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

To find the cumulative distribution function $F(1.5)$, we sum the probabilities for all $x$ values up to 1.5: $F(1.5) = P(X=0) + P(X=1)$ $F(1.5) = 0.4 + 0.3 = 0.7$

To show that the variance $Var(X)=1$, we first calculate $E(X^2)$: $E(X^2) = 0^2 imes 0.4 + 1^2 imes 0.3 + 2^2 imes 0.2 + 3^2 imes 0.1$ $E(X^2) = 0 + 0.3 + 0.8 + 0.9 = 2.0$

Now, using the variance formula: $Var(X) = E(X^2) - (E(X))^2$ $Var(X) = 2.0 - (1.0)^2 = 2.0 - 1.0 = 1.0$

To find $Var(5 - 3X)$, we use the formula for variance of a linear transformation: $Var(aX + b) = a^2 Var(X)$ Here, $a = -3$ and $b = 5$: $Var(5 - 3X) = (-3)^2 Var(X) = 9 imes 1 = 9$

Rohit wins the prize if the total score after 5 games is at least 10. He has 6 points after 3 games, so he needs at least 4 points in the next 2 games.

The possible scores for $X$ are:

• Scoring 4 points requires: (2, 2)
• Scoring 5 points requires: (3, 2) or (2, 3)

Probability calculations are as follows:

1. ( P(X=2) = 0.2 ) and both games need to score 2:
   • Probability = ( 0.2 imes 0.2 = 0.04 )
2. ( P(X=3) = 0.1 ) and any combination of points:
   • Probability for (3, 2) or (2, 3) = ( 0.1 imes 0.2 + 0.2 imes 0.1 = 0.01 + 0.01 = 0.02 )

Thus, total probability that Rohit wins the prize:

$P(win) = 0.04 + 0.02 = 0.06$