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

The expected value $E(X)$ is calculated using the formula:

$E(X) = \sum_{x} x P(X = x)$

Substituting the values:

$E(X) = (-1) \cdot \frac{1}{5} + 0 \cdot \frac{7}{20} + 1 \cdot \frac{1}{10} + 2 \cdot \frac{7}{20} + 3 \cdot \frac{1}{5}$

Calculating each term:

$= -\frac{1}{5} + 0 + \frac{1}{10} + \frac{14}{20} + \frac{3}{5}$

Converting $\frac{3}{5}$ to a tenth denominator:

$= -\frac{2}{10} + 0 + \frac{1}{10} + \frac{14}{20} + \frac{6}{10}$

Combining:

$= \frac{-2 + 1 + 3 + 7}{10} = \frac{9}{10} = 3.1.$

To calculate the variance $Var(X)$, use the formula:

$Var(X) = E(X^2) - (E(X))^2$

First, we find $E(X^2)$:

$E(X^2) = \sum_{x} x^2 P(X = x)$

Calculating:

$= (-1)^2 \cdot \frac{1}{5} + 0^2 \cdot \frac{7}{20} + 1^2 \cdot \frac{1}{10} + 2^2 \cdot \frac{7}{20} + 3^2 \cdot \frac{1}{5}$

$= 1 \cdot \frac{1}{5} + 0 + \frac{1}{10} + 4 \cdot \frac{7}{20} + 9 \cdot \frac{1}{5}$

Calculating each term:

$= \frac{1}{5} + 0 + \frac{1}{10} + \frac{28}{20} + \frac{9}{5}$

Converting $\frac{9}{5}$ and $\frac{1}{5}$ to a tenths denominator:

$= \frac{2 + 0 + 1 + 14 + 18}{10} = \frac{35}{10} = 3.5$

Now, substituting back into the variance formula:

$Var(X) = E(X^2) - (E(X))^2 = 3.5 - (3.1)^2 = 3.5 - 9.61 = 2.21.$

The variance of $Y$, where $Y = 6 - 2X$, can be found using the transformation rule for variance. Since the variance of a constant multiplied by a random variable is:

$Var(aX) = a^2 Var(X)$

we have:

$Var(Y) = Var(6 - 2X) = 4 Var(X) = 4 \cdot 2.21 = 8.84.$

To find the probability $P(X ≥ Y)$, first consider the cases when $Y = 6 - 2X$ is equal or less than $X$:

1. If $X = -1$, then $Y = 8$.
2. If $X = 0$, then $Y = 6$.
3. If $X = 1$, then $Y = 4$.
4. If $X = 2$, then $Y = 2$.
5. If $X = 3$, then $Y = 0$.

The valid cases for $X ≥ Y$ are:

• When $X = 2$ and $Y ≤ 2$, which has:
  • Probability of $P(X = 2) = a = \frac{7}{20}$.
• When $X = 3$, which is always true since $Y$ would never exceed $X$.
  • Probability of $P(X = 3) = \frac{1}{5} = \frac{2}{10}$.

Therefore, the total probability is:

$P(X ≥ Y) = P(X = 2) + P(X = 3) = \frac{7}{20} + \frac{1}{5} = \frac{7}{20} + \frac{4}{20} = \frac{11}{20}.$