Tutorial 2 - Question 2
Experiment: A die is rolled continually until a 6 appears, at
which point the experiment stops.
I Sample space:
S = {(x1 , x2 , ..., xn−1 , 6) : xi ∈ {1, 2, 3, 4, 5}; n ≥ 1}
I Event En : n rolls are necessary to complete the experiment
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En = {(x1 , x2 , ..., xn−1 , 6) : xi ∈ {1, 2, 3, 4, 5}}
( 1 En ) is the event that 6 appears
S
c
( ∞
1 En ) is the event that 6 does not appear
S∞
Tutorial 2 - Question 6
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Insurance coding:
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1: the patient has insurance
0: the patient does not have insurance
Condition coding: good (g), fair (f) and serious (s)
Joint coding for a patient will have the form: (, )
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The sample space: {(1, g ); (1, f ); (1, s ); (0, g ); (0, f ); (0, s )}
Event that the patient is in serious condition:
A = {(1, s ); (0, s )}
Event that the patient is unisured:
B = {(0, g ); (0, f ); (0, s )}
B c ∪ A = {(1, g ); (1, f ); (1, s ); (0, s )}
Tutorial 2 - Question 12
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Firstly, draw the Venn diagram for this problem.
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P(a randomly chosen student is not in any of 3 classes) =
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P(a randomly chosen student is taking exactly 1 class) =
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2 students are chosen randomly:
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50
100 = 0.5
14+10+8
= 0.5
100
P(at least 1 is taking a class) = 1 - P (both 2 are not taking any of
(50
49
2)
3 classes) = 1 − 100
= 1 − 198
(2)
Tutorial 2 - Question 16
5 dice are rolled simultaneously, each can result in 6 possible
outcomes. Total number of possible outcomes of 5 dice: 65 .
I a) P(no two alike) = ?
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1st die can result in 6 possible outcomes: 1, 2, 3, 4, 5, 6
2nd die can result in only 5 possible outcomes, since “no
two alike”. Similarly for the other dice
P(no two alike) = 6x 5x645x 3x 2 ' 0.0926
b)P(one pair) = ?
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The pair can be any from {11, 22, 33, 44, 55, 66} ⇒ 6
choices
Choose 2 from 5 dice to be the pair: 52
The remaining 3 dice must be different from the pair and
different from each
other: 5x 4x 3 choices
6x (52)x (5x 4x 3)
' 0.463
P(one pair) =
65
Tutorial 2 - Question 16 (cont’d)
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c) P(two pair) = ?
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The values of the two pair can
be any 2 different from
{11, 22, 33, 44, 55, 66} ⇒ 62 choices
Choose 2 from 5 dice to be the smaller pair: 52
Choose
2 from the remaining 3 dice to be the larger pair:
3
2
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The remaning die must be different from the two chosen
pairs: 4 choices
6x 5x (52)x (32)x 4
P(two pair) =
' 0.2315
65
d)P(three alike) = ?
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The “three alike” can be any from
{111, 222, 333, 444, 555, 666} ⇒ 6 choices
Choose 3 from 5 dice to be the “three alike”: 53
The remaining 2 dice must be different from the “three
alike” and different 5from each other: 5x 4 choices
6x (3)x (5x 4)
' 0.1543
P(three alike) =
65
Tutorial 2 - Question 16 (cont’d)
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d)P(full house) = P(three alike & one pair) = ?
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d)P(four alike) = ?
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The “three alike” can be any from
{111, 222, 333, 444, 555, 666} ⇒ 6 choices
Choose 3 from 5 dice to be the “three alike”: 53
The remaining pair must be different from the “three
alike”: 5 choices 5
6x ( )x 5
P(full house) = 635 ' 0.0386
The “four alike” can be any from
{1111, 2222, 3333, 4444, 5555, 6666} ⇒ 6 choices
Choose 4 from 5 dice to be the “four alike”: 54
The remaining die must be different from the “four alike”:
5 choices
6x (5)x 5
P(three alike) = 645 ' 0.0193
e)P(five alike) =
6
65
' 0.0008
Tutorial 2 - Question 20
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Let A = {you are dealt a blackjack}, B = {the dealer is
dealt a blackjack}
To get a blackjack, we must have 1 from the 4 Ace cards
and 1 from the 16 cards “10”, “J”, “Q”, “K”
P(A) = P(B) = 4x5216
(2)
P(A∩B) =
(4x 16)x (3x 15)
(522)x (502)
P(neither you or the dealer is dealt a blackjack) =
P (Ac ∩ B c ) = P (A ∪ B )c = 1 − P (A ∪ B ) =
1 − (P (A) + P (B ) − P (A ∩ B )) ' 0.9052
Tutorial 2 - Question 23
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First way: list all possible outcomes of the two dice
(1, 1) (1, 2) . . . (1, 6)
(2, 1) (2, 2) . . . (2, 6)
..
..
..
..
.
.
.
.
(6, 1) (6, 2) . . . (6, 6)
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P(the 2nd die > the 1st die) =
5+4+3+2+1
36
=
15
36
=
5
12
Second way:
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1 = P(the 2nd is higher) + P(the 1st is higher) + P(the 2
dice are equal)
P(the 2nd is higher) = P(the 1st is higher)
6
= 61
P(the 2 dice are equal) = 36
P(the 2nd is higher) =
1− 61
2
=
5
12
Tutorial 2 - Question 42
Two dice are thrown n times, each time the double 6 appears
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with probability 36
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P(double 6 appears at least once) = 1 - P(double 6 does not appear)
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n
P(double 6 appears at least once) = 1 − ( 35
36 )
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P(double 6 appears at least once) ≥
35 n
)
36
35
( )n
36
1−(
n
1
2
if and only if:
1
2
1
≤
2
≥ 25
≥
Tutorial 2 -Question 45
n keys, one of them will open the door. The woman opens the
door on her k th try
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try keys at random, discarding those that do not work:
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The k th try is success, the (k − 1) previous ones are fail
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P(the 1st try is fail) = n−
n
2
P(the 2nd try is fail) = nn−
−1
Similarly, P(the (k − 1)th try is fail) =
P(the k th try is success) =
1
n−(k −2)−1
n−(k −2)
n−(k −1)
P(the woman opens the door on her k th try)
k +1
1
1
n−1 n−2
. . . nn−
n n−1
−k +2 n−k +1 = n
if the woman does not discard previously tried keys:
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1 k −1 1
P(the woman opens the door on her k th try) = ( n−
)
n
n
Tutorial 2 -Question 46
Let n be the number of people in the room.
Total number of possible combinations of the months of their birthdays: 12n
P(at least 2 have birthday in the same month) = 1 - P(all months are different) ≥ 21
In order to have “all months are diffferent”:
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n ≤ 12
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The 1st person’s birthday can fall into any of the 12 month: 12 choices
The 2nd person’s birthday cannot be in the same month as the 1st:
(12-1) choices
I Similarly, for the nth person: 12 − (n − 1) choices
I Total number of possible combinations:
12x (12 − 1)x . . . x (12 − (n − 1))
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P(all months are different) ≤
1
2
if and only if
12x (12 − 1)x . . . x (12 − (n − 1))
1
≤
n
12
2
n ≥ 5
Tutorial 2 - Question 52
10 pairs of shoes, 8 shoes randomly selected:
I a) No complete pair:
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20
8
outcomes
Each chosen shoe is corresponding
to 1 pair
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Choose 8 from 10 pairs: 8 choices
For each chosen pair, we choose the left or the right shoe: 2
choices
8
Total number of possible choices: 10
8 x2
(108)x 28
P(no complete pair) = 20
(8)
b) exactly 1 complete pair:
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Choose the complete pair first: 10 choices
Choose 6 shoes from the remaining 9 pairs such that there
is no complete pairs: 96 x 26
10x (96)x 26
P(exactly 1 complete pair) =
(208)
Tutorial 2 - Question 4
Show that A = (
S∞
1
Ei )F =
S∞
E F) = B
1( i
Approach: show the followings
I If x ∈ A ⇒ x ∈ B
I If x ∈ B ⇒ x ∈ A
I Then we can conclude that A = B
Firstly, show that: if x ∈ A ⇒ x ∈ B
I x ∈A⇒x ∈F
S
I x ∈ A ⇒ x ∈ ( ∞ Ei )
1
S∞
I x ∈(
1 Ei ) ⇒ x ∈ one of the Ei , say for example x ∈ Ej
I x ∈ (Ej F )
S
I x ∈ ∞ (Ei F ) = B
1
Similarly we can show that if x ∈ B ⇒ x ∈ A. Hence, A = B .
Tutorial 2 - Question 6
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Firstly, draw the Venn diagram
a) EF c G c
b) EGF c
c) E ∪ F ∪ G
d) EF ∪ FG ∪ EG
e) EFG
f) E c F c G c
g) E c F c G c ∪ EF c G c ∪ E c FG c ∪ E c F c G
h) (EFG )c
i) EFG c ∪ EGF c ∪ FGE c
j) whole sample space
Tutorial 2 - Question 12
Firstly, it’s quite obvious from the Venn diagram that
P (EF c ∪ E c F ) = P (E ) + P (F ) − 2xP (EF )
Another way:
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P (EF c ∪ E c F ) = P (EF c ) + P (E c F ) since EF c and E c F disjoint
P (E ) = P (ES ) = P (E (F c ∪ F )) = P (EF c ∪ EF ) = P (EF c ) + P (EF )
since EF c and EF disjoint
Similary P (F ) = P (E c F ) + P (EF )
Hence, P (EF c ∪ E c F ) = P (E ) + P (F ) − 2xP (EF )
Tutorial 2 - Question 16
P (E1 E2 . . . En ) ≥ P (E1 ) + P (E2 ) + . . . + P (En ) − (n − 1) (*)
Firstly note that (*) is true for n = 1, n = 2:
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n=1: (*) becomes P (E1 ) ≥ P (E1 ) (true for any set E1 )
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n=2: (*) becomes
P (E1 E2 )
≥
1
≥
P (E1 ) + P (E2 ) − (2 − 1)
P (E1 ) + P (E2 ) − P (E1 E2 ) = P (E1 ∪ E2 )
The last inequality is true for any E1 and E2
Tutorial 2 - Question 16 (cont’d)
Now suppose (*) is true for n = k , we need to show that (*) is true for
n = k +1
Let A = (E1 E2 . . . Ek ), then P (E1 E2 . . . Ek Ek +1 ) = P (AEk +1 ) (1)
Since (*) is true for n = 2 sets, we have:
P (AEk +1 ) ≥ P (A) + P (Ek +1 ) − (2 − 1) (2)
Since (*) is true for n = k we have:
P (A) = P (E1 E2 . . . Ek ) ≥ P (E1 ) + P (E2 ) + . . . + P (Ek ) − (k − 1) (3)
From (1), (2) and (3) we have:
P (E1 E2 . . . Ek +1 ) ≥ P (E1 ) + P (E2 ) + . . . + P (Ek ) + P (Ek +1 ) − (k + 1 − 1)
that is, (*) is true for n = k + 1.
Hence we can conclude that (*) is true for any n ≥ 1.
Tutorial 2 - Question 17
Let denote the N men by M1 , M2 , . . . , MN and their hats by h1 , h2 , . . . , hN .
Consider the M1 : he cannot choose his own hat, so he has (N − 1) choices.
For simplicity, suppose he chooses h2 (other situations are similar).
After the choice of the M1 , we have (M2 , M3 . . . , MN ) and (h1 , h3 , . . . , hN ).
Now there are 2 possible situations for the M2 :
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the M2 chooses the h1 : so the remainings are (M3 , M4 . . . , MN ) and
(h3 , h4 , . . . , hN ). Then the problem becomes “(N − 2) men and their
(N − 2) hats”. So the number of ways is AN −2 .
the M2 does not like the h1 : this situation is exactly the same as the
situation when we have h2 , h3 , . . . , hN because M2 does not like h2
either. So when we replace h1 by h2 , the problem becomes
“(M2 , M3 . . . , MN ) and their (h2 , h3 , . . . , hN ) hats”. So the number of
ways is AN −1
Overall, we have
AN = (N − 1)(AN −1 + AN −2 )
where A1 = 0, A2 = 1