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A group of 5 boys and 3 girls is to be photographed
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120 = 720 ways
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Left (8-2 =) 6 people may be seated with no restriction:
5
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7
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5 = 1680

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How many ways can be arranged if the girls occupy the even places? (All the chairs in one row)
The boys can be seated in P(5,5) = 5! = 120 ways, and the girls in P(3,3) = 3! = 6 ways
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Then there are 7 people altogether and they can be
arranged in 6! = 720 ways
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Therefore, the number of
ways to arranging 8 people with 2 particular people together is = 6! 2! = 1440
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How many possibilities for 2 boys and 3 girls
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(3) = 30 possibilities
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C(3,0) = 10
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C(3,1) = 10
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1

C(8,3) = 8
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6 = 56 ways
3
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1

12 Boys & 10 Girls
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What is the probability that at least 2 Girls are chosen?
1 – [Pr(no Girls) + Pr(1 G)] = 1 -

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C(12,7)  C(12,6)C(10,1)

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11
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21
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C(10,3)  C(12,5)
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C(10,1)  C(12,7)

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How many different poker hands are there?
C(52,5) = 2598960 hands

({1,2,3,4,5} = {5,3,4,2,1})

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How many different poker hands consist entirely of aces and kings?
Number of aces + kings = 8; C(8,5) = 56 hands
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C(4,3)
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6 = 24

# red cards = 26; C(26,5) = 65780 hands
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Therefore there are 4
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3C(13,4) = 111540 ways
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Therefore there are 4
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3C(13,3) = 267696 ways
c) Total = 111540 + 267696 = 379236 ways

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How many ways all the cards from the same suit?
Select a suit, there are C(4,1) = 4 ways to do this
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Final = 4
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How many ways 3 from one suit and 2 from another?
Select a suit, there are C(4,1) = 4 ways to do this
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First 3 from 1 suit there are 4
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C(13,2) = 78
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C(13,3)
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How many ways 2 aces, 2 cards of another denomination, and 1 card of a 3rd denomination
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Therefore, there are 12
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C(4,1) = 44
The outcomes: 6
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(44) = 19008 hands
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Select 2 cards of 1 denomination = C(13,2) = 78 ways
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C(4,1) = 44
# of poker hands = 78
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6
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3

Let the event A: card is a spade, and B is the face card
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What is the
probability that none of the 13 cards is the ace of spades?
Pr( none) x Pr( random 1 is not found |none)
Pr(none| 1 is not) =
Pr(none)xP r(1 is not found |none)  Pr(1 is)xPr(Ace is not | ace is)



3 x1
4
3 x1 ( 1 x 12 )
4
4 13

 13
17

c) Repeat previous experiment 10 times (replacing the card), and the ace of spades is not seen
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13

4

COIN
- 3 coins to be tossed
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a) How many different outcomes are possible?

b) How many different outcomes have exactly 4 heads?

C(10,4) = 210 outcomes

c) How many different outcomes at the most 2 heads? C(10,0) + C(10,1) + C(10,2) = 56
d) How many different outcomes at least 3 heads?
All outcomes – (at most 2 H) = 1024 – 56 = 968
e) What is the probability of obtaining exactly 4 heads?
P  4H  

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C (10,4)
210

 210 
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Determine the player’s mathematical expectation?
36 possible pairs with 2 dice
8 possible for 7 or 11
Pr(7 or 11) = 8 and Pr (not (7 or11)) = 1- 8 = 28
36

36

36

E = 7x 8 + (-2) 28 = 0
36
36
Events E= 1st die is a 3, F = 2nd die is 6

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