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By Samantha Freeman,2014-11-28 01:50
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Topic 9A. Venn diagrams, tree diagrams, tables

?Q1. (a) ??? (b) ??? (c) ???

~????????? (d) ??? (e) ???? ??? (f) ?????

Q2. (a) 2 (b) 38 ，?? (c) (d) ??：：

Q3.

Find this first … Then this … Then this.

C J A B Q4. (c) P(B′) = 0.6 23 8 15 0.2 0.1 0.3 5 4 7 0.4 ：?：???44 Q5. (a) (b) (c) 14 ：，?：，?：，?

S

nQ6. Explain the terms mutually exclusive and independent, and construct Venn diagrams to illustrate

each concept.

Q7. (a) the set B A'.

15 35 15 (b) n(B A′) = 35 35 (c) P( B A′ ) = 0.35

nQ8. The Venn diagram shows a sample space U and events A and B.

n(U) = 36, n(A) = 11, n(B) = 6 and n(A B)′ = 21.

B (a) On the diagram, shade the region (A B)′. A

(b) Find (i) n(A B); (ii) P(A B).

(c) Explain why events A and B are not mutually exclusive.

nQ9. A bag contains 10 red balls, 10 green balls and 6 white balls.

Two balls are drawn at random from the bag without replacement.

What is the probability that they are of different colours?

nQ10. A coin is tossed three times. (a) How many outcomes are in the sample space?

(b) What is the probability of at least two heads?

Q11. Males Females Totals

Unemployed 20 40 60

Employed 90 50 140 (i) 40/200 (ii) 110/200

Totals 110 90 200

Q12.

Boys Girls Total

Television 13 25 38

Sport 33 29 62

Total 46 54 100

(a) 38/100 (b) 71/100 (c) 13/100

Q13.

Pass Fail Total

Machine A 48 12 60

Machine B 36 4 40

Total 84 16 100

(a) P( passes inspection) = 0.84

(b) The company would like the probability that a box passes inspection to be 0.87.

Find the percentage of boxes that should be made by machine B to achieve this.

Pass Fail Total

Machine A x

Machine B ?？？？??

Total 87 100

？；：， ? ？；；?？？？?，? ? ：，

？；：， ? ；？?？；；， ? ：，

?？；？， ? ?！ ，?！？ so B makes 70% of boxes.

Q14. 11/36

Third apple Outcomes First apple Second apple

Red

Red Green

Red Red

Green Green

Red

Red

Green Green

Red Green

Green nQ15. A box contains 22 red apples

and 3 green apples.

Three apples are selected at random,

one after the other, without replacement.

(a) The first two apples are green. What is the probability that the third apple is red? ?；，，；，：；????(b) What is the probability that exactly two of the three apples are red? = ，?；，?；，?，???

9B. Addition Principle and Multiplication Principle

nQ1. A class has 12 boys and 8 girls. There are 4 Japanese boys and 3 Japanese girls.

If a student is selected at random, find (a) p(they are Japanese or a girl)

(b) p(they are not Japanese or a girl)

nQ2. (a) For the events A and B, p(A) = 0.7, p(B) = 0.9 and p(A ? B) = 0.6. p( A?B ) = 1

(b) For the events C and D, p(C) = 0.2, p(D) = 0.8 and p(C ? D) = 0.1. p( C?D ) = 0.9

(c) For the events E and F, p(E) = 0.6, p(F) = 0.5 and p(E ? F) = 0.4. p( E?F ) = 0.6+0.5-0.4 = 0.7

nQ3. For the events A and B, p(A) = 0.5, p(B) = 0.4 and p(A ? B) = 0.9. Show that A and B are mutually exclusive.

nQ4. (a) For the events C and D, p(C) = , p(D) = ，， and p(C ? D) = 0.1. Find x, given that p(C ? D) = 0.7.

(b) For the events E and F, p(E) = ？；，, p(F) = and p(E ? F) = 0.1. Find x, given that p(E ? F) = 0.9. 0.6

Multiplication Principle

nQ5. A class has 4 boys and 8 girls. Three students are selected at random.

(a) If there is no replacement find the probability of getting three boys.

(b) If there is replacement find the probability of getting three boys. ??nQ6. For events A and B, the probabilities are p(A) = , p (B) = . ：：：：

(a) Calculate the value of p(A B) if events A and B are independent. ? (b) Calculate the value of p (A B) if p(B/A) = . ：：

nQ7. (a) (i) For events M and N, p(M) = 0.4 and p(N) = 0.9. Calculate p(M/N) given p(M N) = 0.6.

(ii) Are events M and N independent or mutually exclusive? Explain.

(b) (i) For events A and J, p(A) = 0.2 and p(J) = 0.6. Calculate p(J/A) given p(A J) = 0.12. 0.6

(ii) independent? (Yes! p(A).p(J) = p(A J) ) mutually exclusive? (No! p(A J) ?？?

cQ8. (a) A class contains 13 girls and 11 boys. The teacher randomly selects four students.

Determine the probability that all four students selected are girls.

(b) A painter has 12 tins of paint. Seven tins are red and five tins are yellow.

Two tins are chosen at random. Calculate the probability that both are the same colour.

?????，?： P(same) = p(R and R or Y and Y) = p(R).p(R/R) + p(Y).p(Y/Y) = + = = ：，：：：，：：：?，?? nQ9. Events E and F are independent, with p(E) = and p(E F) = . Calculate p(F) and p(E F). ?? p(E F) = =p(E).p(F) = . p(F) so p(F) = ??? p(E F) = p(E) + p(F) - p(E F) = + ? = ???

nQ10. Two events A and B are such that p(A) = 0.5, p(B) = 0.4 and p(A B) = 0.3.

(a) Find p(A/B) and p(B/A)

(b) Hence, determine if the events are independent.

?????? Combining the Addition and Multiplication Principles ????? ??? ??? ??????

nQ11. Toss a coin and roll a die. Find the probability of not a '6' or a tail.

nQ12. p(rain in Spain) is 0.4 and p(rain in KL) is 0.8. Find p(rain in at least one of those cities).

nQ13. Let A and B be independent events, where p(A) = 0.6 and p(B) = x.

(a) Write down an expression for p(A ? B). 0.6

(b) (i) ??? ?? ? ？；： ? ？；； ?， ? ？；；， ？；，， ?？；，， ， ?；(

(ii) p(A ? B) =0.6(0.5) = 0.3

(c) Explain why A and B are not mutually exclusive. They have overlap-lah!! Why you so bodoh.

cQ14*. Events A and B are independent. Given p(B) = 2p(A), and p(A B) = 0.52, find p(B).

nQ15* My friend and I are going for a Driver's Licence test. We will toss a coin to see who goes first.

We both have a 90% chance of passing. However my friend will be unsettled if I go first and fail.

He would then only have a 60% chance of passing. I don't care about him.

Find the probability at least one of us passes. (0.975)

9C. Conditional probability

cQ1. For the events A and B, p(A) = 0.5, p(B) = 0.4 and p(A B) = 0.3.

?；， Find (a) p( A B' ) = 0.2 (b) p(A given B' ). = = ?；??

nQ2. Consider events A, B such that P (A) 0, P (A) 1, P (B) 0, and P (B) 1.

In each of the situations (a), (b), (c) below state whether A and B are

mutually exclusive (M); independent (I); neither (N).

(a) P(A|B) = P(A) (b) P(A B) = 0 (c) P(A B) = P(A)

nQ3.

Sample Arrive at Alarm rings space school

? ? WL L ！，

? W ，？ ??? ( W( ！，? ?? ? )( (，？ L ? ?) ? ?，？??( )( ? ( ???? (b) Calculate the probability that Duncan will be late for school. ? = ?，??：??

?????， (c) p(woken / late) = = ?????：??

nQ4. The events B and C are dependent, where C is the event “a student takes Chemistry”, and B is the

event “a student takes Biology”. It is known that

P(C) = 0.4, P(B | C) = 0.6, P(B | C) = 0.5.

(a) Complete the following tree diagram.

ChemistryBiology

B

C0.4

B

B

C

B

(b) Calculate the probability that a student takes Biology.

(c) Given that a student takes Biology, what is the probability that the student takes Chemistry?

cQ5. In a school of 88 boys, 32 study economics (E), 28 study history (H) and 39 do not study either

subject. This information is represented in the following Venn diagram.

(88)U

E(32)H(28)

abc

39

(a) a, b, c. 21, 11, 17 ：：，：(b) (i) p( studies both economics and history) = (ii) p(not history given economics) = ???，

137cQ6. Let A and B be events such that P(A) = , P(B) = and P(A B) =. 248

(a) Calculate P(A B). (b) Calculate P(A?B).

(c) Are the events A and B independent? Give a reason for your answer.

cQ7. A packet of seeds contains 40% red seeds and 60% yellow seeds. The probability that a red seed

grows is 0.9, and that a yellow seed grows is 0.8. A seed is chosen at random from the packet.

(a) Complete the probability tree diagram below.

Grows0.9

Red0.4

Does not grow

Grows

Yellow

Does not grow (b) (i) Calculate the probability that the chosen seed is red and grows.

(ii) Calculate the probability that the chosen seed grows.

(iii) Given that the seed grows, calculate the probability that it is red.

cQ8.

Year 1 Year 2 Totals

History 50 35 85

Science 15 30 45

Art 45 35 80

Totals 110 100 210

：？！(?(a) (i) p(A) = ? (ii) p(Year 2 Art student) = ，？？，？？??：????(iii) Are the events A and B independent? NO. p(A).p(B) = . ?p(A???? ，：?，：?，：?

??：：?；：??：??；：：?，?? (b) (c) + = ??，：?；，??，：?；，?????

：：????：：：：??nQ9. (a) p(B) = ? = (b) p(A B) = ? ? = (c) p(A B) = = ????????????

nQ10. In a class, 40 students take chemistry only, 30 take physics only, 20 take both chemistry and

physics, and 60 take neither.

(a) Find the probability that a student takes physics given that the student takes chemistry.

(b) Find the probability that a student takes physics given that the student does not take chemistry.

(c) State whether the events “taking chemistry” and “taking physics” are mutually exclusive,

20(a) P(P(C) = A1 20;40

1 = A1 N1 3

30(b) P(P(C) = A1 30;60

1 = A1 N1 3

(c) Investigating conditions, or some relevant calculations (M1)

P is independent of C, with valid reason A1 N2

eg P(P(C) = P(P(C), P(P(C) = P(P),

205060 (ie P(P C) = P(P) P(C)) 150150150

cQ11. Two restaurants, Center and New, sell fish rolls and salads.

Let F be the event a customer chooses a fish roll.

Let S be the event a customer chooses a salad.

Let N be the event a customer chooses neither a fish roll nor a salad. In the Center restaurant P(F) = 0.31, P(S) = 0.62, P(N) = 0.14.

(a) Show that P(F S) = 0.07.

(b) Given that a customer chooses a salad, find the probability they also chooses a fish roll.

At New restaurant, P(N) = 0.14. Twice as many customers choose a salad as choose a fish roll. Choosing a fish roll is independent of choosing a salad.

(d) Find the probability that a fish roll is chosen.

(a) P(F S) = 1 0.14 (= 0.86) (A1)

Choosing an appropriate formula (M1) eg P(A B) = P(A) + P(B) P(A B)

Correct substitution 1 2 3 4 5 6 eg P(F S) = 0.93 0.86 A1

P(F S) = 0.07 AG N0

Notes:There are several valid approaches. Award(A1)(M1)A1 for relevant working using any

appropriate strategy eg formula, Venn Diagram, or table.

Award no marks for the incorrect solution

P(F S) = 1 P(F) + P(S) = 1 0.93 = 0.07

(b) Using conditional probability (M1)

?P(FS)?eg P(F S) ?P(S)~?

0.07P(F S) = (A1) 0.62

= 0.113 A1 N3

(c) F and S are not independent A1 N1

EITHER

If independent P(F S) = P(F), 0.113 0.31 R1R1 N2

OR

If independent P(F S) = P(F) P(S), 0.07 0.31 0.62 (= 0.1922) R1R1 N2

(d) Let P(F) = x

P(S) = 2P(F) (= 2x) (A1)

2For independence P(F S) = P(F)P(S) (= 2x) (R1)

Attempt to set up a quadratic equation (M1)

2eg P(F S) = P(F)P(S) P(F)P(S), 0.86 = x + 2x 2x

22x 3x + 0.86 = 0 A2

x = 0.386, x = 1.11 (A1)

P(F) = 0.386 (A1) N5

[16]

nQ12. Two fair dice are thrown and 1 2 3 4 5 6 7

the number showing on each is noted. 2 3 4 5 6 7 8

The sum of these two numbers is S. 3 4 5 6 7 8 9

5 4 6 7 8 9 10 ，：5 6 7 8 9 10 11 (a) p(S is less than 8) = ??：：6 7 8 9 10 11 12 (b) at least one die shows a 3; ??

(c) at least one die shows a 3, ? given that S is less than 8. ，：

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