STAT213 Tutorial sheet #3 solutions

 

Question 1

a)      The total number of choices for the denominator is 64 = 1296.

The events in the sample space are:

§  all choose the same combo

§  three choose 1 combo; one chooses another

§  two choose 1 combo; two choose another

§  two choose 1 combo; two choose 2 separate combos (3 combos in total)

§  all choose different combos

Now to compute the probabilities:

All the same:

Choose the combo: 6C1 = 6. Total = 6; probability = 6/1296 = 0.46%

Three choose 1 combo; one chooses another:

Choose combo for group of 3: 6C1 = 6

Choose combo for group of 1: 5C1 = 5

Arrange combos: AAAB = 4!/3! = 4

Total = 6 x 5 x 4 = 120; probability = 120/1296 = 9.26%

Two choose 1 combo; two choose another:

Choose combos for groups of 2: 6C2 = 15

Arrange combos: AABB = 4!/2!2! = 4C2 = 6

Total = 15 x 6 = 90; probability = 90/1296 = 6.94%

Two choose 1 combo; two choose 2 separate combos (3 combos in total):

Choose combo for group of 2: 6C1 = 6

Choose combos for groups of 1: 5C2 = 10

Arrange combos: AABC = 4!/2! = 12

Total = 6 x 10 x 12 = 720; probability = 720/1296 = 55.56%

All choose different combos:

Permute 6 combos among 4 people: 6P4 = 360; probability = 360/1296 = 27.78%

b)      We can create a frequency table of the number of combos chosen:

# combo

Frequency

%

Cum. %

1

6

0.46%

0.46%

2

120 + 90

= 210

16.2%

16.67%

3

720

55.56%

72.22%

4

360

27.78%

100%

Then P(X ³ 3 | X ³ 2) = P(X ³ 3)/P(X ³ 2) = [1 – P(X £ 2)]/[1 – P(X £ 1)] = 83.33/99.54 = 83.72%

 

Question 2

a)      The total number of choices is 24C5 = 42,504. As with poker, the sample space consists of:

1: none the same

2: pair

3: 2 pair

4: 3 of a kind

5: full house

6: 4 of a kind

The strategy is to choose the faces based on how many groups of cards there are and then to choose the suits for each face.

1: none the same - 5 groups of 1 card

6C5 x (4C1)5 = 6 x 1024 = 6,144

2: pair - 1 group of 2 cards; 3 groups of 1 card

6C1 x 5C3 x 4C2 x (4C1)3 = 6 x 10 x 6 x 64 = 23,040

3: 2 pair - 2 groups of 2 cards; 1 group of 1 card

6C2 x 4C1 x (4C2)2 x 4C1 = 15 x 4 x 36 x 4 = 8,640

4: 3 of a kind - 1 group of 3 cards; 2 groups of 1 card

6C1 x 5C2 x 4C3 x (4C1)2 = 6 x 10 x 4 x 16 = 3,840

5: full house - 1 group of 3 cards; 1 group of 2 cards

6C1 x 5C1 x 4C3 x 4C2 = 6 x 5 x 4 x 6 = 720

6: 4 of a kind - 1 group of 4 cards; 1 group of 1 card

6C1 x 5C1 x 4C4 x 4C1 = 6 x 5 x 1 x 4 = 120

And now the frequency table:

Event

Frequency

Percent

none same

6144

14.46%

pair

23040

54.21%

2 pair

8640

20.33%

3 of kind

3840

9.03%

full house

720

1.69%

4 of kind

120

0.28%

Total

42504

100.00%

b)      P(3 spades or 2 jacks) = P(3 spades) + P(2 jacks) – P(3 spades and 2 jacks)

P(3 spades) = (6C3)(18C2)/(24C5) = (20 x 153)/42,504 = 3,060/42,504

P(2 jacks) = (4C2)(20C3)/(24C5) = (6 x 1140)/42,504 = 6,840/42,504

For P(3 spades and 2 jacks), there are 2 cases: either we include the jack of spades or exclude it.

Include jack of spades:

Jack of spades: 1C1 = 1

One other jack: 3C1 = 3

Two other spades: 5C2 = 10

One other card: 15C1 = 15

Total of this case = 1 x 3 x 10 x 15 = 450

Exclude jack of spades:

Two jacks: 3C2 = 3

Three spades: 5C3 = 10

Total of this case = 3 x 10 = 30

Total = 450 + 30 = 480

P(3 spades and 2 jacks) = 480/42,540

So, P(3 spades or 2 jacks) = (3,060 + 6,850 – 480)/42,504 = 9,430/42,504 = 22.19%

c)      The total number of choices is 4C2 x 20C3 = 6 x 1140 = 6,840.

The sample space consists of pair, two pair and full house.

Pair:

Choose face for remaining cards: 5C3 = 10

Choose suits for pair: 4C2 = 6

Choose suits for remaining cards: (4C1)3 = 43 = 64

Total = 10 x 6 x 64 = 3,840

Two pair:

Choose face for other pair: 5C1 = 5

Choose face for last card: 4C1 = 4

Choose suits for pairs: (4C2)2 = 62 = 36

Choose suit for last card: 4C1 = 4

Total = 5 x 4 x 36 x 4 = 2,880

Full house:

Choose face for 3 of a kind: 5C1 = 5

Choose suits for pair: 4C2 = 6

Choose suits for 3 of a kind: 4C3 = 4

Total = 5 x 6 x 4 = 120

And now the frequency table:

Event

Frequency

Percent

pair

3840

56.14%

2 pair

2880

42.11%

full house

120

1.75%

Total

6840

100.00%

 

Question 3

a)      (4C1)(4C1)(4C1)/(12C3) = 64/220 = 29.09%

b)      P(b = 0) = (4C0)(8C3)/(12C3) = 56/220

P(b ³ 1) = 1 – 56/220 = 74.55%

c)      Since there is an equal number of each colour, we know P(r ³ 1) = 164/220 = 74.55%.

There are 2 approaches we can take.

Approach 1:

P(r = 2) = (4C2)(8C1)/(12C3) = 48/220

P(r = 3) = (4C3)(8C0)/(12C3) = 4/220

P(r ³ 2) = 48/220 + 4/220 = 52/220

So P(r ³ 2 | r ³ 1) = 52/164 = 31.71%

Approach 2:

P(r = 0) = (4C0)(8C3)/(12C3) = 56/220

P(r = 1) = (4C1)(8C2)/(12C3) = 112/220

Then P(r ³ 2) = 1 –  (56/220 + 112/220) = 1 – 168/220 = 52/220

And the rest follows.

 

Question 4

a)      14P4 = 24,024. The other method is 14x13x12x11 = 24,024

b)      14C3 = 364

c)      The number of ways in which no senior person is chosen is (6C0)(8C3) = 56. Then, the number of ways in which at least one senior person is chosen is 364 – 56 = 308. The long way is to compute the number of ways in which 1, 2 and 3 seniors are chosen which are 168, 120 and 20 respectively and add them to get 308.

 

Question 5

a)      36P4 = 1,413,720. The other method is 36x35x34x33 = 1,413,720

b)      26x35x34x33 = 1,021,020. The other method is 26 x 35P3 = 26 x 39,270 = 1,021,020