Two fair 4-sided dice each has its faces labelled '1', '2', '3' and '4'. The two dice are thrown and the absolute difference in score on their bottom faces is denoted by XX.

  1. (i)

    Find P(X=x)\mathrm{P}(X = x) for all possible values of xx.

  2. (ii)

    Find E(X)\mathrm{E}(X) and Var(X)\mathrm{Var}(X).

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Worked solution

5(i)

1. Tabulate the absolute differences for all possible outcomes of rolling two 4-sided dice.

Let the scores on the two dice be D1D_1 and D2D_2. The possible values for X=∣D1−D2∣X = |D_1 - D_2| are shown in the table:
D1∖D2123410123210123210143210\begin{array}{|c|cccc|} \hline D_1 \setminus D_2 & 1 & 2 & 3 & 4 \\ \hline 1 & 0 & 1 & 2 & 3 \\ 2 & 1 & 0 & 1 & 2 \\ 3 & 2 & 1 & 0 & 1 \\ 4 & 3 & 2 & 1 & 0 \\ \hline \end{array}
There are 4×4=164 \times 4 = 16 equally likely outcomes.

2. Calculate the probability $\mathrm{P}(X=x)$ for each possible value of $X$.

Counting the frequencies of each possible absolute difference:

  • X=0X = 0 appears 44 times: P(X=0)=416=14\mathrm{P}(X=0) = \frac{4}{16} = \frac{1}{4}
  • X=1X = 1 appears 66 times: P(X=1)=616=38\mathrm{P}(X=1) = \frac{6}{16} = \frac{3}{8}
  • X=2X = 2 appears 44 times: P(X=2)=416=14\mathrm{P}(X=2) = \frac{4}{16} = \frac{1}{4}
  • X=3X = 3 appears 22 times: P(X=3)=216=18\mathrm{P}(X=3) = \frac{2}{16} = \frac{1}{8}
Answer

P(X=0)=14\mathrm{P}(X=0) = \frac{1}{4}
P(X=1)=38\mathrm{P}(X=1) = \frac{3}{8}
P(X=2)=14\mathrm{P}(X=2) = \frac{1}{4}
P(X=3)=18\mathrm{P}(X=3) = \frac{1}{8}

5(ii)

1. Recall or calculate the probability distribution of $X$ from part (i).

The possible outcomes for two 4-sided dice give a total of 4×4=164 \times 4 = 16 equally likely pairs. The absolute differences XX and their probabilities are:

  • P(X=0)=416=14\mathrm{P}(X=0) = \frac{4}{16} = \frac{1}{4}
  • P(X=1)=616=38\mathrm{P}(X=1) = \frac{6}{16} = \frac{3}{8}
  • P(X=2)=416=14\mathrm{P}(X=2) = \frac{4}{16} = \frac{1}{4}
  • P(X=3)=216=18\mathrm{P}(X=3) = \frac{2}{16} = \frac{1}{8}

2. Calculate the expectation $\mathrm{E}(X)$.

E(X)=∑xP(X=x)=0(14)+1(38)+2(14)+3(18)=0+38+48+38=108=54 (or 1.25)\begin{align*} \mathrm{E}(X) &= \sum x \mathrm{P}(X=x) \\ &= 0\left(\frac{1}{4}\right) + 1\left(\frac{3}{8}\right) + 2\left(\frac{1}{4}\right) + 3\left(\frac{1}{8}\right) \\ &= 0 + \frac{3}{8} + \frac{4}{8} + \frac{3}{8} \\ &= \frac{10}{8} = \frac{5}{4} \text{ (or } 1.25\text{)} \end{align*}

3. Calculate the variance $\mathrm{Var}(X)$.

First, calculate E(X2)\mathrm{E}(X^2):

E(X2)=∑x2P(X=x)=02(14)+12(38)+22(14)+32(18)=0+38+4(28)+98=208=52 \begin{align*} \mathrm{E}(X^2) &= \sum x^2 \mathrm{P}(X=x) \\ &= 0^2\left(\frac{1}{4}\right) + 1^2\left(\frac{3}{8}\right) + 2^2\left(\frac{1}{4}\right) + 3^2\left(\frac{1}{8}\right) \\ &= 0 + \frac{3}{8} + 4\left(\frac{2}{8}\right) + \frac{9}{8} = \frac{20}{8} = \frac{5}{2} \end{align*}

Then, use the variance formula:

Var(X)=E(X2)−[E(X)]2=52−(54)2=4016−2516=1516 (or 0.9375) \begin{align*} \mathrm{Var}(X) &= \mathrm{E}(X^2) - [\mathrm{E}(X)]^2 \\ &= \frac{5}{2} - \left(\frac{5}{4}\right)^2 \\ &= \frac{40}{16} - \frac{25}{16} \\ &= \frac{15}{16} \text{ (or } 0.9375\text{)} \end{align*}
Answer

E(X)=54\mathrm{E}(X) = \frac{5}{4}, Var(X)=1516\mathrm{Var}(X) = \frac{15}{16}

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