Destination Maths Chapter 19 Probability II 1 Probability II Practice Questions 19.1 1. An ice cream shop has vanilla, chocolate or strawberry ice cream and you can have either chocolate or caramel sauce on top. (i) Represent this information on a tree diagram. (ii) Use the tree diagram to answer the following questions: What is the probability that a customer chosen at random will order (a) a chocolate ice cream? Total number of outcomes: 6 Outcomes of Interest Probability Total Outcomes = P Chocolat 2 e ( ) 6 1 3 = = (b) an ice cream with caramel sauce? P Caramel 3 ( ) 6 1 2 = = (c) a strawberry ice cream with caramel sauce? 1 (Strawberry with Caramel) 6 P =
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Destination Maths Chapter 19 Probability II
1
Probability II
Practice Questions 19.1 1. An ice cream shop has vanilla, chocolate or strawberry ice cream
and you can have either chocolate or caramel sauce on top.
(i) Represent this information on a tree diagram.
(ii) Use the tree diagram to answer the following questions:
What is the probability that a customer chosen at random will order
(a) a chocolate ice cream?
Total number of outcomes: 6
Outcomes of InterestProbabilityTotal Outcomes
=
P Chocolat 2e( )613
=
=
(b) an ice cream with caramel sauce?
P Caramel 3( )
612
=
=
(c) a strawberry ice cream with caramel sauce?
1
(Strawberry with Caramel)6
P =
Destination Maths Chapter 19 Probability II
2
2. You are buying a jumper from a shop. You have to choose a size, colour
and style.
You can choose between small, medium or large for size; blue, pink, red
or brown for colour; and round-neck or V-neck for style.
(i) Draw a tree diagram to represent the sample space.
(ii) Use the tree diagram to answer the following questions:
(a) How many different options do you have?
By counting from the tree diagram: 24 options in total
Using FPoC: Size × Colour × Style
3 × 4 × 2
= 24 options in total
Destination Maths Chapter 19 Probability II
3
(b) What is the probability that you pick a brown jumper?
Outcomes of InterestProbabilityTotal Outcomes
=
6P(Brown)2414
=
=
(c) What is the probability that you pick a jumper that has a V-neck?
12P(V neck)2412
=
=
(d) What is the probability that you pick a large, red, round-neck jumper?
1
P(Large and Red and Round)24
=
(e) What is the probability that the jumper you pick doesn’t have a round
neck?
P(no round) = 1 – P(Round)
1
1
– 12
2
=
=
Destination Maths Chapter 19 Probability II
4
3. A fair die is thrown and a fair coin is flipped at the same time.
(i) Draw a two-way table to represent this sample space.
Coin H T
Die
1 1H 1T
2 2H 2T
3 3H 3T
4 4H 4T
5 5H 5T
6 6H 6T
(ii) Use the two-way table to calculate the probability of getting:
Total number of outcomes = 12
Outcomes of Interest
Probability :Total Outcomes
(a) a head
6P(Head)
1212
=
=
(b) a 3
2P(3)
1216
=
=
Destination Maths Chapter 19 Probability II
5
(c) a head and a 3
There are two ways to calculate this probability:
From the outcomes in the tree diagram: 1P(Head and 3)12
=
By multiplying along the branches: P(H) and P(3)
P(H) × P(3)
1 12 6
112
×
=
(d) a tail and an even number
There are two ways to calculate this probability:
From the outcomes in the tree diagram:
3P(Tail and Even)1214
=
=
By multiplying along the branches:
P T P Even1 12
( ) (
14
)
2
×
= ×
=
(e) a tail or a prime number.
Prime numbers are 2, 3 and 5.
P(Tail or Prime)
Count the numbers of outcomes that contain a tail or a prime number
Outcomes that contain a tail or a prime number = 9
P(Tail or Prime)
191234
=
=
Destination Maths Chapter 19 Probability II
6
4. Suppose we flip a coin and spin a spinner with four coloured sections of equal size
at the same time.
(i) What is the sample space for the coin?
Coin = {head, tail}
(ii) What is the sample space for the spinner?
Spinner = {red, green, yellow and blue}
(iii) Draw a sample space diagram and use it to write out the sample space for
when we throw both the coin and spin the spinner.
Coin
H T
Spin
ner
Red RH RT
Green GH GT
Yellow YH YT
Blue BH BT
The sample is {RH, RT, GH, GT, YH, YT, BH, BT}
(iv) Using your sample space diagram, calculate the probability of getting:
(a) a tail on the coin
4( )P T il
1
a8
2
=
=
Destination Maths Chapter 19 Probability II
7
(b) red on the spinner
2(Red)P
814
=
=
(c) a tail on the coin and red on the spinner
1
( ) and P(RedP ail )8
T =
(d) a tail on the coin or red on the spinner
5
( ) or P(RedP ail )8
T =
* be careful not to the RT twice
(e) the colour on the spinner is not red
P(not Red) 1 P(Red)1
14
34
= −
= −
=
(f) green on the spinner
2P(Green)
814
=
=
(g) a head on the coin and either blue or yellow on the spinner.
P(head, blue) or P(head, yellow)
1 18 82814
= +
=
=
Destination Maths Chapter 19 Probability II
8
5. A spinner with four equal segments, numbered 1 to 4, is spun twice.
(i) Represent the sample space for this experiment using a
two-way table.
2nd Spin
1 2 3 4
1st S
pin
1 1, 1 1, 2 1, 3 1, 4
2 2, 1 2, 2 2, 3 2, 4
3 3, 1 3, 2 3, 3 3, 4
4 4, 1 4, 2 4, 3 4, 4
(ii) Use your two-way table to calculate the probability that:
(a) the same number comes up on both spins
Total Outcomes = 16
P(Same number) = {(1, 1), (2, 2), (3, 3), (4, 4)}
41614
=
=
(b) both numbers are odd
P(Both odd) = {(1, 1), (1, 3), (3, 1), (3, 3)}
41614
=
=
(c) the number on both the first and second spinner is even
(iv) Anita chooses a multiple of two and Rebecca chooses a multiple of three Multiples of two: 2, 4, 6, 8, 10 Multiples of three: 3, 6, 9 These outcomes are shown below in yellow:
3. Twenty-six people were surveyed about their choice of mobile phones. The survey finds
that 14 people have Apple iPhones, 10 have Samsungs and five have Nokias. Four have
Destination Maths Chapter 19 Probability II
65
Apple iPhones and Samsungs, three have Apple iPhones and Nokias and one has a Samsung and a Nokia. No one has all three types of phone. Represent this information on a Venn diagram and use it to calculate the probability that a person chosen at random has:
26 people 14 iPhones, 10 Samsungs, 5 Nokias 4 iPhones + Samsungs, 3 iPhones + Nokia; 1 Samsung + Nokia 0 All 3.
iPhone only
14 – (4 + 3) = 14 – 7 = 7
Samsung only
10 – (4 + 1)
= 10 – 5
= 5
Nokia only
5 – (1 + 3)
= 5 – 4
= 1
None
26 – (7 + 4 + 3 + 1 + 5 + 1)
= 26 – 21
= 5
(i) an iPhone
Destination Maths Chapter 19 Probability II
66
14P(iPhone)
267
13
=
=
(ii) a Samsung or a Nokia
5 1 1 3 4P(Samsung or Nokia)26
14267
13
+ + + +=
=
=
(iii) two phones
P iPhone and Samsung or P iPhone and Nokia or P Samsung and Nok
P(Two phones)( ) ( )( )
3 4 126
8264
a
13
i
+ +=
=
=
=
(iv) no phone
5P(no phone)26
=
(v) no iPhone.
5 1 1 5P(no iPhone)26
12266
13
+ + +=
=
=
Destination Maths Chapter 19 Probability II
67
4. A group of 20 people are waiting at a bus stop one morning. Nine of them
have an umbrella, six have a raincoat and three have both an umbrella and
a raincoat.
Represent this information on a Venn diagram and use it to calculate the
probability that a person chosen at random has:
20 people: 9 Umbrella, 6 raincoat, 3 both.
Umbrella only
= 9 – 3
= 6
Raincoat only
= 6 – 3
= 3
None
20 – (6 + 3 + 3)
= 20 – 12
= 8
(i) an umbrella
9P(Umbrella)20
=
(ii) an umbrella and a raincoat
( )P Umbrella and Raincoat 320
=
Destination Maths Chapter 19 Probability II
68
(iii) an umbrella or a raincoat P(Umbrella or Raincoat) = P(Umbrella) + P(Raincoat) − P(Umbrella) and Raincoat)
9 6 320 20 20122035
= + −
=
=
(iv) neither.
8P(neither)2025
=
=
5. A group of students were asked if they preferred soccer, rugby or Gaelic football. The results are shown on the Venn diagram.
Calculate the probability that a randomly chosen students likes: (i) soccer or rugby
Total : 7 14 12 9 15 6 20 19 102
P(Soccer) or P(Rugby) = Rugby + Soccer
7 14 12 9 15 6= 102
631022134
+ + + + + + + =
+ + + + +
=
=
Destination Maths Chapter 19 Probability II
69
(ii) rugby or Gaelic football
P(Rugby) or P(Gaelic)Rugby Gaelic20 9 15 6 14 12
10276
1023851
= ++ + + + +
=
=
=
(iv) soccer and Gaelic football but not rugby
P(Soccer and Gaelic but not Rugby)
7 9 20102
361026
17
+ +=
=
=
(iv) only Gaelic football
20P(Only Gaelic)
1021051
=
=
(v) none of these sports.
19
P(None)102
=
6. Two events A and B have the following probabilities: P(A) = 0·4 P(B) = 0·6 P(A ∩ B) = 0·2 Draw a Venn diagram to represent these probabilities and use it to calculate:
P(A\B)= 0·4 – 0·2 P(B\A)= 0·6 – 0·2
= 0·2 = 0·4
Destination Maths Chapter 19 Probability II
70
(i) P(A ∪ B)
P(A ∪ B)
= P(A) + P(B) − P(A ∩ B)
= 0·6 + 0·4 − 0·2
= 0·8
(ii) P(A ∪ B)′
P(A ∪ B)′ = 0·2
7. Let A and B be independent events, where P(A) = 0·4 and P(B) = 0·7.
(i) Find P(A ∩ B)
P(A ∩ B) = P(A and B)
= P(A) × P(B)
= 0·4 × 0·7
= 0·28
(ii) Find P(A ∪ B)
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
= 0·4 + 0·7 – 0·28
= 0·82
(iii) Draw a Venn diagram and shade the region that represents A ∩ B′
A ∩ B′ = A\B
Destination Maths Chapter 19 Probability II
71
(iv) Find P(A ∩ B′).
P(A ∩ B′) = P(A/B)
= P(A only)
= 0·4 – 0·28
= 0·12
8. A driving test consists of a practical test and theory test. One day everyone who took the
test passed at least one section. 64% passed the practical section and 78% passed the
theory section.
64% Practical (Pr)
78% Theory (T)
64 + 78 = 142%
Pr ∩ T = 142% − 100%
= 42%
(i) Represent this information on a Venn diagram showing the probabilities of
candidates in each section of the diagram.
Pr′ = 64 – 42
= 22%
T′ = 78 – 42
= 36%
One person is chosen at random from all the people who took the test that day. What is
the probability that this person:
(ii) passed the practical section and the theory section?
P(Theory and Practical) = 42% (both = Pr ∩ T)
Destination Maths Chapter 19 Probability II
72
(iii) passed the theory section only?
P(Theory only) = 36%
Practice Questions 19.5
1. Jamie is playing a game where he must toss a coin until he gets a head. What is the
probability that he gets a head for the first time on:
(i) the first throw?
Success = Flipping a head
1 1 1P(Success) P(Failure) 1
2 2 21
P(Head on first flip)2
= = −
=
(ii) the second throw?
1 1P(FS)
2 214
= ×
=
(iii) the third throw?
1 1 1P(FFS)
2 2 218
= × ×
=
(iv) his fifth throw?
44 1 1P(F) P(S)
2 2132
× = ×
=
Destination Maths Chapter 19 Probability II
73
2. Ciaran is playing a game using a fair die. He must throw the die until he gets a 6.
What is the probability that he throws a 6 in:
Success = throwing a 6
1P(S) P(F) 1 P(S)6
1 16
5 6
= = −
−
(i) one throw?
1
P(S)6
=
(ii) two throws?
5 1P(FS)
6 65
36
= ×
=
(iii) three throws?
5 5 1P(FFS)
6 6 625
216
= × ×
=
(iv) his tenth throw?
P(Success on 10th throw)
9
9
P(F) P(S)
5 16 6
0 0323
= ×
= ×
= ⋅
Destination Maths Chapter 19 Probability II
74
3. A fair die is thrown three times.
What is the probability that a 2 or a 3 is thrown:
2 1Success 2 or 3 P(S)6 3
P(F) 1 P(S)1 = 13
2 3
= = =
= −
−
=
(i) for the first time on the third throw?
P(FFS)2 2 13 3 34
27
= × ×
=
(ii) once in three throws?
P(FFS) or P(SFF) or P(FSF)4 1 2 2 2 1 2
27 3 3 3 3 3 349
= + × × + × ×
=
4. A spinner has five equal sections.
Three of them are green and two of them are yellow. Chloe is playing a
game where she has to spin the spinner until it lands on yellow.
What is the probability that she lands on yellow for the first time on:
3 2 2P(Green) P(Yellow) P(S) P(Yellow)5 5 5
3P(F) P(Green)5
= = = =
= =
Destination Maths Chapter 19 Probability II
75
(i) the first spin?
P(Yellow) = P(S) = 25
(ii) the third spin?
P(GGY)=P(FFS)
3 3 25 5 518
125
= × ×
=
(iii) the sixth spin?
5
5
P(F) P(S)
3 25 5486
15625
×
= ×
=
(iv) the ninth spin?
Give your answer in scientific notation.
P(F)8 × P(S)
8
3
3 25 5
6 718464 10−
= ×
= ⋅ ×
5. Zoe is the best free-throw taker on her basketball team. Her average probability
of scoring from a free shot is 0·78.
If she takes three free throws in a game, what is the probability that she will:
10. Suppose a student takes a multiple choice test.
The test has 10 questions, each of which has four possible answers (only one correct).
(i) If the student guesses the answer to each question, do the questions form a
sequence of Bernoulli trials? Explain your answer.
Yes. Each question is independent and has two options; success or failure.
Therefore, the test forms a series of Bernoulli trials.
Destination Maths Chapter 19 Probability II
81
(ii) List the possible outcomes.
Possible outcomes:
1. Success – the guess is correct
2. Failure – the guess is incorrect
(iii) Write the probability associated with each outcome.
1 1P(Success) P(Failure) 14 4
3 4
= = −
=
Practice Questions 19.6
1. A goalkeeper has a probability of 35
of saving a penalty. How many penalties would you
expect him to save out of 50 penalties taken?
3P(Save) 50 Penalties5
Expected3 50530 Expect to save
n
p n
= =
= ×
= ×
=
2. A batch of 1,600 items is examined. The probability that an item from this batch is
defective is 0·04. How many items from this batch are defective?
1600 items = n P(Defective) = 0·04
Expected = p × n
= 0·04 × 1600
= 64 Expect defective
Destination Maths Chapter 19 Probability II
82
3. In an experiment, a standard six-sided die was rolled 72 times. The results are shown in
the table.
Which number on the die was obtained the expected number of times?
Number on the die 1 2 3 4 5 6
Frequency 11 8 12 15 16 10
1P(One number on a die) 72 rolls6
Expected1 72612
n
p n
= =
= ×
= ×
=
Therefore, 3 was rolled the expected number of times
4. In a random survey of the voting intentions of a local electorate, the
following results were obtained:
Politician A Politician B Politician C
165 87 48
Total votes = 165 + 87 + 48
= 300
(i) Calculate the probability that a randomly selected voter will vote for:
(a) Politician A
165P(A)3001120
=
=
Destination Maths Chapter 19 Probability II
83
(b) Politician B
87P(B)30029
100
=
=
(c) Politician C
48P(C)300425
=
=
(ii) If there are 7,500 people in the electorate, how many votes would you expect to be
for:
7,500 people = n
Expected = p × n
(a) Politician A?
11Expected 7,500204125
= ×
=
(b) Politician B?
29Expected 7,5001002175
= ×
=
(c) Politician C?
4Expected 7,500251200
= ×
=
Destination Maths Chapter 19 Probability II
84
5. In a raffle, 250 tickets are sold at €1 each for three prizes of €100, €50 and €10.
You buy one ticket.
250 tickets sold at €1 prizes: €100, €50, €10
(i) What is the expected value of this raffle.
Expected value: Σx.p(x)
Outcome (x) p(x) x.p(x)
1st €100 1250
1 2
100250 5
× =
2nd €50 1250
1 1
50250 5
× =
3rd €10 1250
1 1
10250 25
× =
Lose 0 247250
247
0 0250
× =
2 1 1( )5 5
.25
16250 64
x p x = + +
=
= ⋅
∑
(ii) Does it represent a good investment of €1?
Explain your answer.
It does not represent a good investment of €1 as the expected value is 0·64
which means if a large number of people bought tickets the average win is
below the pay in amount. You would expect to lose 0·36.
Net “winnings” = 0·64 – 1
= – 0·36
Destination Maths Chapter 19 Probability II
85
6. Jennifer is playing a game at an amusement park.
There is a 0·1 probability that she will score 10 points, a 0·2 probability that she
will score 20 points, and a 0·7 probability that she will score 30 points.
How many points can Jennifer expect to receive by playing the game?
Expected value = Σx.p(x)
x p(x) x.p(x)
10 0·1 10 × 0·1 = 1
20 0·2 20 × 0·2 = 4
30 0·7 30 × 0·7 = 21
Σx.p(x) = 1 + 4 + 21
= 26 points
7. You take out a fire insurance policy on your home.
The annual premium is €300.
In case of fire, the insurance company will pay you €200,000.
The probability of a house fire in your area is 0·0002.
X p(x) x.p(x)
Fire 200,000 0·0002 200,000 × 0·0002 = 40
No Fire 0 [1 – 0·002] 0·9998 0 × 0·9998 = 0
(i) What is the expected value?
Expected value = Σx·p(x)
= 40 + 0
= 40
Destination Maths Chapter 19 Probability II
86
(ii) What is the insurance company’s expected value?
The expected value is the same for the insurance company and customer.
However, the perspective is changed. The customer is down €260 since they paid
€300 for the policy, the company is up €260.
(iii) Suppose the insurance company sells 100,000 of these policies. What can the
company expect to earn?
Profit on 1 policy: = €260
100,000 policies : 100,000 × €260 = €26,000,000
8. A hundred tickets are sold for a movie at the cost of €10 each. Some tickets have cash
prizes as a part of a promotional campaign: one prize of €50, three prizes of €25 and five
prizes of €20.
What is the expected value if you buy one ticket?
Expected value = Σx.p(x)
x p(x) x.p(x)
€50 1100
1 150
100 2× =
€25 3100
3 325
100 4× =
€20 5100
520 1
100× =
0 91100
910 0
100× =
1 3. ( ) 1 02 4
9 4
€2 ·25 Expected value.
Σ = + + +
=
=
x p x
You pay €10 so “nett winnings” = 2·25 – 10 = − €7∙75
9. You pay €10 to play the following game of chance.
Destination Maths Chapter 19 Probability II
87
There is a bag containing 12 balls, five are red, three are green and the rest are yellow.
You are to draw one ball from the bag.
You will win €14 if you draw a red ball and you will win €12 if you draw a green ball.
How much do you expect to win or lose if you play this game 100 times?
Outcome x Pay Out p(x) x.p(x)
Red €14 512
5 3514
12 6× =
Green €12 312
312 3
12× =
Yellow 0 412
40 0
12× =
35. ( ) 36
536
8 83
Σ = +
=
= ⋅
x p x
Pay €10 – Expected value €8·83 so “nett winnings” = 8 ⋅83 – 10 6
– 7=
7Loss6
100 Games 1 167 10
€11
0350
36 67 nett loss
= −
= −
⋅
⋅ ×
=
= −
= −
10. A sports club decides to hold a field day to raise money for local
charities. One of the games involves spinning the wheel shown on
the right.
Destination Maths Chapter 19 Probability II
88
You win the amount the pointer lands on. It costs €5 to play the game.
(i) What is the expected value of the game? Expected value: Σx.p(x)
x Outcome x p(x) x.p(x) Green €3 1
4
34
Yellow €7 18
78
Blue €10 18
10 58 4=
White €5 14
54
Pink €2 14
2 14 2=
3 7 5 5 1( ).4 8 4 4 2378
4 625
Σ = + + + +
=
= ⋅
p x x
€4·63 is the expected value of the game.
(ii) How could they adjust the wheel to ensure they make more money for the charities? In order to make more money for the charities they could lower the amount that is won in each/some segment. or
Otherwise you could enlarge the €2 segment and make the €5 segment smaller
11. A child asks his parents for some money. The parents make the following offers.
Father’s offer: The child flips a coin. If the coin lands heads-up, the father will give the
child €20. If the coin lands tails-up, the father will give the child nothing.
Destination Maths Chapter 19 Probability II
89
Mother’s offer: The child rolls a 6-sided die. The mother will give the child €3 for each
dot on the up side of the die.
(i) Which offer has the greater expected value?
Expected Value : Σx.p(x)
Father’s offer
x p(x) x.p(x)
Heads €20 12
20 102=
Tails 0 12
0
) 1. €( 0Σ =x p x
Destination Maths Chapter 19 Probability II
90
Mother’s offer
x p(x) x.p(x)
1 = €3 16
1 3 136 6 2× = =
2 = €6 16
1 66 16 6× = =
3 = €9 16
1 9 396 6 2× = =
4 = €12 16
1 1212 26 6× = =
5 = €15 16
1 15 5156 6 2× = =
6 = €18 16
1 1818 36 6× = =
1 3 5. ( ) 1 2 32 2 2212
01
1 05
€0
5
x p xΣ = + + + + +
=
=⋅
⋅=
€10·50 > €10
Therefore, the mother’s offer is greater.
(ii) Which offer would you choose if you were the child? Justify your answer.
I would choose the mother’s offer as over many times the expected outcome
would be higher.
Destination Maths Chapter 19 Probability II
91
12. (i) What is the expected value for each of the following investment packages for a €1,000 investment?
5. Judy is playing a game in which she rolls a fair dice three times and tries to get ‘6’ as many times as she can.
(i) Using the options ‘6’ and ‘not a 6’, draw a tree diagram to represent this game. Fill
in the probabilities on the branches of the tree diagram.
Tree Diagram:
(ii) Calculate the probability that Judy gets a ‘6’ on all three tries.
P(666) = 1 1 16 6 6
× ×
=1
216
(iii) Calculate the probability that Judy gets a ‘6’ on at least two of her three tries.
P(At least 2 6s) = P(6, 6, Not 6) or P(6, Not 6, 6) or P(666) or P(Not 6, 6, 6)
1 1 5 1 5 1 1 5 1 16 6 6 6 6 6 216 6 6 65 5 1 5
216 216 216 216162162
27
= × × + × × + + × ×
= + + +
=
=
Destination Maths Chapter 19 Probability II
114
6. A bag contains 10 counters with the letters of the word STATISTICS written on them. A counter is chosen at random and not replaced before choosing another one.
10 counters STATISTICS No replacement
(i) Draw a tree diagram to represent this situation. Tree Diagram S, T, C = consonants = 7 letters
A, I = Vowels = 3 letters
(ii) What is the probability of getting two consonants?
P(2 consonants) = P(SS) or P(ST) or P(SC) or P(TS) or P(TT) or P(TC) or
P(CC) or P(CT) or P(CS)
3 2 3 3 3 1 3 3 3 210 9 10 9 10 9 10 9 10 9
3 1 1 1 3 1 3 3 010 9 10 10 9 10 9 90
6 9 3 9 6 3 390 90 90 90 90 90 9042907
15
= × + × + × + × + × + × + × + × + × +
= + + + + + +
=
=
or
P(consonant, consonant) = 7 6
10 9×
42907
15
=
=
Destination Maths Chapter 19 Probability II
115
(iii) What is the probability of getting a vowel at least once?
P(1 vowel at least) = P(vowel, vowel) or P(vowel, consonant) or
P(consonant, vowel)
3 2 3 7 7 310 9 10 9 10 96 21 21
90 90 9048908
15
= × + × + ×
= + +
=
=
(iv) What is the probability of getting exactly one vowel?
P(1 vowel) = P(vowel, consonant) or P(consonant, vowel)
21 2190 9042907
15
= +
=
=
(v) What is the probability of not getting exactly one vowel?
P(Not 1 vowel) = 1 – P(1 vowel)
7115
815
= −
=
Destination Maths Chapter 19 Probability II
116
7. In a certain game, you pick a card from a standard deck of 52 cards. If the card is a heart,
you win.
If the card is not a heart, you replace the card to the deck, reshuffle, and draw again.
What is the probability that you will pick the first heart on the third draw?
P(success) = P(Heart)
= 14
P(Failure) = 1 – P(Success)
P(not heart) = 114
−
= 34
P(not heart and not heart and heart) = P(FFS)
P(FFS) = 3 3 14 4 4× ×
964
=
8. A biased die is used in a game. The probabilities for each outcome
are given below.
Outcome 1 2 3 4 5 6
Probability 0·22 0·12 0·25 0·14 0·18 0·09
(i) The die is thrown once. What is the probability of getting: