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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. nn marbles are randomly drawn from the box, one by one and with replacement. The number of red marbles drawn is denoted by RR.

(iii) Given that the mean of RR is 4.5, find nn and P(R>4)P(R>4). [3]

(iv) Given instead that P(R=0 or 1)<0.01P(R=0\text{ or }1)<0.01, write down an inequality for nn and find the least value of nn. [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.

Count 0: 0.00355; outside eventCount 1: 0.0256; outside eventCount 2: 0.0844; outside eventCount 3: 0.169; outside eventCount 4: 0.228; outside eventCount 5: 0.219; in eventCount 6: 0.153; in eventCount 7: 0.0788; in eventCount 8: 0.0295; in eventCount 9: 0.00788; in eventCount 10: 0.00142; in eventCount 11: 0.000155; in eventCount 12: 0.00000773; in event
Count X ranges from 0 to 12. Use the inspector for individual values.

Highlighted bars form the requested event.

Solution

Review practice · 2 of 4

Compare an ordered third success with exactly three successes

A firm sells two types of electrical components, AA and BB.

Type AA components are packed in boxes of 50. On average, 2% of the Type AA components are faulty.

(i) State, in context, two assumptions needed for the number of faulty components of Type AA in a box to be well-modelled by a binomial distribution. [2]

You are now given that the number of faulty components of type AA 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 AA 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.

Count 0: 0.364; outside eventCount 1: 0.372; outside eventCount 2: 0.186; in eventCount 3: 0.0607; in eventCount 4: 0.0145; in eventCount 5: 0.00273; in eventCount 6: 0.000418; in eventCount 7: 0.0000536; in eventCount 8: 0.00000588; in eventCount 9: 5.60e-7; in eventCount 10: 4.69e-8; in eventCount 11: 3.48e-9; in eventCount 12: 2.31e-10; in eventCount 13: 1.38e-11; in eventCount 14: 7.42e-13; in eventCount 15: 3.64e-14; in eventCount 16: 1.62e-15; in eventCount 17: 6.63e-17; in eventCount 18: 2.48e-18; in eventCount 19: 8.52e-20; in eventCount 20: 2.70e-21; in eventCount 21: 7.86e-23; in eventCount 22: 2.11e-24; in eventCount 23: 5.25e-26; in eventCount 24: 1.21e-27; in eventCount 25: 2.56e-29; in eventCount 26: 5.02e-31; in eventCount 27: 9.11e-33; in eventCount 28: 1.53e-34; in eventCount 29: 2.36e-36; in eventCount 30: 3.38e-38; in eventCount 31: 4.45e-40; in eventCount 32: 5.39e-42; in eventCount 33: 6.00e-44; in eventCount 34: 6.12e-46; in eventCount 35: 5.71e-48; in eventCount 36: 4.86e-50; in eventCount 37: 3.75e-52; in eventCount 38: 2.62e-54; in eventCount 39: 1.64e-56; in eventCount 40: 9.23e-59; in eventCount 41: 4.59e-61; in eventCount 42: 2.01e-63; in eventCount 43: 7.63e-66; in eventCount 44: 2.48e-68; in eventCount 45: 6.74e-71; in eventCount 46: 1.49e-73; in eventCount 47: 2.60e-76; in eventCount 48: 3.31e-79; in eventCount 49: 2.76e-82; in eventCount 50: 1.13e-85; in event
Count X ranges from 0 to 50. Use the inspector for individual values.

Highlighted bars form the requested event.

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, pp, 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 pp and solve for pp. [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 p=0.10p=0.10.

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.

Count 0: 0.00378; in eventCount 1: 0.0236; in eventCount 2: 0.0708; in eventCount 3: 0.136; outside eventCount 4: 0.187; outside eventCount 5: 0.196; outside eventCount 6: 0.163; outside eventCount 7: 0.111; outside eventCount 8: 0.0624; outside eventCount 9: 0.0295; outside eventCount 10: 0.0118; outside eventCount 11: 0.00402; outside eventCount 12: 0.00117; outside eventCount 13: 0.000293; outside eventCount 14: 0.0000628; outside eventCount 15: 0.0000115; outside eventCount 16: 0.00000180; outside eventCount 17: 2.38e-7; outside eventCount 18: 2.65e-8; outside eventCount 19: 2.44e-9; outside eventCount 20: 1.83e-10; outside eventCount 21: 1.09e-11; outside eventCount 22: 4.95e-13; outside eventCount 23: 1.61e-14; outside eventCount 24: 3.36e-16; outside eventCount 25: 3.36e-18; outside event
Count X ranges from 0 to 25. Use the inspector for individual values.

Highlighted bars form the requested event.

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 XX be the number of passengers who do not turn up for that flight.

(i) State, in context, two assumptions needed for XX 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.

Count 0: 0.000158; in eventCount 1: 0.00144; in eventCount 2: 0.00651; in eventCount 3: 0.0194; in eventCount 4: 0.0431; in eventCount 5: 0.0757; outside eventCount 6: 0.110; outside eventCount 7: 0.135; outside eventCount 8: 0.144; outside eventCount 9: 0.135; outside eventCount 10: 0.112; outside eventCount 11: 0.0844; outside eventCount 12: 0.0575; outside eventCount 13: 0.0358; outside eventCount 14: 0.0204; outside eventCount 15: 0.0108; outside eventCount 16: 0.00528; outside eventCount 17: 0.00240; outside eventCount 18: 0.00102; outside eventCount 19: 0.000406; outside eventCount 20: 0.000152; outside eventCount 21: 0.0000535; outside eventCount 22: 0.0000178; outside eventCount 23: 0.00000557; outside eventCount 24: 0.00000166; outside eventCount 25: 4.67e-7; outside eventCount 26: 1.25e-7; outside eventCount 27: 3.18e-8; outside eventCount 28: 7.69e-9; outside eventCount 29: 1.78e-9; outside eventCount 30: 3.91e-10; outside eventCount 31: 8.23e-11; outside eventCount 32: 1.66e-11; outside eventCount 33: 3.18e-12; outside eventCount 34: 5.86e-13; outside eventCount 35: 1.03e-13; outside eventCount 36: 1.75e-14; outside eventCount 37: 2.84e-15; outside eventCount 38: 4.41e-16; outside eventCount 39: 6.59e-17; outside eventCount 40: 9.46e-18; outside eventCount 41: 1.30e-18; outside eventCount 42: 1.73e-19; outside eventCount 43: 2.20e-20; outside eventCount 44: 2.70e-21; outside eventCount 45: 3.18e-22; outside eventCount 46: 3.61e-23; outside eventCount 47: 3.94e-24; outside eventCount 48: 4.14e-25; outside eventCount 49: 4.18e-26; outside eventCount 50: 4.07e-27; outside eventCount 51: 3.82e-28; outside eventCount 52: 3.45e-29; outside eventCount 53: 3.00e-30; outside eventCount 54: 2.51e-31; outside eventCount 55: 2.03e-32; outside eventCount 56: 1.57e-33; outside eventCount 57: 1.18e-34; outside eventCount 58: 8.46e-36; outside eventCount 59: 5.86e-37; outside eventCount 60: 3.91e-38; outside eventCount 61: 2.51e-39; outside eventCount 62: 1.55e-40; outside eventCount 63: 9.18e-42; outside eventCount 64: 5.24e-43; outside eventCount 65: 2.87e-44; outside eventCount 66: 1.51e-45; outside eventCount 67: 7.66e-47; outside eventCount 68: 3.72e-48; outside eventCount 69: 1.74e-49; outside eventCount 70: 7.77e-51; outside eventCount 71: 3.33e-52; outside eventCount 72: 1.37e-53; outside eventCount 73: 5.37e-55; outside eventCount 74: 2.02e-56; outside eventCount 75: 7.26e-58; outside eventCount 76: 2.49e-59; outside eventCount 77: 8.16e-61; outside eventCount 78: 2.55e-62; outside eventCount 79: 7.57e-64; outside eventCount 80: 2.14e-65; outside eventCount 81: 5.74e-67; outside eventCount 82: 1.46e-68; outside eventCount 83: 3.52e-70; outside eventCount 84: 8.02e-72; outside eventCount 85: 1.72e-73; outside eventCount 86: 3.49e-75; outside eventCount 87: 6.62e-77; outside eventCount 88: 1.18e-78; outside eventCount 89: 1.96e-80; outside eventCount 90: 3.02e-82; outside eventCount 91: 4.33e-84; outside eventCount 92: 5.73e-86; outside eventCount 93: 6.97e-88; outside eventCount 94: 7.74e-90; outside eventCount 95: 7.79e-92; outside eventCount 96: 7.06e-94; outside eventCount 97: 5.69e-96; outside eventCount 98: 4.04e-98; outside eventCount 99: 2.48e-100; outside eventCount 100: 1.30e-102; outside eventCount 101: 5.58e-105; outside eventCount 102: 1.90e-107; outside eventCount 103: 4.82e-110; outside eventCount 104: 8.06e-113; outside eventCount 105: 6.67e-116; outside event
Count X ranges from 0 to 105. Use the inspector for individual values.

Highlighted bars form the requested event.

Solution

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