Exam-style Questions: Probability
1.Ten pupils are placed at random in a straight line. Find:
a) how many different ways there are of them lining up
(1 marks)
b) the number of ways of arranging the 10 pupils if the 2 youngest are to stand next to each other
(3 marks)
c) how many different ways the first 4 places can be filled if they were to have a race.
(2 marks)
d) the number of combinations of selecting 2 representatives from the 10
(2 marks)
(Marks available: 8)
Answer outline and marking scheme for question: 1
a) 10! = 3,628,800
(1 mark)
b) the trick is to treat the 2 youngest as 1 item (M1). This leaves us with arranging 9 pupils, ie 9! (A1) Ways.
BUT the 2 youngest can be arranged in 2! ways
so total ways = 2! x9! = 725760
(3 marks)
c) | = |
10! |
= 5040 |
(2 marks)
d) | = |
10! 8! 2! |
= 45 |
(2 marks)
(Marks available: 8)
2. On my way to work the probability that I have to stop at the first set of traffic lights is 0.3. The probability of stopping at the second is 0.75.
Find the probability that on one morning:
a) I stop at both sets of lights
(2 marks)
b) I stop only once at the second set of lights
(2 marks)
c) I stop at least once
(2 marks)
(Marks available: 6)
Answer outline and marking scheme for question: 2
Although it isn't essential in this case it is advisable to draw a tree diagram to help with the calculations.
a) P(stop and stop) = 0.3 x 0.75 = 0.225
(2 marks)
b) P(go and stop) = 0.7 x 0.75 = 0.525
(2 marks)
c) P(stop at least once) = 1 - P(no stop)
= 1 - 0.175 = 0.825
(2 marks)
(Marks available: 6)
3. In a penalty shoot out competition 2 players take a penalty to score a goal. Each player has a probability of 0.7 to score.
Calculate the probability of:
a) only one penalty being scored
on further inspection it appears that if the 1st penalty taker misses the probability of the 2nd penalty taker scoring is increased to 0.85
(2 marks)
b) draw a tree diagram to show all the events of the 2 penalties using this new evidence
(2 marks)
c) find the probability that the second penalty is missed
(2 marks)
d) find the probability that the 1st penalty is missed given that the second is missed.
(3 marks)
(Marks available: 9)
Answer outline and marking scheme for question: 3
a) P(only 1 scored) = P(score and miss) or P(miss and score)
= 0.7 x 0.3 + 0.3 x 0.7 = 0.42
(2 marks)
b) see ans sheet 2 for tree diagram
(2 marks)
c) P(2nd miss) = P(miss and miss) or P(score and miss)
= 0.3 x 0.85 + 0.7 x 0.3 = 0.255
(2 marks)
d) P(1st miss|2nd miss) = P(1st and 2nd miss) = 0.045 = 0.176 (3sf)
P(2nd miss) 0.255
(3 marks)
(Marks available: 9)