Solutions to Supplementary Exercise (Probability) 1 N05/II/10 (AOM 8174)

(a) ( ) ( )( )

PP

PA B

A BB∩

=

( ) ( ) ( ) ( )P P P P0.4 0.3 0.640.06

A B A B A B∩ = + − ∪

= + −=

( ) ( )( )

PP

P0.060.3

0.2

A BA B

B∩

=

=

=

(b) ( ) ( ) ( ) ( )P P P PC D C D C D∪ = + − ∩

( ) ( ) ( ) ( ) ( )P P P P P0.7 0.6 0.6

0.4 0.10.25

C D C D C Dx x

xx

∪ = + − ×

= + −==

2 N09/I/7 (H1) (i) ( ) ( ) ( ) ( )P P P PA B A B A B∩ = + − ∪

1 2 173 5 3016

= + −

=

(ii) ( ) ( )( )

PP

PA B

A BB∩

=

( )

1 56 25 1 P

12 3A

= ×

= ≠ =

Hence A and B are not independent. (shown) (iii) ( )P A B′∪ = light blue + yellow + blue

( )( ) ( )

17 2130 5

56

P A B P B′= ∪ +

= − +

=

2 very useful formulas to remember:

conditional probability ( ) ( )( )

PP

PA B

A BB∩

=

and ( ) ( ) ( ) ( )P P P PA B A B A B∪ = + − ∩

( ) ( ) ( )P P PC D C D∩ = × only when C and D are independent

Check for independence by checking whether ( ) ( )P PA B A=

3 N99/II/6 (a) (i) :K H∪ The card taken is either a King or a Heart

( ) 2 3 1 3 3 2 7P30 15

K H + + + + +∪ = =

(ii) :J H′ ′∩ The card taken is not a Jack and not a Heart

( ) 2 3 1 5 3 2 8P30 15

J H + + + + +′ ′∩ = =

(iii) ( ) ( )( )

n diamondP diamond

nK

KK

∩=

1 12 3 1 3 9

= =+ + +

(b) ( )8

230

2

28P both Jacks 0.064435

CC

= = = (3 d.p.)

(c) ( )9 13 8

3 3 330

3

P 3 Kings 3 Queens 3 Jacks C C CC

+ +∪ ∪ =

84 286 56 0.1054060+ +

= = (3 d.p.)

4 N87/II/6

(i) P(A wins on 2nd draw) = P(RRR) + P(RGR)

3 2 1 3 5 2 38 7 6 8 7 6 28

= × × + × × =

(ii) P(A wins | wins on 2nd draw) nd

nd nd

P( wins on 2 draw)P( wins on 2 draw) P( wins on 2 draw)

AA B

=+

R

G

R R

R

R

R

R

G

G

G

G

G

G

R G R G R G R G R G R G

A’s 1st draw

A’s 2nd draw

B’s 1st

draw B’s 2

nd

draw

( ) ( )( )

( )( )

P nP

P nA B A B

A BB B∩ ∩

= =

if outcomes are equally likely

3 mutually exclusive cases

Drawing a tree diagram helps in visualizing the information given. Red box – A wins Orange box – B wins Green box – neither wins after 2 draws and A draws red on 1st draw

( ) ( ) ( )

328

3 P RRGR P GRRR +P GRGR28

328

3 3 2 5 1 5 3 2 1 5 3 4 228 8 7 6 5 8 7 6 5 8 7 6 5

3128

3 30 30 120 228 1680

=+ +

=+ × × × + × × × + × × ×

= =+ +

+

(iii) P(neither wins after 2 draws | A draws red on 1st draw)

= [P(RRGG) + P(RGGR) + P(RGGG)] / P(R) 3 2 5 4 3 5 4 2 3 5 4 3 3/8 7 6 5 8 7 6 5 8 7 6 5 8

2 5 4 5 4 2 5 4 37 6 5

40 40 60210

23

= × × × + × × × + × × × × × + × × + × ×

=× ×

+ +=

=

5 N88/II/6

(a) P(all different) 10 9 8 710 10 10 10

= × × ×

63 0.504125

= =

(b) P(exactly 3 different) 10 3

3 1

4

4!2!

10

C C× ×=

54 0.432125

= =

(c) P(have X or Y) = 1 – P(no X, no Y) 481

10369 0.5904625

= −

= =

P(n more cards needed) > 0.99 11 9 1 9 9 1 9 1 0.99

10 10 10 10 10 10 10 10

n− + × + × × + + × >

May use probability method

Or P&C method: Choose 3 cards from the 10 available, then amongst the 3, choose 1 to be repeated, then count the number of arrangements possible.

Use complement as much fewer cases this way

A

B

C

Y

W

Y

W

Y

W

0.25

0.35

0.4

0.2

0.3

0.8

0.7

1 9110 10

0.999110

91 0.9910

9 0.0110

9ln ln 0.0110

ln 0.01 43.7ln 0.9

n

n

n

n

n

− >−

− >

<

<

> =

least 44n∴ =

6 [AJC Prelim 2008/H2 Q7] (a)

(i) ( ) ( )3!P(at least 2 from B) P BBB' P BBB 2!

= × +

2 30.35 (0.65)3 0.35

0.282= +=

(ii) 3P(at least 1 is yellow) 1 P( none are yellow) 1 0.355 0.732 = − = − = (iii) ( )P at least 1 is yellow all from A

( )3

1 P none are yellow all from A

= 1 0.8 0.488

= −

− =

Good to write out the separate cases to illustrate your thought process

When direct listing involves too many cases, it is more advisable to use complement method when the complement case is straightforward.

Note: 3 balls are picked randomly (i.e. no order involved); But P(BBB') suggests order and thus, it is necessary to compensate

order by multiplying 3!2!

(i.e. for

the case of 2 balls from B, the ball NOT from B can be the 1st, 2nd or 3rd ball, 3 possibilities.)

Drawing a tree diagram helps in organizing the information given

Found using the idea of reduced sample space which is identified by the box-up region in the tree diagram.

Identify G.P.

Remember to change inequality sign when dividing by negative number

Alternatively, ( )

y w w y y w y y y

P at least 1 is yellow all from A

P(at least 1 yellow all from A)= P(all from A)

3P(A A A )+3P(A A A ) P(A A A ) =

P(AAA)

∩

+

2 2 3

3

3x(0.25x0.2)(0.25x0.8) 3x(0.25x0.2) (0.25x0.8)+(0.25x0.2)= 0.4880.25

+=

b)

( )

P(white balls that are faulty)=P(ball is faulty | it is white )

P ball is faulty and white=

P(ball is white )

0.1 0.25 0.8

1 0.3550.0310

× ×=

−=

7 [CJC Prelim 2008/H1 Q7]

(a) P [the 3 guests are seated separately] = 125

!9x!6 3

7

=P

(b)

P[Late] = 0.4x0.05 + 0.5x0.06 + 0.1x0.01 = 0.051

P[MRT’ | Late]

=( P( )+P( ) (0.5x0.06) (0.1x0.01) 31)P( ) (0.4x0.05) (0.5x0.06) (0.1x0.01) 51

Bus Late Car LateLate

∩ ∩ += =

+ +

Note: P&C can be used to determine probability

Alternative: 1-P[MRT | Late]

Direct listing of cases works, but it can be very tedious at times!

P[Late for second time on last day] = P[Late once during the first 4 days and late on the last day] = 4C1(0.051)(0.949)3(0.051) = 0.00889

8 [HCI Prelim 2008/H2 Q9] i)

P(player wins the grand prize) = P (WWW) = 2 4 6

8 10 12

= 120

ii) P (player wins a consolation prize) = P (BWW or WBW or WWB)

= 6 2 4 2 6 4 2 4 68 10 12 8 10 12 8 10 12

+ +

= 320

iii) P (player wins a consolation prize/first draw was white) =

P (player wins a consolation prize and 1st draw was white)P (1st draw was white)

=

= P (WWB or WBW)P (W)

= 110

14

= 25

4C1 as John can be late on either day 1, day 2, day 3 or day 4

Alternative: We deal with the reduced sample space when the 1st draw is white P (player wins a consolation prize)

= P (WB or BW) = 4 6 6 4

10 12 10 12× + × = 2

5

P (player wins at least 2 grand prizes in his 4 attempts) = 1 − P (player did not win any grand prize) – P (player wins only 1 grand prize)

= 4 319 1 191 4

20 20 20 − −

= 0.0140

Alternative: Let random variable X be the number of grand prizes a player wins in his 4 attempts at the

game. Then 1~ 4,20

X B

.

P (player wins at least 2 grand prizes in his 4 attempts) = P ( 2)X ≥ = 1 − P ( 1)X ≤ = 1 − 0.98598 = 0.0140

9 [IJC Prelim 2008/IJC/JC2/H2 Q7] i) ( )P Y

(Ace,King,Queen, Jack drawn from1st Pack) (score from 2nd Pack <5)(others drawn from1st Pack) (totalscoreon both dice <5)

16 12 36 652 36 52 361778

P PP P

= ×+ ×

= × + ×

=

ii)

( ') ( ) ( ') ( ')

9 17 36 30(1 )13 78 52 363539

P X Y P X P Y P X Y∪ = + − ∩

= + − − ×

=

Use complement method as direct listing has more cases

Taught under Binomial Distribution, application of probability

Good to list the cases

( ) ( ') 1P Y P Y+ =

Visualize with a Venn diagram ( ') :X Y∪ Light blue + red + yellow X : Red + yellow

' :Y Red + light blue ( ') :X Y∩ : Red

10 [JJC Prelim 2008/H2 Q7]

(i) P(game ends in a draw) = P(A & B mouse) +P( A & B elephant)+ P(A & B cat)

= 0.3x0.2 + 0.4x0.3 + 0.3x0.5 = 0.330 (ii) P(A wins the first game) = P(A cat, B mouse) +P( A mouse, B elephant)+ P(A elephant, B cat) = 0.3x0.3 + 0.4x0.2 + 0.3x05 = 0.320 (iii) P(B wins 1st game∩ 1st game not a draw) = 1 – P(A wins) – P(draw) = 1- 0.320 – 0.330 = 0.350 P(1st game not a draw) = 1 – P(1st game ends in a draw) = 1 – 0.330= 0.670 P( B wins 1st game, given that 1st game is not a draw)

= 0.3500.522

0.670=

(iv) P(B will win the contest) =P(B wins 1st game)+ P(B wins 2nd game, prior games draw)+… =P(B wins 1st game)+ P(B wins 2nd game)P(1st game draw)+… = 0.350 + (0.330)(0.350) + 2 3(0.330) (0.350) (0.330) (0.350) .....+ +

= 0.3500.522

1 0.330=

−

11 [NYJC Prelim 2008/H2 Q6] (i) ( ) 1 1 1P Mattew is 1st and Mary is 6th in the queue =

10 9 90 =

8! 1Or = 10! 90

(ii) P(either Alice is first or Mark is second (or both) in the queue)

( ) ( ) ( )= P Alice first + P Mark second P Alice first and Mark Second1 1 1= +

10 10 9017= 90

−

−

(iii) P(there is only one girl chosen if five persons are chosen at random from this group) 5 5

1 410

5

C C 25= = C 252

(iv) P(all five girls are next to each other | at least four boys stand next to each other)

Good to list the cases. Note: A & B are independent of each other

Easier to use complement method as complement cases are calculated

P&C method

Probability method

P&C method

Infinite G.P. sum, apply S∞ formula. Separate topics can appear together in 1 question

Apply ( ) ( ) ( ) ( )P A B P A P B P A B∪ = + − ∩

( )( )4

15 6

4 2

P 5 Girls together at least 4 boys together=

P at least

!5!

4 boys together

5! 5 C= C 4! C 6!2! 5!

1= 9

∩

+

12 [NJC Prelim 2008/H2 Q13] (a)

(i) ( ) ( ) ( ) ( ) ( )2 1 1( ) ( )3 4 4

5( )9

P A B P A P B P A P B

P B P B

P B

∪ = + −

= + −

=

( ) 1 5 5P x 04 9 36

A B∩ = = ≠ Thus, A and B are not mutually exclusive.

(ii) ( )1 5xP(( ) ( )) P( ) 54 9P ( ) ( ) 2P( ) P( ) 24

3

A B A B A BA B A BA B A B

∩ ∩ ∪ ∩∩ ∪ = = = =

∪ ∪

(b) Let J and S be events to denote doing a Javanese and Swedish massage respectively.

0.6

0.3

0.7

0.7

0.3

0.7

0.4

0.6

0.4

0.2

0.8

0.4 S

J

0.6

0.3

J

J

S

S

J

J

J

J

S

S

S

S

Case 1: 4 boys Case 2: 5 boys

5!: arranging 5 girls, 5!: arranging 5 boys 4C1: Positioning the girls among the boys

4 ways to slot in the girls

( ) ( ) ( )P A B P A P B∩ = for independent events A and B

Since A B∩ is a subset of ( )A B∪

A tree diagram is useful for organizing the information given

(i) P(JJJ) = (0.8)(0.6)(0.6) = 0.288 (ii) P(JSS) + P(SJS) +P(SSJ) = (0.8)(0.4)(0.3) + (0.2)(0.7)(0.4) + (0.2)(0.3)(0.7) = 0.194 (iii) P(JSSJ) + P(SJSJ) +P(SSJJ) = (0.8)(0.4)(0.3)(0.7) + (0.2)(0.7)(0.4)(0.7) + (0.2)(0.3)(0.7)(0.6) = 0.1316 = 0.132 (3 s.f.)

13 [PJC Prelim 2008/H2 Q11] (i) P(scoring exactly 5 points) ( ) ( )P spinning 1, 1, 3 P spinning 1, 2, 2= +

3 3 2 3! 3 2 2 3!8 8 8 2! 8 8 8 2!

= +

45256

=

(ii) P(spins “2” at least once |scoring a total of less than 6 points)

spin "2" at least once and score less than 6 pointsPscore less than 6 points

=

( ) ( )( ) ( ) ( )

P spinning 1, 1, 2 + P spinning 1, 2, 2 P total 3 points P total 4 points P total 5 points

=+ +

3 3 2 3! 3 2 2 3! + 8 8 8 2! 8 8 8 2!3 3 3 3 3 2 3! 458 8 8 8 8 8 2! 256

= + +

27 9 + 256 128

27 27 45512 256 256

=+ +

1019

= (or 0.526 to 3.s.f.)

14 [RJC Prelim 2008/H2 Q10] Given P(win) 1

2= and P(defeat) 1

6= 1 P(draw)

3⇒ =

(i) P(1 draw and 1 defeat) = P(1 draw, 1 defeat, 2 wins)

21 1 1 4!

3 6 2 2! = × × ×

16

= (shown)

(ii) P(wins the first match and goes on to win exactly one other match)

12

= ×P(wins one other match and loses or draws in the other 2)

21 1 1 3! 3

2 2 2 2! 16 = =

3!2!

accounts for the permutations of

1, 1 ,3 and 1, 2, 2

List cases

4!2!

accounts for the different

possible ordering of the draw, defeat and wins

(iii) P(wins exactly one match | obtains four pts)

P(wins exactly one match and obtains four points)

P(obtains four points)=

P(1 win, 1 draw, 2 defeats)

P(1 win, 1 draw, 2 defeats) P(4 draws)=

+

2

2 4

1 1 1 4!2 3 6 2!

1 1 1 4! 12 3 6 2! 3

× × × =

× × × +

1918

1 1 1118 81

= =+

List cases