The binomial distribution
Practice section 6 of 7
Questions 31–36 in this collection
Practice questions
Review practice · 1 of 6
Translate birthday counts and find a strict minimum sample size
It may be assumed that the dates of birth in a large population are distributed throughout the year so that the probability of a randomly chosen person’s date of birth being in any particular month may be taken as .
(i) Find the probability that, of 6 people chosen at random, exactly 2 will have birthdays in January. [3]
(ii) Find the probability that, of 8 people chosen at random, at least one will have a birthday in January. [3]
(iii) people are chosen at random. Find the least value of such that the probability that at least one will have a birthday in January exceeds 0.9. [3]
(iv) Find the probability that, of 100 people chosen at random, at least 40 will have birthdays in May, June, July or Agusut. [4]
[J84/I/12]
CJC · Review Problems · Question 9
See which counts the words include
(i) Exactly two January birthdays
n = 6, p = 0.08333333333333333. Included counts: 2–2.
Highlighted bars form the requested event.
0.073548619 ≈ 0.0735 (3 s.f.)
Solution
Review practice · 2 of 6
Derive a cookie success rate and compare two minimum-packet methods
Temasek Bakery produces a large number of “lucky” cookies during the festive season. of these cookies contains a winning slip. The cookies are sold by packets. Each packet consists of 8 randomly chosen cookies and the mean number of cookies with a winning slip per packet is 3.2.
(i) Find the probability that a randomly chosen packet contains at most 4 cookies with a winning slip. [2]
(ii) Anand buys 12 packets of cookies. Find the probability that there are more than 10 packets that contain at most 4 cookies with a winning slip. [2]
(iii) Beng decides to buy packets of cookies. Find the least value of such that the probability that there is at least one packet that contain exactly 4 cookies with a winning slip is more than 0.90. [4]
[2017/TJC/MYE/II/5]
CJC · Review Problems · Question 10
See which counts the words include
(i) At most four slips
n = 8, p = 0.4. Included counts: 0–4.
Highlighted bars form the requested event.
0.82632960 ≈ 0.826 (3 s.f.)
Solution
Review practice · 3 of 6
Convert weeks to trials and use a variance restriction
According to the school rules, a student who arrives at school after 0730 hours is considered late.
(i) During the school term, a boy cycles to school on 5 days each week. On any given day, the probability that he arrives after 0730 hours is 0.1. For a period of 4 weeks, calculate the probability that he is late on at least one day but not more than 3 days. [2]
When the boy has cycled to school for the th time, the probability that he arrived at school late at least once is greater than 0.99. Calculate the least value of . [2]
(ii) A girl travels to school by bus on 5 days a week. Over a long period of time, the variance of the number of days per week on which she is late is 0.8. Given that denotes the probability that she arrives late and that , evaluate . [3]
[2007/SRJC/II/10(a)]
CJC · Review Problems · Question 11
See which counts the words include
(i) Between one and three late days
n = 20, p = 0.1. Included counts: 1–3.
Highlighted bars form the requested event.
0.74547002 ≈ 0.745 (3 s.f.)
Solution
Review practice · 4 of 6
Read a binomial mode and an expected number of qualifying samples
Samples, each of 12 articles, are taken at random from a large consignment in which 25% of articles are defective.
Find the number of defective articles which is most likely to occur in a single sample, and find the probability of obtaining this number.
If 100 samples of 12 articles are to be examined, calculate the number of samples in which you would expect to find 4 or more defective articles.
CJC · Review Problems · Question 12
See which counts the words include
Most likely count: three defectives
n = 12, p = 0.25. Included counts: 3–3.
Highlighted bars form the requested event.
0.25810361 ≈ 0.258 (3 s.f.)
Solution
Review practice · 5 of 6
Use a noninteger mean, compare modal counts and audit finite-population assumptions
One question in a survey is as follows:
“Do you use Facebook during office hours at least once a day? Yes or No?”
It is known that on average, 7 out of 9 staff members answered “yes” to the above question.
For a random sample of 60 staff members,
(i) find the standard deviation of the number of staff members who answer “yes”, [1]
(ii) find the probability that the number of staff members who answer “yes” exceeds the mean, [3]
(iii) find the most probable number of staff members who answer “yes”, [2]
(iv) state, with a reason, if the above answers are valid if the total number of staff members in the company is small. [1]
[2013/DHS/10(b)(modified)]
CJC · Review Problems · Question 13
See which counts the words include
(ii) More yes responses than the mean
n = 60, p = 0.7777777777777778. Included counts: 47–60.
Highlighted bars form the requested event.
0.53212913 ≈ 0.532 (3 s.f.)
Solution
Review practice · 6 of 6
Find the smallest binomial distribution with two tied modes
The random variable is the number of successes in independent trials of an experiment in which the probability of a success at any trial is .
Denoting by , it is given that , .
A distribution is said to be bimodal if it has two modes.
Find the least value of , and the corresponding modes of , given that is bimodal and that . [3]
[2017/ACJC/MYE/I/8]
CJC · Review Problems · Question 14
See which counts the words include
First tied mode: 17
n = 24, p = 0.72. Included counts: 17–17.
Highlighted bars form the requested event.
0.17537403 ≈ 0.175 (3 s.f.)