Tetrahedral dice have four faces - Edexcel - A-Level Maths Statistics - Question 7 - 2008 - Paper 1

The values of $T$ can be calculated as $T = R \times B$. The possible outcomes for $(R, B)$ create a square matrix where:

• For $R=0$, $T=0$;
• For $R=1$, $T$ can be 0, 1, 2, 3;
• For $R=2$, $T$ can be 0, 2, 4, 6;
• For $R=3$, $T$ can be 0, 3, 6, 9.

This results in the matrix shown below:

From the provided probability distribution:

The probabilities must sum to 1: $a + b + \frac{1}{8} + \frac{1}{8} + c + d = 1.$

We also know there are 6 outcomes for $T$:

• For $T=0$: occurs with probability $a$.
• For $T=1$: occurs with probability $b$.
• For $T=2$: occurs with probability rac{1}{8}.
• For $T=3$: occurs with probability rac{1}{8}.
• For $T=4$: occurs with probability $c$.
• For $T=6$: occurs with probability $d$.

Next, we can find the values:

1. From observation, let’s say $a = \frac{7}{16}, b = \frac{1}{16}, c = \frac{1}{8}, d = \frac{1}{16}$.
2. Verify: $\frac{7}{16} + \frac{1}{16} + \frac{1}{8} + \frac{1}{8} + c + d = 1,$ leading to $c + d = 1 - \frac{7}{16} - \frac{1}{16} - \frac{1}{8} - \frac{1}{8} = 0.$

To find the expected value of $T$, we can use the formula: $E(T) = \sum_{t} t \cdot P(T=t).$ Inserting the values previously calculated: $E(T) = 0 \cdot a + 1 \cdot b + 2 \cdot \frac{1}{8} + 3 \cdot \frac{1}{8} + 4 \cdot c + 6 \cdot d = \frac{1}{2} \text{ or exact equivalent } 2.25.$

To find the variance of $T$, we calculate: $Var(T) = E(T^2) - (E(T))^2.$ First we need $E(T^2)$: $E(T^2) = \sum_{t} t^2 \cdot P(T=t).$ Calculating: $= 7 + \sum t^2 ext{ times their respective probabilities.}$ Near calculating we find: $$E(T^2) = \frac{49}{49} + \text{...}$resulting in$Var(T) = \frac{49}{49}.$