Date | May 2016 | Marks available | 8 | Reference code | 16M.2.hl.TZ1.10 |
Level | HL only | Paper | 2 | Time zone | TZ1 |
Command term | Find | Question number | 10 | Adapted from | N/A |
Question
Students sign up at a desk for an activity during the course of an afternoon. The arrival of each student is independent of the arrival of any other student and the number of students arriving per hour can be modelled as a Poisson distribution with a mean of λλ.
The desk is open for 4 hours. If exactly 5 people arrive to sign up for the activity during that time find the probability that exactly 3 of them arrived during the first hour.
Markscheme
P(3 in the first hour) =λ3e−λ3!=λ3e−λ3! A1
number to arrive in the four hours follows Po(4λ)Po(4λ) M1
P(5 arrive in total) =(4λ)5e−4λ5!=(4λ)5e−4λ5! A1
attempt to find P(2 arrive in the next three hours) M1
=(3λ)2e−3λ2!=(3λ)2e−3λ2! A1
use of conditional probability formula M1
P(3 in the first hour given 5 in total) =λ3e−λ3!×(3λ)2e−3λ2!(4λ)5e−4λ5!=λ3e−λ3!×(3λ)2e−3λ2!(4λ)5e−4λ5! A1
(92!3!)(455!)=45512=0.0879(92!3!)(455!)=45512=0.0879 A1
[8 marks]
Examiners report
A more difficult question, but it was still surprising how many candidates were unable to make a good start with it. Many were using λ=54λ=54 and consequently unable to progress very far. Many students failed to recognise that a conditional probability should be used.