WorkSHEET 9.1 Probability distributions Name: · 2019-10-13 · random variable. (c) Not a random variable, since match has already occurred. (d) Although infinite, still a discrete

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.

2 You roll a die 3 times. What is the probability you get at least 2 sixes?

Using Yr 10 Probability Rules:

𝑃(> 2𝑠𝑖𝑥𝑒𝑠) = 𝑃(2𝑠𝑖𝑥𝑒𝑠) + 𝑃(3𝑠𝑖𝑥𝑒𝑠) = 𝑃(6,6,60) + 𝑃(6,60, 6) + 𝑃(60, 6,6) + 𝑃(6,6,6)

=16 ×

16 ×

56 +

16 ×

56 ×

16 +

56 ×

16 ×

16 +

16 ×

16 ×

16

=227

3 Validate that last answer using Binomial Probability.

Using Binomial:

𝐵𝑖𝑛𝑜𝑚𝑖𝑎𝑙𝐶𝐷 =2, 3, 3,16>

= 0.074074074

=227

4 You are guessing the answers to a 3 question multiple choice test where each question has 5 possible answers. What is the probability you pass?

Using Yr 10 Probability Rules: 𝑃(> 2𝑐𝑜𝑟𝑟𝑒𝑐𝑡) = 𝑃(2𝑐𝑜𝑟𝑟𝑒𝑐𝑡) + 𝑃(3𝑐𝑜𝑟𝑟𝑒𝑐𝑡) = 𝑃(⇃, ⇃,×) + 𝑃(⇃,×, ⇃) + 𝑃(×, ⇃, ⇃) + 𝑃(⇃, ⇃, ⇃)

=15 ×

15 ×

45 +

15 ×

45 ×

15 +

45 ×

15 ×

15 +

15 ×

15 ×

15

=13125


5 Validate that last answer using Binomial Probability.

Using Binomial:

𝐵𝑖𝑛𝑜𝑚𝑖𝑎𝑙𝐶𝐷 =2, 3, 3,15>

= 0.104

=13125

6 Consider the following probability table:

𝑥 0 1 2 3 𝑃(𝑋 = 𝑥) 0.1 0.2 0.3 𝑘

Determine the value of 𝑘.

Because;

H𝑃(𝑋 = 𝑥) = 1

0.1 + 0.2 + 0.3 + 𝑘 = 1

𝑘 = 0.4


𝑥 2 3 4 5 𝑃(𝑋 = 𝑥) 2𝑘 𝑘 3𝑘 4𝑘

Determine the value of 𝑘, hence determine the probability table values.

Because;

H𝑃(𝑋 = 𝑥) = 1

2𝑘 + 𝑘 + 3𝑘 + 4𝑘 = 1

𝑘 =110

𝑥 2 3 4 5 𝑃(𝑋 = 𝑥) 0.2 0.1 0.3 0.4


𝑥 0 1 2 3 𝑃(𝑋 = 𝑥) 0.1 0.2 0.3 0.4

Determine 𝐸(𝑥).

𝐸(𝑥) =H𝑥 × 𝑃(𝑋 = 𝑥)

0 × 0.1 + 1 × 0.2 + 2 × 0.3 + 3 × 0.4

∴ 𝐸(𝑥) = 2



𝑥 2 3 4 5 𝑃(𝑋 = 𝑥) 0.2 0.1 0.3 0.4

Determine 𝐸(𝑥).

𝐸(𝑥) =H𝑥 × 𝑃(𝑋 = 𝑥)

2 × 0.2 + 3 × 0.1 + 4 × 0.3 + 5 × 0.4

∴ 𝐸(𝑥) = 3.9


10 I throw a coin twice. How many Heads do I expect to get?

NO, you can’t just say you expect to get 1 heads! We need a Probability Table and then calculate the Expected value! As we are doing something 2 times, the values for the probability table can come from a year 8 two-way table!

H T

H HH HT

T TH TT

State the use of 𝑃 = LMNM

Hence we arrive at the Probability Table

𝑥 0 1 2 𝑃(𝑋 = 𝑥) 0.25 0.5 0.25

𝐸(𝑥) =H𝑥 × 𝑃(𝑋 = 𝑥)

0 × 0.25 + 1 × 0.5 + 2 × 0.25

∴ 𝐸(𝑥) = 1

Phew! We did know it was going to be 1J I love it when different mathematical techniques get the same correct answer!


11 I throw a coin thrice (three times). How many Heads do I expect to get?

Use the Binomial Probability Rule to get your values for the Probability Table;

O𝐶P =12>

P

=12>

O

=18

O𝐶R =12>

R

=12>

S

=38

O𝐶S =12>

S

=12>

R

=38

O𝐶O =12>

O

=12>

P

=18

Hence we arrive at the Probability Table

𝑥 0 1 2 3 𝑃(𝑋 = 𝑥) 1

8 38

38

18

𝐸(𝑥) =H𝑥 × 𝑃(𝑋 = 𝑥)

0 ×18 + 1 ×

38 + 2 ×

38 + 3 ×

18

∴ 𝐸(𝑥) = 1.5


12 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) = =

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

** Silly textbook … this is like throwing 1 coin 5 times, so the values in the table below come from using the Binomial Probability rule! **

(b)

521

321

321

325

3210

3210

325

321)Pr(

543210

x

x

3213223233243253263273283293210

X= 0 1 2 3 4 5


13 Calculate the Variance of the following Probability Table.

𝑥 0 1 2 3 𝑃(𝑋 = 𝑥) 0.1 0.2 0.3 0.4

Using the rule on the Formula Sheet;

𝑉𝑎𝑟(𝑋) =H𝑝(𝑥 − 𝜇)S Need 𝜇:

𝐸(𝑥) =H𝑥 × 𝑃(𝑋 = 𝑥)

= 0 × 0.1 + 1 × 0.2 + 2 × 0.3 + 3 × 0.4

𝐸(𝑥) = 2

∴ 𝜇 = 2 𝑉𝑎𝑟(𝑋) = 0.1(0 − 2)S + 0.2(1 − 2)S

+ 0.3(2 − 2)S + 0.4(3 − 2)S

∴ 𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = 1


𝑥 0 1 2 3 𝑃(𝑋 = 𝑥) 0.1 0.2 0.3 0.4

Using other rule. 𝑉𝑎𝑟(𝑋) = 𝐸(𝑋S) − X𝐸(𝑋)YS

𝐸(𝑋) =H𝑥 × 𝑃(𝑋 = 𝑥)

= 0 × 0.1 + 1 × 0.2 + 2 × 0.3 + 3 × 0.4

𝐸(𝑥) = 2

𝐸(𝑋S) =H𝑥S × 𝑃(𝑋 = 𝑥)

= 0S × 0.1 + 1S × 0.2 + 2S × 0.3 + 3S × 0.4

𝐸(𝑥) = 5

𝑉𝑎𝑟(𝑋) = 5 − 2S




𝑥 0 1 2 3 𝑃(𝑋 = 𝑥) 0.1 0.2 0.3 0.4

Using the Casio … input List 1 and List 2 ensure frequency is set to List 2 calculate 1-VAR

𝐸(𝑋) = 2 𝐸(𝑋S) = 5

Use rule: 𝑉𝑎𝑟(𝑋) = 𝐸(𝑋S) − X𝐸(𝑋)YS

𝑉𝑎𝑟(𝑋) = 5 − 2S


16 Calculate the Standard Deviation of the following Probability Table.

𝑥 0 1 2 3 𝑃(𝑋 = 𝑥) 0.1 0.2 0.3 0.4

𝑆𝐷 = [𝑉𝑎𝑟(𝑋) Refer to previous calculations for Variance, hence

𝑆𝐷 = √1 = 1

17 Calculate the Standard Deviation of the following Probability Table.

𝑥 0 1 2 3 𝑃(𝑋 = 𝑥) 0.1 0.2 0.3 0.4

Using the Casio … input List 1 and List 2 ensure frequency is set to List 2 calculate 1-VAR

𝜎 = 1

18 Any chance you get, when you see a Probability Table, calculate the variance and SD in different ways. You need lots of practice!


19 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.

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,

+ + + + + x = 1

Putting over a common denominator,

= 1

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

x =

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

(b)

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

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

Pr(double) = ´ + ´ + ´ +

´ + ´ + ´

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.

xxxx

x

2361

61

61)Pr(

654321

61

61

61

3x

2x

6632111 xxx +++++

113

223

333

111

11

3

22

3

11

1

6

1

6

1

6

1)Pr(

654321

x

x

61

61

61

61

61

61

111

111

223

223

113

113


20 Show that p(x) = , for x = 1,2, … 6

is a probability distribution. State Pr (2 < x < 6)

Set up a table of probabilities

Calculate the sum of the probabilities.

Sum = = 1

Because ∑𝑃(𝑋 = 𝑥) = 1, we can confirm that

the given 𝑝(𝑥) IS a probability distribution.

Obtaining data from the table;

Pr (2 < x < 6) = + +

=

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

Use formula E(X) = SxP(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

6634 -x

6621

6617

6613

669

665

661)Pr(

654321

x

x

66211713951 +++++

669

6613

6617

6639

45.005.025.015.01.0)Pr(54321

xXx=


22 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.

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

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

E(gain) = 1( ) –2( ) + 4( ) –2( )

+ 1( ) –2( ) +1( ) + 4( )

+ 1( ) –2( ) –2( )

E(Gain) =

E(Gain) = – =

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

about $0.33 every time you play!

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

Let y = profit/loss for x = 10.

E(Gain) = SGain(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

361

362

363

364

365

366

365

364

363

362

361)Pr(

12111098765432xx

22141212421Gain361

362

363

364

365

366

365

364

363

362

361)Pr(

12111098765432

-----

xx

361

362

363

364

365

366

365

364

363

362

361

3631655121 +++++

36241284 ++++

-

3642

3630

3612

1285243Gain13.021.009.016.025.006.01.0)Pr(10987654

---= xXx

69.1513.004.2

-=-

=y


24 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.

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

07.07.11.2.17.12.21.05.)Pr(54321012

xXx=

--


25 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.

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

40961

20481

102414

5121

2561

12813

641

321

1612

81

41

211

CarlaBettyAliceRound

Documents

WorkSHEET 9.1 Probability distributions Name: · 2019-10-13 · random variable. (c) Not a random variable, since match has already occurred. (d) Although infinite, still a discrete