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CM141A – Probability and Statistics I Solutions

to exercise sheet 3

1. We know that a die is selected at random from two twenty-faced dice on which the

symbols 1-10 are written with non-uniform frequency as follows:

Symbol 1 2 3 4 5 6 7 8 9 10

Number of faces of die A 6 4 3 2 1 1 1 1 1 0

Number of faces of die B 3 3 2 2 2 2 2 2 1 1

The data set we are given is the following results of 7 die experiments: 5, 3, 9, 3, 8, 4, 7.

Define the two events A, B corresponding to the selection of die A or B respectively.

Also, let us denote the data D in the following way D ={C

i

}

i 1,

= 

,7 ={5,3,9,3,8,4,7}.

Using that notation we are interested in P A D()= P A C({ } i

)

In order to obtain the probability that the die is die A, we will use Bayes’ theorem:

P({C i

} A P A) ( ) P({C i

} A P A) ( )

P A C({ }i ) = P({ }Ci ) = P({ }CiA P A) ( )+ P({ }CiB P B) ( )

Obviously,. Also

P({ }CiA)=∏i 17= P C A( i)= P C A( 1 ) P C A( 7 ) = 20 20 20 20 20 20 20 1 3 1 3

1 21 = 20187

P({ }CiB) =∏i 17= P C B(i)= P C B( 1 ) P C B( 7 ) = 20 20 20 20 20 20 20 2 2 1 2

2 22 = 20647

Therefore, we obtain

P A C({ } i

) = 18

7

647 =

=

= .2195,

20 20

so around 22%.

2. a. Given P A B ( | )=1, we know that

P A B ( | ) = P A B ( P B ( )



)

= 1 ⇒ P A B P B ( ) = ( ) ⇒ B A ⊂

that is, B is a subset of A - you might want to look at the corresponding Venn diagram for

that (the shaded area is the intersection):

It is therefore clear that for the complementary events A , B opposite is true, i.e. A ⊂ B ,

and thus A  B = A ⇒ P A (  B ) = P A ( ) ⇒ P B A == QED.

b. Given P A B ( | ) = P A B ( | ), we can use the definition of the conditional probability to

get the equation

or after multiplication by P B P B ( ) ( ): P A (  B P B ) ( )= P A (  B P B ) ( ).

Now, note the following facts: P B ()= − 1 P B ( ) and P A ( ) = P A (  B )+ P A (  B )

Thus we can rewrite P A (  B )⎡

⎣

1 − P B ( )⎤ ⎡

⎦ ⎣

= P A ( )− P A (  B )⎤

⎦

P B ( )

Or P A (  B )− P A (  B P B ) ( ) = P A P B ( ) ( )− P A (  B P B ) ( )

Which boils down to saying that A and B are independent since P A B P A P B ( ) = ( ) (

).

c. We know that both P A C P B C ( | )> ( | ) and P A C ( | ) > P B C ( | ) are true.

Let’s multiply these inequalities by P C ( ) and P C ( ) respectively (both are nonnegative)

P A C P C ( | ) ( ) > P B C P C ( | ) ( )

P A C P C ( | ) ( ) > P B C P C ( | ) ( )

A

B

( )

( )

()

| 1



PA B

PA

( )

( )

()

( )

()

( )

| |

= = =





PA B

PA B

PAB PAB

PB PB

summing up the two gives

P A C P C P A C P C P B C P C P B C P C ( | ) ( )+ ( | ) ( )> ( | ) ( )+ ( |

) ( ).

We can now identify the left hand side as P A ( ) (the law of total probability), and

similarly identify the right hand side as P B ( ), from which follows P A P B ( )> ( ) QED.

3. Let B i

denote the event that the ball i is black, and let R i

= B

i

. Then

P R ( 2 | B P B 1 ) ( 1 )

P B ( 1 | R 2 )= =

P R (

2

| B P B

1

) (

1

)+ P R (

2

| R P R

1

) (

1

)

r b.

=

b + + + r c b r

b r b r + c r b + r + c

+

b + + + r c b rb + + + r c b r

4. Let Y be a random variable giving the number of heads minus the number of tails in

three tosses of a coin.

a. Here is the list of the elements of the sample space S for the three tosses of the coin

and their assigned values of Y:

HHH HHT HTH HTT THH THT TTH TTT

3 1 1 - 1 1 - 1 - 1 - 3

b. Assuming that the coin is biased so that a head is twice as likely to occur as a tail,

means that the probability for head is , while that of tail is. Therefore:

P HHH () =( ) 233 = 278

P HHT ( )= P HTH ( )= P THH () =( )

2

3

2

( )

1

3

=

27

4

P HTT () = P THT () = P TTH () =( )

2

3

( )

1

3

2

=

27

2

P TTT ()=( ) 1 3 3 = 271

Summarizing these results gives the probability distribution of the RV Y:

y 3 1 - 1 - 3

p(y) 8 27 12 27 6 27 1 27

The same information is represented in the following Histogram:

c. μ= Y = ∑ yp y ( ) = 3 × + × + − × + − × 278 1 1227 ( 1 ) 276 (

3 ) 271 = 24 12 6 3+ − − 27 = 1

y ∈ − −{ 3, 1,1,3}

In order to calculate the standard deviation σ, we will use the formula σ

2

= Y

2

− Y

2

Y 2 = ∑ y p y 2 ( )= 32 × + × + − 278 12 1227 ( 1 ) 2 × + − 276 (

3 ) 2 × 271 = 72 12 6 9+ + + 27 = = 1133 23 ,

y ∈ − −{ 3, 1,1,3}

and thus: σ 2 = Y 2 − Y 2 = − 113 ( 1 ) 2 = 83 ⇒ σ=

8

3

(d) The Cumulative Distribution Function is:

⎧ 0 y <− 3

⎪

⎪

⎪

t ≤ y ⎪

⎪

⎪

⎩

1 3 ≤ y

0

• 3 - 1 1 3

y

y

0.

0.

0.

0.

F y

⎛ ⎞ 10

5. There are ⎜ ⎟

⎝ ⎠ 4 ways of selecting any 4 CDs from 10. We can select x jazz CDs from 5

⎛ ⎞⎛ 5 5 ⎞

and then select ( 4 − x ) from the remaining CDs in p x ( ) =⎜ ⎟⎜⎝ ⎠⎝

x 4 − x

⎟

⎠

ways. Hence

⎛ ⎞⎛ 5 5 ⎞

⎝ ⎠⎝

p y ( )= ⎜ ⎟⎜ y 4 − y ⎟⎠ with y = 0,1,2,3,

⎛ ⎞ 10

⎜ ⎟⎝ ⎠ 4

Summarizing these results gives the probability distribution of the RV Y:

y 0 1 2 3 4

p(y) 1/42 10/42 20/42 10/42 1/

The same information can be represented by the following histogram:

6. We know that the proportion of people who respond to a certain mail-order solicitation

is a continuous random variable X that has the density function

⎧

⎪

p x ( ) =

⎨

⎪⎩ 0 elsewhere

(a) We need to show that P ( 0 < X < 1 )=1:

1 11

P ( 0 < < = X 1

)

∫ p x dx ( ) =∫ 5

2

( x + 2 ) dx =

5

2

(

1

2

x

2

x

)

0 =

5

2

(

1

2

• = = 2 ) 5 2

2 5

QED

0 0

0

0 1 2 3 4

x

(b) The probability that more than 1/4 but fewer than 1/2 of the people contacted will

respond to this type of solicitation can be formally written as

12 12

12

P (

1

4

< X <

1

2

) =∫ p x dx ( )=∫ 5

2

( x + 2 ) dx =

5 2

2

(

1

x

2

x

)

14 =

14 14

1 11

(c) μ= X =∫ xp x dx ( ) =∫ x 52 ( x + 2 ) dx = 5 3 2 ( 1 x 3 + 2 12 x 2 ) 0 = 5 3 2 ( 1 + = 1 ) 158

0 0

In order to calculate the standard deviation σ, we will use the formula σ

2

= X

2

− X

2

1 11

X 2 =∫ x p x dx 2 ( )=∫ x 2 5 2 ( x + 2 ) dx = 52 ( 14 x 4 + 2 13 x 3 ) 0 = 52 ( 14 + 32 )= 3011 ,

0 0

and thus: σ 2 = X 2 − X 2 = 1130 −( ) 158 2 = 1130 − 22564 =

(d) The Cumulative Distribution Function:

⎧ 0 x < 0 0 x

<

0

⎧

x ⎪ x

−∞ ⎪ 0 ⎪

⎩

1 x

1 ⎩ 1 x > 1 ⎪

FHxL

1.0 0.5 0.5 1.0 1.5 2.

x

0.

0.

0.

0.

x2 0 x 1

1112 2 x 3

≤ <

≤ <

7. a. Intuitively X > Y since the RV X gives more weight to a big group while Y gives the

same weight to everyone. This may sound unclear – so go over the calculation and give it

another thought. I find it important to develop such a quantitative intuition out of detailed

calculations.

b. Let’s calculate the probability distribution for both X and Y. Actually, both X and Y

assume exactly the same values. The only difference is in the assigned probabilities:

x/y 25 33 40 50

p(x) 25 148 33 148 40 148 50 148

p(y) 1 4 1 4 1 4 1 4

Thus:

X

xp x

x

Y yp y

y

This confirms the guess that X > Y.

c. In order to calculate the variances we will use the formula Var X ( ) = X

2

− X

2

X 2 =∑ x p x 2 ( )= 252 × 14825 + 332 × 14833 + 402 × 14840 + 502 × 14850 = 240562148 1625.4 ,

x

and thus: Var X ( )= X 2 − X 2 = 240562148 −( 5814148 ) 2 82.

Similarly, Y

2

=∑ y p y

2

( ) =

1

4

( 25

2

2

2

2

)=

5814

4

=1453.5,

y

and thus: Var Y ( )= Y

2

− Y

2

=

5814

4

− 37

2

8. (a) Let’s start by plotting F x( )

⎧ 0 x < 0

⎪

⎪⎪

F x( )=⎨2 3 1 ≤ <x 2

⎪

⎪

⎪

⎩

1 3 ≤ x

(b) (i) P X 1 P X 1 F

, where ε is arbitrarily small.

.

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x

0.

0.

0.

0.

F x