List the pairs of scores that give a sum of 6.

The probability distribution for the sum of the scores on the two dice is shown below.

Find the value of p , of q , and of r .

Fred plays a game. He throws two fair 4-sided dice four times. He wins a prize if the sum is 5 on three or more throws.

Find the probability that Fred wins a prize.

three correct pairs     A1A1A1     N3

e.g. (2, 4), (3, 3), (4, 2) , R2G4, R3G3, R4G2

[3 marks]

\(p = \frac{1}{{16}}\) , \(q = \frac{2}{{16}}\) , \(r = \frac{2}{{16}}\)     A1A1A1     N3

[3 marks]

let X be the number of times the sum of the dice is 5

evidence of valid approach     (M1)

e.g. \(X \sim {\rm{B}}(n{\text{, }}p)\) , tree diagram, 5 sets of outcomes produce a win

one correct parameter     (A1)

e.g. \(n = 4\) , \(p = 0.25\) , \(q = 0.75\)

Fred wins prize is \({\rm{P}}(X \ge 3)\)     (A1)

appropriate approach to find probability     M1

e.g. complement, summing probabilities, using a CDF function

correct substitution     (A1)

e.g. \(1 - 0.949 \ldots \) , \(1 - \frac{{243}}{{256}}\) , \(0.046875 + 0.00390625\) , \(\frac{{12}}{{256}} + \frac{1}{{256}}\)

\({\text{probability of winning}} = 0.0508\) \(\left( {\frac{{13}}{{256}}} \right)\)     A1     N3

[6 marks]

All but the weakest candidates managed to score full marks for parts (a) and (b). An occasional error in part (a) was including additional pair(s) or listing (3, 3) twice.

All but the weakest candidates managed to score full marks for parts (a) and (b).

Many candidates found part (c) challenging, as they failed to recognize the binomial probability. Successful candidates generally used either the binomial CDF function or the sum of two binomial probabilities. Some used approaches like multiplying probabilities or tree diagrams, but these were less successful.