The binomial distribution
Practice section 7 of 7
Questions 37–40 in this collection
Practice questions
Review practice · 1 of 4
Choose hypergeometric or binomial sampling from replacement rules
A jar contains 10 blue and 8 red marbles. Five marbles are randomly drawn from the box, one by one and without replacement.
(i) Explain why it is inappropriate to model the number of blue marbles by a binomial distribution. [1]
(ii) Find the probability that exactly three marbles are blue. [3]
Another jar contains 20 blue and 12 red marbles. marbles are randomly drawn from the box, one by one and with replacement. The number of red marbles drawn is denoted by .
(iii) Given that the mean of is 4.5, find and . [3]
(iv) Given instead that , write down an inequality for and find the least value of . [3]
[2018/JJC/II/8]
CJC · Review Problems · Question 15
See which counts the words include
(iii) More than four red marbles
n = 12, p = 0.375. Included counts: 5–12.
Highlighted bars form the requested event.
0.48971663 ≈ 0.490 (3 s.f.)
Solution
Review practice · 2 of 4
Compare an ordered third success with exactly three successes
A firm sells two types of electrical components, and .
Type components are packed in boxes of 50. On average, 2% of the Type components are faulty.
(i) State, in context, two assumptions needed for the number of faulty components of Type in a box to be well-modelled by a binomial distribution. [2]
You are now given that the number of faulty components of type in a box follows a binomial distribution.
(ii) Find the probability that a box of these components contains more than 1 faulty component. [2]
Five boxes of Type components are selected at random.
(iii) Find the probability that the fifth box is the third box selected which has more than one faulty component. [2]
(iv) Find the probability that there are 3 boxes with more than 1 faulty component each. [2]
(v) Explain why the answer to part (iv) is greater than the answer to part (iii). [1]
[2021/ASRJC/II/10]
CJC · Review Problems · Question 16
See which counts the words include
(ii) More than one faulty component
n = 50, p = 0.02. Included counts: 2–50.
Highlighted bars form the requested event.
0.26422861 ≈ 0.264 (3 s.f.)
Solution
Review practice · 3 of 4
Separate infection counts from a later conditional-test model
During a pandemic, an unknown infectious disease spread through a population. Researchers wanted to study the probability, , that a person is infected. They collected some data from a random sample of 25 people.
(i) State, in the context of this question, two assumptions needed for the number of infected people to be well modelled by a binomial distribution. [2]
Assume now that these assumptions do in fact hold.
(ii) It is given that the probability that at most 2 people are infected is 0.0982. Write an equation in and solve for . [2]
A total of 30 random samples of 25 people were chosen from this population.
(iii) Find the probability that the total number of samples with at most 2 infected people is at least 5. [3]
It is now given that .
A diagnostic test for the presence of the disease was carried out on the population. It was found that the percentage of infected people who were tested positive is 85%, while the percentage of non-infected people who were tested negative is 96%.
(iv) Find the probability that a person who tested positive carries the disease. [3]
(v) Hence or otherwise, find the probability that a person who tested positive does not carry the disease. [1]
(vi) Discuss briefly if the diagnostic test is worthwhile. [1]
[2021/SAJC/II/10]
CJC · Review Problems · Question 17
See which counts the words include
(ii) At most two in a sample of 25
n = 25, p = 0.2000123866059771. Included counts: 0–2.
Highlighted bars form the requested event.
0.098200000 ≈ 0.0982 (3 s.f.)
Solution
Review practice · 4 of 4
Reverse an overbooking event into a conditional no-show count
An airline company sold 105 tickets for a particular flight. Based on past experience, 8% of passengers do not turn up for the flight. Let be the number of passengers who do not turn up for that flight.
(i) State, in context, two assumptions needed for to be well modelled by a binomial distribution. [2]
(ii) Find the probability that fewer than 5 do not turn up for the flight. [2]
“Overbooking” occurs when there are more passengers who turn up than the number of seats available for the flight. When this happens, the airline will re-accommodate the excess passengers, that is, arrangements will be made for them to take another flight.
(iii) It is given that the particular flight has only 100 seats. Find the probability that at least 3 passengers have to be re-accommodated given that “overbooking” has occurred. [3]
[2017/MJC/MYE/I/9]
CJC · Review Problems · Question 18
See which counts the words include
(ii) Fewer than five no-shows
n = 105, p = 0.08. Included counts: 0–4.
Highlighted bars form the requested event.
0.070629383 ≈ 0.0706 (3 s.f.)