Date | November 2018 | Marks available | 7 | Reference code | 18N.1.SL.TZ0.S_9 |
Level | Standard Level | Paper | Paper 1 | Time zone | Time zone 0 |
Command term | Find | Question number | S_9 | Adapted from | N/A |
Question
A bag contains nn marbles, two of which are blue. Hayley plays a game in which she randomly draws marbles out of the bag, one after another, without replacement. The game ends when Hayley draws a blue marble.
Let nn = 5. Find the probability that the game will end on her
Find the probability, in terms of nn, that the game will end on her first draw.
Find the probability, in terms of nn, that the game will end on her second draw.
third draw.
fourth draw.
Hayley plays the game when nn = 5. She pays $20 to play and can earn money back depending on the number of draws it takes to obtain a blue marble. She earns no money back if she obtains a blue marble on her first draw. Let M be the amount of money that she earns back playing the game. This information is shown in the following table.
Find the value of kk so that this is a fair game.
Markscheme
2n2n A1 N1
[1 mark]
correct probability for one of the draws A1
eg P(not blue first) = n−2nn−2n, blue second = 2n−12n−1
valid approach (M1)
eg recognizing loss on first in order to win on second, P(B' then B), P(B') × P(B | B'), tree diagram
correct expression in terms of nn A1 N3
eg n−2n×2n−1n−2n×2n−1, 2n−4n2−n2n−4n2−n, 2(n−2)n(n−1)2(n−2)n(n−1)
[3 marks]
correct working (A1)
eg 35×24×2335×24×23
1260(=15)1260(=15) A1 N2
[2 marks]
correct working (A1)
eg 35×24×13×2235×24×13×22
660(=110)660(=110) A1 N2
[2 marks]
correct probabilities (seen anywhere) (A1)(A1)
eg P(1)=25P(1)=25, P(2)=620P(2)=620 (may be seen on tree diagram)
valid approach to find E (M) or expected winnings using their probabilities (M1)
eg P(1)×(0)+P(2)×(20)+P(3)×(8k)+P(4)×(12k)P(1)×(0)+P(2)×(20)+P(3)×(8k)+P(4)×(12k),
P(1)×(−20)+P(2)×(0)+P(3)×(8k−20)+P(4)×(12k−20)P(1)×(−20)+P(2)×(0)+P(3)×(8k−20)+P(4)×(12k−20)
correct working to find E (M) or expected winnings (A1)
eg 25(0)+310(20)+15(8k)+110(12k)25(0)+310(20)+15(8k)+110(12k),
25(−20)+310(0)+15(8k−20)+110(12k−20)25(−20)+310(0)+15(8k−20)+110(12k−20)
correct equation for fair game A1
eg 310(20)+15(8k)+110(12k)=20, 25(−20)+15(8k−20)+110(12k−20)=0
correct working to combine terms in k (A1)
eg −8+145k−4−2=0, 6+145k=20, 145k=14
k = 5 A1 N0
Note: Do not award the final A1 if the candidate’s FT probabilities do not sum to 1.
[7 marks]