Tim throws two identical fair dice simultaneously. Each die has six faces: two faces numbered 1, two faces numbered 2 and two faces numbered 3. His score is the sum of the two numbers shown on the dice.

(a)     (i)     Calculate the probability that Tim obtains a score of 6.

  (ii)     Calculate the probability that Tim obtains a score of at least 3.

Tim plays a game with his friend Bill, who also has two dice numbered in the same way. Bill’s score is the sum of the two numbers shown on his dice.

(b)     (i)     Calculate the probability that Tim and Bill both obtain a score of 6.

  (ii)     Calculate the probability that Tim and Bill obtain the same score.

(c)     Let X denote the largest number shown on the four dice.

  (i)     Show that \({\text{P}}(X \leqslant 2) = \frac{{16}}{{81}}\).

  (ii)     Copy and complete the following probability distribution table.

 

 

  (iii)     Calculate \({\text{E}}(X)\) and \({\text{E}}({X^2})\) and hence find \({\text{Var}}(X)\).

(d)     Given that X = 3, find the probability that the sum of the numbers shown on the four dice is 8.

 

(a)     let T be Tim’s score

(i)     \({\text{P}}(T = 6) = \frac{{11}}{9}{\text{ }}( = 0.111{\text{ 3 sf)}}\)     A1

 

(ii)     \({\text{P}}(T \geqslant 3) = 1 - {\text{P}}(T \leqslant 2) = 1 - \frac{1}{9} = \frac{8}{9}{\text{ }}( = 0.889{\text{ 3 sf)}}\)     (M1)A1

[3 marks]

 

(b)     let B be Bill’s score

(i)     \({\text{P}}(T = 6{\text{ and }}B = 6) = \frac{1}{9} \times \frac{1}{9} = \frac{1}{{81}}{\text{ }}( = 0.012{\text{ 3 sf)}}\)     (M1)A1

 

(ii)     \({\text{P}}(B = T) = {\text{P}}(2){\text{P}}(2) + {\text{P}}(3){\text{P}}(3) + \ldots + {\text{P}}(6){\text{P}}(6)\)

\( = \frac{1}{9} \times \frac{1}{9} + \frac{2}{9} \times \frac{2}{9} + \frac{3}{9} \times \frac{3}{9} + \frac{2}{9} \times \frac{2}{9} + \frac{1}{9} \times \frac{1}{9}\)     M1

\( = \frac{{19}}{{81}}{\text{ }}( = 0.235{\text{ 3 sf)}}\)     A1

[4 marks]

 

(c)     (i)     EITHER

\({\text{P}}(X \leqslant 2) = \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3}\)     M1A1

because \({\text{P}}(X \leqslant 2) = {\text{P}}\left( {(a,{\text{ }}b,{\text{ }}c,{\text{ }}d)|a,{\text{ }}b,{\text{ }}c,{\text{ }}d = 1,{\text{ }}2} \right)\)     R1

or equivalent

\({\text{P}}(X \leqslant 2) = \frac{{16}}{{81}}\)     AG

OR

there are sixteen possible permutations which are

 

     M1A1

 

Note: This information may be presented in a variety of forms.

 

\({\text{P}}(X \leqslant 2) = \frac{{1 + 4 + 6 + 4 + 1}}{{81}}\)     A1

\( = \frac{{16}}{{81}}\)     AG

 

(ii)         A1A1

 

(iii)     \({\text{E}}(X) = \sum\limits_{x = 1}^3 {x{\text{P}}(X = x)} \)     (M1)

\( = \frac{1}{{81}} + \frac{{30}}{{81}} + \frac{{195}}{{81}}\)

\( = \frac{{226}}{{81}}\,\,\,\,\,\)(2.79 to 3 sf)     A1

\({\text{E}}({X^2}) = \sum\limits_{x = 1}^3 {{x^2}{\text{P}}(X = x)} \)

\( = \frac{1}{{81}} + \frac{{60}}{{81}} + \frac{{585}}{{81}}\)

\( = \frac{{646}}{{81}}\,\,\,\,\,\)(7.98 to 3 sf)     A1

\({\text{Var}}(X) = {\text{E}}({X^2}) - {\left( {{\text{E}}(X)} \right)^2}\)     (M1)

\( = 0.191\) (3 sf)     A1

Note: Award M1A0 for answers obtained using rounded values \(\left( {e.g.{\text{ Var}}(X) = 0.196} \right)\).

[10 marks]

 

(d)    

\({\text{P}}\left( {{\text{total is }}8 \cap (X = 3)} \right) = \frac{{18}}{{81}}\)     M1A1

since \({\text{P}}(X = 3) = \frac{{65}}{{81}}\)

\({\text{P}}\left( {{\text{total is }}8|(X = 3)} \right) = \frac{{{\text{P}}\left( {({\text{total is 8)}} \cap (X = 3)} \right)}}{{{\text{P}}(X = 3)}}\)     M1

\( = \frac{{18}}{{65}}{\text{ }}( = 0.277)\)     A1

[4 marks]

Total [21 marks]

Most candidates with a reasonable understanding of probability managed to answer well parts (a), (b) and some of part (c). However some candidates did not realize that different scores were not equally likely which lead to incorrect answers in several parts. Surprisingly, many candidates completed the table in part c) ii) with values that did not add up to 1. Very few candidates answered part (d) well. The enumeration of possible cases was sometimes attempted but with little success.