Maths Quest Maths B Year 12 for Queensland Chapter 9 Probability Distributions WorkSHEET 9.1 1

WorkSHEET 9.1 Probability distributions Name: _________________________ 1 Which of the following are discrete random

variables? (a) The heights of students in a Year 12

class.

(b) The weights, to the nearest kg, of students in a Year 12 class.

(c) The number of runs scored in a cricket test match in Brisbane in 2002.

(d) The number of consecutive heads obtained when repeatedly tossing a coin.

(e) The price, per litre of petrol, in randomly selected stations in Queensland.

(f) The actual volume of petrol in the 1000 litre ‘unleaded petrol’ tanks at those same stations after being filled by a tanker.

(a) Height is a continuous. (b) Weight, to the nearest kg, is a discrete

random variable. (c) Not a random variable, since match has

already occurred. (d) Although infinite, still a discrete random

variable. (e) Discrete, since price is always quoted to

the nearest $0.001 (f) Varies continuously, even when ‘full’ due

to continuous pressure and temperature variation.

Maths Quest Maths B Year 12 for Queensland Chapter 9 Probability Distributions WorkSHEET 9.1 2

2 Five coins are tossed simultaneously and the number of heads recorded. (a) Tabulate the probability distribution for

the number of heads.

(b) Draw a probability distribution graph of the outcomes.

(a) Pr(5 heads) = Pr(0 heads) = 521 =

321

There are 5 ways to get 4 heads or 1 head. There are 10 ways to get 3 heads or 2 heads.

321

325

3210

3210

325

321)Pr(

543210

x

x

(b)

3213223233243253263273283293210

X= 0 1 2 3 4 5

Maths Quest Maths B Year 12 for Queensland Chapter 9 Probability Distributions WorkSHEET 9.1 3

3 A die is ‘fixed’ so that certain numbers will appear more often. The probability that a 6 appears is twice the probability of a 5 and 3 times the probability of a 4. The probabilities of 3, 2 and 1 are unchanged from a normal die. The probability distribution table is given below.

xxxx

x

2361

61

61)Pr(

654321

Find: (a) The value of x in the probability

distribution and hence complete the probability distribution.

(b) The probability of getting a ‘double’ with two of these dice. Compare with the ‘normal’ probability of getting a double.

(a) Since the sum of the probabilities must be 1,

61 +

61 +

61 +

3x +

2x + x = 1

Putting over a common denominator,

6632111 xxx +++++ = 1

Collect like terms and remove fraction, 3 + 11x = 6

x = 113

Note Pr(5) = 223 , Pr(4) =

333 =

111

(b)

11

3

22

3

11

1

6

1

6

1

6

1)Pr(

654321

x

x

(c) Probability of a ‘double’ is given by

Pr(1) × Pr(1) + Pr(2) × Pr(2) + … +Pr(6) × Pr(6)

Pr(double) = 61× 61× 61× 61 +

61× 61 +

111× 111 +

223× 223 +

113× 113

Convert each product to a decimal.

Pr(double) = 0.02777 + 0.02777 + 0.02777 + 0.00826 + 0.01860 + 0.07438

Pr(double) = 0.1846 The ‘normal’ probability of a double is

0.1666.

Maths Quest Maths B Year 12 for Queensland Chapter 9 Probability Distributions WorkSHEET 9.1 4

4 Show that p(x) =

6634 −x , for x = 1,2, … 6

is a probability distribution. State Pr (2 < x < 5)

Set up a table of probabilities

6621

6617

6613

669

665

661)Pr(

654321

x

x

Calculate the sum of the probabilities.

Sum = 66

211713951 +++++ = 1

Pr (2 < x < 5) = 669 +

6613 +

6617

= 6639

5 Find the expected value of the following discrete probability distribution.

45.005.025.015.01.0)Pr(54321

xXx=

Use formula E(X) = ΣxPr(X = x) E(X) = 1(0.1) + 2(0.15) + 3(0.25)

+ 4(0.05) + 5(0.45) E(X) = 0.1 + 0.3 + 0.75 + 0.2 + 2.25

= 3.6

Maths Quest Maths B Year 12 for Queensland Chapter 9 Probability Distributions WorkSHEET 9.1 5

6 Consider the following gambling game, based on the outcome of the total of 2 dice:

– if the total is a perfect square, you win $4 – if the total is 2, 6, 8 or 10, you win $1 – otherwise, you lose $2.

(a) Find the expected value of this game. (b) Determine if it is a fair game.

(a) Set up the probability distribution table.

361

362

363

364

365

366

365

364

363

362

361)Pr(

12111098765432xx

Add a row, which indicates win (+) or loss(–).

22141212421Gain361

362

363

364

365

366

365

364

363

362

361)Pr(

12111098765432

−−−−−

xx

Use formula E(gain) = ΣGain(x) Pr(X = x)

E(gain) = 1(361 ) –2(

362 ) + 4(

363 ) –2(

364 )

+ 1(365 ) –2(

366 ) +1(

365 ) + 4(

364 )

+ 1(363 ) –2(

362 ) –2(

361 )

E(Gain) = 36

31655121 +++++

36

241284 ++++−

E(Gain) = 3642 –

3630 =

3612

E(Gain) = 0.333 (b) This game is ‘unfair’ — you stand to gain

about $0.33 every time you play!

7 Find the missing profit (or loss) so that the following probability table has an expected value of 0.

1285243Gain13.021.009.016.025.006.01.0)Pr(10987654

−−−

= xXx

Let y = profit/loss for x = 10.

E(Gain) = ΣGain(x) Pr(X = x) E(Gain) = –3(0.1) + 4(0.06) – 2(0.25) + 5(0.16) –

8(0.09) + 12(0.21) + y(0.13) Simplify and set E(gain) = 0 –0.3 + 0.24 – 0.5 + 0.8 – 0.72 + 2.52 + 0.13y = 0 2.04 + 0.13y = 0 Solve for y

69.1513.004.2

−=−

=y

Maths Quest Maths B Year 12 for Queensland Chapter 9 Probability Distributions WorkSHEET 9.1 6

8 For the following probability distribution calculate: (a) E(X) (b) E(2X)

(c) E(X + 2) (d) E(X2)

(e) E(X2) – [E(X)]2.

07.07.11.2.17.12.21.05.)Pr(54321012

xXx=

−−

Using a table of values (or the Maths Quest spreadsheet ‘Prob distribution’): (a) E(X) = 1.22

(b) E(2X) = 2.44

(c) E(X + 2) = 3.22

(d) E(X2) = 5.24

(e) E(X2) – [E(X)]2 = 3.7516

Maths Quest Maths B Year 12 for Queensland Chapter 9 Probability Distributions WorkSHEET 9.1 7

9 Three players play the following game for a prize pool of $210. Alice tosses a coin — if it is heads she wins. If not, then Betty tosses the coin — if it is heads she wins. If not, then Carla tosses the coin — if it is heads she wins. If not, then Alice tosses the coin again, winning if it is a head … and so on. Find the expected value of each person in this game.

Because, in theory, this game could go on forever, determine (relative) probabilities as follows. In round 1,

Alice has a 12 chance of winning, while Betty has a

12 ×

12 chance, and Carla has a

12 ×

12 ×

12 chance.

In Round 2,

Alice has a 12 ×

12 ×

12 ×

12 chance … and so on.

These probabilities are tabulated below.

40961

20481

102414

5121

2561

12813

641

321

1612

81

41

211

CarlaBettyAliceRound

By looking at each row, the probabilities are in the ratio of 4 : 2 : 1 Thus Alice has 4 ‘chances’, Betty has 2 and Carla has 1.

E(Alice) = 47 (210) = $120

E(Betty) = 27 (210) = $60

E(Carla) = 17 (210) = $30