When Rohit plays a game, the number of points he receives is given by the discrete random variable $X$ with the following probability distribution - Edexcel - A-Level Maths Statistics - Question 3 - 2009 - Paper 1

The cumulative distribution function $F(x)$ is found by summing the probabilities of all outcomes up to $x$.

Thus,

$F(1.5) = P(X \leq 1) = P(X = 0) + P(X = 1) = 0.4 + 0.3 = 0.7$

To find the variance, first compute the expected value $E(X^2)$:

$E(X^2) = 0^2 \times 0.4 + 1^2 \times 0.3 + 2^2 \times 0.2 + 3^2 \times 0.1$

Calculating we get:

$E(X^2) = 0 + 0.3 + 0.8 + 0.9 = 2.0$

Then, variance is given by:

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

Thus,

$Var(X) = 2.0 - (1.0)^2 = 2.0 - 1 = 1.0$

Using the formula for the variance of a linear transformation, we have:

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

Where $a = -3$ and $b = 5$. Thus:

$Var(5 - 3X) = (-3)^2 Var(X) = 9 \times 1 = 9$

Rohit needs at least $10 - 6 = 4$ points from 2 games. We can consider the cases where $X_1 + X_2 \geq 4$:

1. Case where $X_1 + X_2 = 4$:

   • The only combination is (2,2): Probability = $0.2 \times 0.2 = 0.04$.

2. Case where $X_1 + X_2 = 5$:

   • Combinations (2,3) and (3,2): Probability = $(0.2 \times 0.1) + (0.1 \times 0.2) = 0.02 + 0.02 = 0.04$.

3. Case where $X_1 + X_2 = 6$:

   • Combination (3,3): Probability = $0.1 \times 0.1 = 0.01$.

Summing the probabilities:

Total probability = $0.04 + 0.04 + 0.01 = 0.09$

Thus, the probability that Rohit wins the prize is $0.09$.