IB Math – High Level Year 1 – Probability Practice 1 Alei - Desert Academy Macintosh HD:Users:bobalei:Dropbox:Desert:HL1:6StatProb:Practice:HLProbPractice1.docx on 12/06/2015 at 1:43 PM Page 1 of 4 Probability Practice 1 1. A bag contains 2 red balls, 3 blue balls and 4 green balls. A ball is chosen at random from the bag and is not replaced. A second ball is chosen. Find the probability of choosing one green ball and one blue ball in any order. Working: Answer: ………………………………………….. (Total 4 marks) 2. In a bilingual school there is a class of 21 pupils. In this class, 15 of the pupils speak Spanish as their first language and 12 of these 15 pupils are Argentine. The other 6 pupils in the class speak English as their first language and 3 of these 6 pupils are Argentine. A pupil is selected at random from the class and is found to be Argentine. Find the probability that the pupil speaks Spanish as his/her first language. Working: Answer: ………………………………………….. (Total 4 marks) 3. A new blood test has been shown to be effective in the early detection of a disease. The probability that the blood test correctly identifies someone with this disease is 0.99, and the probability that the blood test correctly identifies someone without that disease is 0.95. The incidence of this disease in the general population is 0.0001. A doctor administered the blood test to a patient and the test result indicated that this patient had the disease. What is the probability that the patient has the disease? (Total 6 marks) 4. Given that events A and B are independent with P(A ∩ B) = 0.3 and P(A ∩ B′) = 0.3, find P(A ∪ B). Working: Answer: .................................................................. (Total 3 marks) 5. A girl walks to school every day. If it is not raining, the probability that she is late is . If it is raining, the probability that she is late is . The probability that it rains on a particular day is . On one particular day the girl is late. Find the probability that it was raining on that day. Working: Answer: ………………………………………….. (Total 3 marks) 5 1 3 2 4 1
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ProbabilityPractice11. A bag contains 2 red balls, 3 blue balls and 4 green balls. A ball is chosen at random from the bag
and is not replaced. A second ball is chosen. Find the probability of choosing one green ball and one blue ball in any order.
Working:
Answer: …………………………………………..
(Total 4 marks) 2. In a bilingual school there is a class of 21 pupils. In this class, 15 of the pupils speak Spanish as their
first language and 12 of these 15 pupils are Argentine. The other 6 pupils in the class speak English as their first language and 3 of these 6 pupils are Argentine.
A pupil is selected at random from the class and is found to be Argentine. Find the probability that the pupil speaks Spanish as his/her first language.
Working:
Answer: …………………………………………..
(Total 4 marks) 3. A new blood test has been shown to be effective in the early detection of a disease. The probability
that the blood test correctly identifies someone with this disease is 0.99, and the probability that the blood test correctly identifies someone without that disease is 0.95. The incidence of this disease in the general population is 0.0001.
A doctor administered the blood test to a patient and the test result indicated that this patient had the disease. What is the probability that the patient has the disease?
(Total 6 marks) 4. Given that events A and B are independent with P(A ∩ B) = 0.3 and P(A ∩ B′) = 0.3,
10. The probability that it rains during a summer’s day in a certain town is 0.2. In this town, the probability that the daily maximum temperature exceeds 25°C is 0.3 when it rains and 0.6 when it does not rain. Given that the maximum daily temperature exceeded 25°C on a particular summer’s day, find the probability that it rained on that day.
probability, P(B), that the building will be completed on time is 0.60. The probability that the materials arrive on time and that the building is completed on time is 0.55. (i) Show that events A and B are not independent. (ii) All the materials arrive on time. Find the probability that the building will not be
completed on time. (5)
(b) There was a team of ten people working on the building, including three electricians and two plumbers. The architect called a meeting with five of the team, and randomly selected people to attend. Calculate the probability that exactly two electricians and one plumber were called to the meeting.
(2) (c) The number of hours a week the people in the team work is normally distributed with a mean
of 42 hours. 10% of the team work 48 hours or more a week. Find the probability that both plumbers work more than 40 hours in a given week.
(8) (Total 15 marks)
14. The random variable X has a Poisson distribution with mean λ.
(a) Given that P(X = 4) = P(X = 2) + P(X = 3), find the value of λ. (3)
(b) Given that λ = 3.2, find the value of (i) P(X ≥ 2); (ii) P(X ≤ 3 | X ≥ 2).
(5) (Total 8 marks)
15. Robert travels to work by train every weekday from Monday to Friday. The probability that he
catches the 08.00 train on Monday is 0.66. The probability that he catches the 08.00 train on any other weekday is 0.75. A weekday is chosen at random. (a) Find the probability that he catches the train on that day. (b) Given that he catches the 08.00 train on that day, find the probability that the chosen day is
[4] 3. Let D be the event that the patient has the disease and S be the event that
the new blood test shows that the patient has the disease. Let D′ be the complement of D, ie the patient does not have the disease. Now the given probabilities can be written as p(S | D) = 0.99, p(D) = 0.0001, p(S | D′) = 0.05. (A1)(A1)(A1)
Since the blood test shows that the patient has the disease,
we are required to find p(D | S). By Bayes’ theorem,
p(D | S) = (M1)
= (M1)
= 0.001976...= 0.00198 (3 sf) (A1) OR
(A3) Note: Award (A1) for 0.99, (A1) for 0.0001, (A1) for 0.05
[13] 12. (a) The number of multiples of 4 is 250. (M1)
Required probability = 0.25. (A1)(C2) (b) The number of multiples of 4 and 6 is (M1)
the number of multiples of 12 (A1) = 83. (A1) Required probability = 0.083 (A1) (C4)
[6] 13. (a) (i) To be independent P(A ∩ B) = P(A) × P(B) (R1)
P(A) × P(B) = (0.85)(0.60) = 0.51 but P(A ∩ B) = 0.55 (A1) P(A ∩ B) ≠ P(A) × P(B) Hence A and B are not independent. (AG)
(ii)
P(B′ | A) = (M1)
= (M1)
= (= 0.353) (A1) 5
(b) Probability of 2 electricians and 1 plumber = (M1)
= (A1)
OR Probability of 2 electricians and 1 plumber = (M1)
= (= 0.238) (A1) 2
(c) X = number of hours worked. X ~ N (42, σ2) P(X ≥ 48) = 0.10 (AG) P(X < 48) = 0.90 (M1) Φ(z) = 0.90 z = 1.28 (z = 1.28155) (A1) (Answers given to more than 3 significant figures will be accepted.)
= 0.665 (A1) OR P(X > 40) = 0.665 (G2) Therefore, the probability that one plumber works more than 40 hours per week is 0.665. The probability that both plumbers work more than 40 hours per week = (0.665)2 (M1) = 0.443 (Accept 0.442 or 0.444) (A1) 8