imes rac{1}{4} + (3) imes rac{1}{5}
Calculating gives:
E(X) = -rac{1}{5} + 0 + rac{1}{10} + rac{2}{4} + rac{3}{5}
Combining the values leads to:

E(X) = -rac{1}{5} + rac{1}{10} + rac{1}{2} + rac{3}{5} = rac{3.1}{1}.

Variance is given by:
Var(X) = E(X^2) - [E(X)]^2.

First, calculate E(X²):
$E(X^2) = (-1)^2 imes P(X = -1) + (0)^2 imes P(X = 0) + (1)^2 imes P(X = 1) + (2)^2 imes P(X = 2) + (3)^2 imes P(X = 3)$
E(X^2) = 1 imes rac{1}{5} + 0 + 1 imes rac{1}{10} + 4 imes rac{1}{4} + 9 imes rac{1}{5}
Calculating leads us to:

E(X^2) = rac{1}{5} + rac{1}{10} + 1 + rac{9}{5} = rac{23}{10}.
Then,

Var(X) = rac{23}{10} - (3.1)^2 = rac{23}{10} - rac{61}{100} = rac{21.0}{5}.

To find Var(Y), we use the formula:
Var(Y) = Var(-2X) = (-2)^2 Var(X) = 4 Var(X). After calculating:
Var(Y) = 4 imes rac{21}{5} = rac{84}{5}.

We need to evaluate the cases when X = -1, 0, 1 for Y = 6 - 2X:

1. When $X = -1$, $Y = 8$.
2. When $X = 0$, $Y = 6$.
3. When $X = 1$, $Y = 4$.
4. When $X = 2$, $Y = 2$.
5. When $X = 3$, $Y = 0$.
   To find P(X ≥ Y):
   P(X ≥ Y) is equivalent to the sum of the probabilities for X taking values where it is greater than or equal to Y. After identifying, the relevant probabilities sum to: P(X ≥ Y) = rac{1}{5} + rac{1}{4} + rac{1}{5} = rac{9}{20}.