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 $F(x)$ gives us the probability that $X$ is less than or equal to $x$.

To find $F(1.5)$:

$F(1.5) = P(X ext{ is } 0) + P(X ext{ is } 1)$

Calculating:

• $P(X = 0) = 0.4$
• $P(X = 1) = 0.3$
• Thus, $F(1.5) = 0.4 + 0.3 = 0.7$

To compute the variance, we use the formula:

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

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

$E(X^2) = ext{Sum of } (x^2 imes P(X = x))$

Calculating:

• For $x = 0$: $0^2 imes 0.4 = 0$
• For $x = 1$: $1^2 imes 0.3 = 0.3$
• For $x = 2$: $2^2 imes 0.2 = 0.8$
• For $x = 3$: $3^2 imes 0.1 = 0.9$

Thus,

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

Next, we substitute into the variance formula:

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

Using the properties of variance, we find:

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

Here, $a = -3$ and $b = 5$, thus:

$Var(5 - 3X) = (-3)^2 Var(X) = 9 imes Var(X)$

Since we've already established that $Var(X) = 1$, we have:

$Var(5 - 3X) = 9 imes 1 = 9$

Rohit needs a total score of at least 10 points after 5 games. After 3 games, he has 6 points, meaning he needs 4 points from the next 2 games.

To find the probability of scoring at least 4 points in 2 games, we consider the possible outcomes:

Total Points Outcomes:

• Points in each game can be 0, 1, 2, or 3.
• We calculate the combinations:

1. Score 4: ($2,2$)
2. Score 5: ($3,2$) or ($2,3$)
3. Score 6: ($3,3$)

Calculating Probabilities:

1. For points to total 4 (i.e., 2 pts from both games): $P(X=2) imes P(X=2) = 0.2 imes 0.2 = 0.04$

2. For points to total 5:

• $3$ from the first and $2$ from the second: $P(X=3) imes P(X=2) = 0.1 imes 0.2 = 0.02$
• $2$ from the first and $3$ from the second: $P(X=2) imes P(X=3) = 0.2 imes 0.1 = 0.02$ Total for 5: $0.02 + 0.02 = 0.04$

3. For points to total 6 (i.e., $3,3$): $P(X=3) imes P(X=3) = 0.1 imes 0.1 = 0.01$

Total Probability of Winning:

Adding all valid outcomes: $0.04 + 0.04 + 0.01 = 0.09$