Download presentation

Presentation is loading. Please wait.

1
**Permutations and Combinations**

Sigma solutions Permutations and Combinations

2
Ex. 8.04 Page 161

3
Sigma: Page 161 Ex 8.04 1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated. (a)How many different PIN numbers are possible? Without replacement and order matters - arranging (not just selecting). \ permutations. ? ? ? ?

4
Sigma: Page 161 Ex 8.04 1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated. (a)How many different PIN numbers are possible? Without replacement and order matters - arranging (not just selecting). \ permutations. ? ? ? ?

5
Sigma: Page 161 Ex 8.04 1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated. (a)How many different PIN numbers are possible? Without replacement and order matters - arranging (not just selecting). \ permutations. 9 possibilities for the first digit 9 ? ? ?

6
Sigma: Page 161 Ex 8.04 1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated. (a)How many different PIN numbers are possible? Without replacement and order matters - arranging (not just selecting). \ permutations. 9 ? ? ?

7
Sigma: Page 161 Ex 8.04 1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated. (a)How many different PIN numbers are possible? Without replacement and order matters - arranging (not just selecting). \ permutations. 8 possibilities for the second digit – one used up 9 8 ? ?

8
Sigma: Page 161 Ex 8.04 1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated. (a)How many different PIN numbers are possible? Without replacement and order matters - arranging (not just selecting). \ permutations. 9 8 ? ?

9
Sigma: Page 161 Ex 8.04 1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated. (a)How many different PIN numbers are possible? Without replacement and order matters - arranging (not just selecting). \ permutations. 7 possibilities for the third digit – 2 used up 9 8 7 ?

10
Sigma: Page 161 Ex 8.04 1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated. (a)How many different PIN numbers are possible? Without replacement and order matters - arranging (not just selecting). \ permutations. 9 8 7 ?

11
Sigma: Page 161 Ex 8.04 1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated. (a)How many different PIN numbers are possible? Without replacement and order matters - arranging (not just selecting). \ permutations. 6 possibilities for the third digit – 3 used up 9 8 7 6

12
Sigma: Page 161 Ex 8.04 1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated. (a)How many different PIN numbers are possible? Without replacement and order matters - arranging (not just selecting). \ permutations. Same as saying: 9 × 8 × 7 × 6 = different PINS are possible. 9 8 7 6

13
Sigma: Page 161 Ex 8.04 1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated. … (b)How many of the PIN numbers start with an odd digit? ? ? ? ?

14
Sigma: Page 161 Ex 8.04 1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated. … (b)How many of the PIN numbers start with an odd digit? Number of possibilities for first digit = 5 ? ? ? ?

15
Sigma: Page 161 Ex 8.04 1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated. … (b)How many of the PIN numbers start with an odd digit? Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9) 5 ? ? ?

16
Sigma: Page 161 Ex 8.04 1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated. … (b)How many of the PIN numbers start with an odd digit? Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9) 5 ? ? ?

17
Sigma: Page 161 Ex 8.04 1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated. … (b)How many of the PIN numbers start with an odd digit? Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9) Then there are 3 remaining positions to fill. 5 ? ? ?

18
Sigma: Page 161 Ex 8.04 1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated. … (b)How many of the PIN numbers start with an odd digit? Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9) Then there are 3 remaining positions to fill. 8 unused digits left to choose from without replacement. 5 ? ? ?

19
Sigma: Page 161 Ex 8.04 1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated. … (b)How many of the PIN numbers start with an odd digit? Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9) Then there are 3 remaining positions to fill. 8 unused digits left to choose from without replacement. 5 ? ? ?

20
Sigma: Page 161 Ex 8.04 1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated. … (b)How many of the PIN numbers start with an odd digit? Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9) Then there are 3 remaining positions to fill. 8 unused digits left to choose from without replacement. 5 8 ? ?

21
Sigma: Page 161 Ex 8.04 1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated. … (b)How many of the PIN numbers start with an odd digit? Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9) Then there are 3 remaining positions to fill. 8 unused digits left to choose from without replacement. 5 8 ? ?

22
Sigma: Page 161 Ex 8.04 1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated. … (b)How many of the PIN numbers start with an odd digit? Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9) Then there are 3 remaining positions to fill. 8 unused digits left to choose from without replacement. 5 8 7 ?

23
Sigma: Page 161 Ex 8.04 1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated. … (b)How many of the PIN numbers start with an odd digit? Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9) Then there are 3 remaining positions to fill. 8 unused digits left to choose from without replacement. 5 8 7 ?

24
Sigma: Page 161 Ex 8.04 1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated. … (b)How many of the PIN numbers start with an odd digit? Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9) Then there are 3 remaining positions to fill. 8 unused digits left to choose from without replacement. 5 8 7 6

25
Sigma: Page 161 Ex 8.04 1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated. … (b)How many of the PIN numbers start with an odd digit? Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9) Then there are 3 remaining positions to fill. 8 unused digits left to choose from without replacement. Order matters so still permutations. 5 8 7 6

26
Sigma: Page 161 Ex 8.04 1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated. … (b)How many of the PIN numbers start with an odd digit? Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9) Then there are 3 remaining positions to fill. 8 unused digits left to choose from without replacement. Order matters so still permutations. i.e. 8P3 Nbr poss. PINS starting with an odd digit = 5 × 8P3 5 8 7 6

27
Sigma: Page 161 Ex 8.04 1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated. … (b)How many of the PIN numbers start with an odd digit? Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9) Then there are 3 remaining positions to fill. 8 unused digits left to choose from without replacement. Order matters so still permutations. i.e. 8P3 Nbr poss. PINS starting with an odd digit = 5 × 8P3 or 5 × 8 × 7 × 6 5 8 7 6

28
Sigma: Page 161 Ex 8.04 1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated. … (b)How many of the PIN numbers start with an odd digit? Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9) Then there are 3 remaining positions to fill. 8 unused digits left to choose from without replacement. Order matters so still permutations. i.e. 8P3 Nbr poss. PINS starting with an odd digit = 5 × 8P3 or 5 × 8 × 7 × 6 = 1680 PINS. 5 8 7 6

29
Sigma: Page Ex 8.04 1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated. … (c) Calculate the probability that a client is given a PIN number starting with an odd digit Total number poss. PIN numbers (from a) = 9P4 = 3024 Nbr poss. PINS starting with an odd digit (from b) = 5 × 8P3 = 1680 P(1st digit is odd) = = = answer

30
Sigma: Page Ex 8.04 2. Company ‘numbers’ its invoices to clients using 1 letter of the alphabet followed by 3 digits from 0 to 9, repeats allowed. (a)How many different ‘numbers’ could be used?

31
Sigma: Page Ex 8.04 2. Company ‘numbers’ its invoices to clients using 1 letter of the alphabet followed by 3 digits from 0 to 9, repeats allowed. (a)How many different ‘numbers’ could be used? Repeats allowed (i.e. with replacement) so the number of possibilities does not reduce after a digit is used. Letter 3 digits from 0 to 9 ? ? ? ?

32
Sigma: Page Ex 8.04 2. Company ‘numbers’ its invoices to clients using 1 letter of the alphabet followed by 3 digits from 0 to 9, repeats allowed. (a)How many different ‘numbers’ could be used? Repeats allowed (i.e. with replacement) so the number of possibilities does not reduce after a digit is used. Letter 3 digits from 0 to 9 ? ? ? ?

33
Sigma: Page Ex 8.04 2. Company ‘numbers’ its invoices to clients using 1 letter of the alphabet followed by 3 digits from 0 to 9, repeats allowed. (a)How many different ‘numbers’ could be used? Repeats allowed (i.e. with replacement) so the number of possibilities does not reduce after a digit is used. Letter 3 digits from 0 to 9 26 ? ? ?

34
Sigma: Page Ex 8.04 2. Company ‘numbers’ its invoices to clients using 1 letter of the alphabet followed by 3 digits from 0 to 9, repeats allowed. (a)How many different ‘numbers’ could be used? Repeats allowed (i.e. with replacement) so the number of possibilities does not reduce after a digit is used. Letter 3 digits from 0 to 9 26 ? ? ?

35
Sigma: Page Ex 8.04 2. Company ‘numbers’ its invoices to clients using 1 letter of the alphabet followed by 3 digits from 0 to 9, repeats allowed. (a)How many different ‘numbers’ could be used? Repeats allowed (i.e. with replacement) so the number of possibilities does not reduce after a digit is used. Letter 3 digits from 0 to 9 26 10 ? ?

36
Sigma: Page Ex 8.04 2. Company ‘numbers’ its invoices to clients using 1 letter of the alphabet followed by 3 digits from 0 to 9, repeats allowed. (a)How many different ‘numbers’ could be used? Repeats allowed (i.e. with replacement) so the number of possibilities does not reduce after a digit is used. Letter 3 digits from 0 to 9 26 10 10 ?

37
Sigma: Page Ex 8.04 2. Company ‘numbers’ its invoices to clients using 1 letter of the alphabet followed by 3 digits from 0 to 9, repeats allowed. (a)How many different ‘numbers’ could be used? Repeats allowed (i.e. with replacement) so the number of possibilities does not reduce after a digit is used. Letter 3 digits from 0 to 9 26 10 10 10

38
Sigma: Page Ex 8.04 2. Company ‘numbers’ its invoices to clients using 1 letter of the alphabet followed by 3 digits from 0 to 9, repeats allowed. (a)How many different ‘numbers’ could be used? Repeats allowed (i.e. with replacement) so the number of possibilities does not reduce after a digit is used. Letter 3 digits from 0 to 9 26 10 10 10 So number of possible ‘numbers’ = 26 × 10 × 10 × 10 or 26 × 103 = different ‘numbers’ possible.

39
Sigma: Page Ex 8.04 2. Company ‘numbers’ its invoices to clients using 1 letter of the alphabet followed by 3 digits from 0 to 9, repeats allowed. … (b)How many would end in the numbers 8 or 9? Repeats still allowed (i.e. with replacement) so the number of possibilities for the other positions stays the same. Letter 2 digits from 0 to 9 Last digit is either 8 or 9 26 10 10 ?

40
Sigma: Page Ex 8.04 2. Company ‘numbers’ its invoices to clients using 1 letter of the alphabet followed by 3 digits from 0 to 9, repeats allowed. … (b)How many would end in the numbers 8 or 9? Repeats still allowed (i.e. with replacement) so the number of possibilities for the other positions stays the same. Letter 2 digits from 0 to 9 Last digit is either 8 or 9 26 10 10 2 2 possibilities So number of possible ‘numbers’ = 26 × 10 × 10 × 2 or 26 × 102 × 2 = different ‘numbers’ possible ending in 8 or 9.

41
Sigma: Page Ex 8.04 2. Company ‘numbers’ its invoices to clients using 1 letter of the alphabet followed by 3 digits from 0 to 9, repeats allowed. (c) Calculate the probability of choosing an invoice that does not contain the number 1 (NCEA Merit level). Total number poss. invoice ‘numbers’ (from a) = 26 × 103 = Nbr poss. invoice numbers not containing a 1 = 26 × 93 = P(does not contain a 1) = = = answer

42
**Sigma: Page 161 - Ex 8.04 = = or 0.6044 (4sf) answer**

5. A cricket team of 11 is to be chosen from a squad of 14, 2 of whom are brothers. Find the probability that: (a) Both brothers are chosen. This is a selection-only problem (note the word ‘chosen’). Order is not important. So we’re dealing with Combinations. Total number poss. combinations for the 11 = 14C11 = 364 There is only one way of selecting both brothers: 2C2 - i.e. 1. Number of ways of selecting the other 9 from the other 12 = 12C9 - i.e. 220. P(selecting both brothers in the 11) = = = or (4sf) answer

43
**Sigma: Page 161 - Ex 8.04 = = or 0.03297 (4sf) answer**

5. A cricket team of 11 is to be chosen from a squad of 14, 2 of whom are brothers. Find the probability that: (b) Neither brother is chosen. Still is a selection-only problem (note the word ‘chosen’). Order is not important. So we’re dealing with Combinations. Total number poss. combinations for the 11 = 14C11 = 364 There is only one way of selecting neither brother: 2C0 - i.e. 1. Nbr ways of selecting all 11 from the 12 non-brothers = 12C11 - i.e. 12. P(selecting neither bro. in the 11) = = = or (4sf) answer

44
**Sigma: Page 161 - Ex 8.04 = = or 0.3626 (4sf) answer**

5. A cricket team of 11 is to be chosen from a squad of 14, 2 of whom are brothers. Find the probability that: (c) Only one brother is chosen. (NCEA Merit level) Still is a selection-only problem (note the word ‘chosen’). Order is not important. So we’re dealing with Combinations. Total number poss. combinations for the 11 = 14C11 = 364 Nbr ways of selecting 1 brother from 2: 2C1 - i.e. 2 (could pick either bro). Nbr ways of selecting 10 from the 12 non-brothers = 12C10 - i.e. 66. P(selecting one bro. in the 11) = = = or (4sf) answer

45
**Sigma: Page 161 - Ex 8.04 = = x 1 2 3 P(X=x)**

6. (a) Complete the probability distribution table for the nbr of red balls obtained when 3 balls are selected at random from an urn containing 6 red and 4 yellow balls. (NCEA Merit level) Let X: number of red balls selected. P(X=0) Nbr ways of selecting 0 of the 6 red balls = 6C0 - i.e. 1. Nbr ways of selecting 3 of the 4 yellow balls = 4C3 - i.e. 4. P(selecting 0 red balls) = = = x 1 2 3 P(X=x)

46
**Sigma: Page 161 - Ex 8.04 = = or x 1 2 3 P(X=x)**

6. (a) Complete the probability distribution table for the nbr of red balls obtained when 3 balls are selected at random from an urn containing 6 red and 4 yellow balls. (NCEA Merit level) Let X: number of red balls selected. P(X=1) Nbr ways of selecting 1 of the 6 red balls = 6C1 - i.e. 6. Nbr ways of selecting 2 of the 4 yellow balls = 4C2 - i.e. 6. P(selecting 1 red ball) = = = or x 1 2 3 P(X=x)

47
**Sigma: Page 161 - Ex 8.04 = = or x 1 2 3 P(X=x)**

6. (a) Complete the probability distribution table for the nbr of red balls obtained when 3 balls are selected at random from an urn containing 6 red and 4 yellow balls. (NCEA Merit level) Let X: number of red balls selected. P(X=2) Nbr ways of selecting 2 of the 6 red balls = 6C2 - i.e. 15. Nbr ways of selecting 1 of the 4 yellow balls = 4C1 - i.e. 4. P(selecting 2 red ball) = = = or x 1 2 3 P(X=x)

48
**Sigma: Page 161 - Ex 8.04 = = or x 1 2 3 P(X=x)**

6. (a) Complete the probability distribution table for the nbr of red balls obtained when 3 balls are selected at random from an urn containing 6 red and 4 yellow balls. (NCEA Merit level) Let X: number of red balls selected. P(X=3) Nbr ways of selecting 3 of the 6 red balls = 6C3 - i.e. 20. Nbr ways of selecting 0 of the 4 yellow balls = 4C0 - i.e. 1. P(selecting 3 red balls) = = = or x 1 2 3 P(X=x)

49
**Sigma: Page 161 - Ex 8.04 = answer x 1 2 3 P(X=x)**

6. (b) If X is the number of red balls obtained, find an expression for P(X=x). The expression should use combinations. (NCEA Excellence level) Let X: number of red balls selected. P(X= x) Nbr ways of selecting x of the 6 red balls = 6Cx Nbr ways of selecting (3- x) of the 4 yellow balls = 4C(3-x) P(selecting x red balls) = = answer x 1 2 3 P(X=x)

50
Sigma: Page Ex 8.04 7. Three people each have a pair of socks, which are all washed one day then returned at random. What is the probability that a particular person gets their own pair of socks back? Selection only. Order doesn’t matter (left sock same as right sock). \ combinations. Total number of ways of selecting any 2 socks from the 6 = 6C2 - i.e. 15. Nbr ways of selecting a specific pair (selecting 2 socks from 2) = 2C2 - i.e. 1. P(selecting a given pair) = = =

51
Sigma: Page 161 Ex 8.04 8. In the card game of Poker, a hand of 5 cards is dealt from a pack of If all 5 cards are from the same suit, the hand is called a ‘flush’. What is the probability of getting a ‘flush’ in the game of Poker? Without replacement. Selecting only so order doesn’t matter. \ combinations. Total nbr of possible selections of ANY 5 cards = 52C5 = Number of possible ‘flushes’ – i.e. selections of 5 of same suit:

52
Sigma: Page 161 Ex 8.04 8. In the card game of Poker, a hand of 5 cards is dealt from a pack of If all 5 cards are from the same suit, the hand is called a ‘flush’. What is the probability of getting a ‘flush’ in the game of Poker? Without replacement. Selecting only so order doesn’t matter. \ combinations. Total nbr of possible selections of ANY 5 cards = 52C5 = Number of possible ‘flushes’ – i.e. selections of 5 of same suit: = 4 × 13C5

53
Sigma: Page 161 Ex 8.04 8. In the card game of Poker, a hand of 5 cards is dealt from a pack of If all 5 cards are from the same suit, the hand is called a ‘flush’. What is the probability of getting a ‘flush’ in the game of Poker? Without replacement. Selecting only so order doesn’t matter. \ combinations. Total nbr of possible selections of ANY 5 cards = 52C5 = Number of possible ‘flushes’ – i.e. selections of 5 of same suit: = 4 × 13C5 (i.e. 4 suits – each has 13 cards and we’re choosing 5 of them) = possible selections of 5 of same suit.

54
8. In the card game of Poker, a hand of 5 cards is dealt from a pack of If all 5 cards are from the same suit, the hand is called a ‘flush’. What is the probability of getting a ‘flush’ in the game of Poker? Without replacement. Selecting only so order doesn’t matter. \ combinations. Total nbr of possible selections of ANY 5 cards = 52C5 = Number of possible ‘flushes’ – i.e. selections of 5 of same suit: = 4 × 13C5 (i.e. 4 suits – each has 13 cards) = possible selections of 5 of same suit. P(getting a flush) = = = or (4sf) answer

55
Sigma: Page 161 Ex 8.04 8. In the card game of Poker, a hand of 5 cards is dealt from a pack of If all 5 cards are from the same suit, the hand is called a ‘flush’. What is the probability of getting a ‘flush’ in the game of Poker? OR can solve the problem without using combinations, just by using the multiplication principle. Total possible number of arrangements of any 5 cards: 52 × 51 × 50 × 49 × 48

56
Sigma: Page 161 Ex 8.04 8. In the card game of Poker, a hand of 5 cards is dealt from a pack of If all 5 cards are from the same suit, the hand is called a ‘flush’. What is the probability of getting a ‘flush’ in the game of Poker? OR can solve the problem without using combinations, just by using the multiplication principle. Total possible number of arrangements of any 5 cards: 52 × 51 × 50 × 49 × 48

57
Sigma: Page 161 Ex 8.04 8. In the card game of Poker, a hand of 5 cards is dealt from a pack of If all 5 cards are from the same suit, the hand is called a ‘flush’. What is the probability of getting a ‘flush’ in the game of Poker? OR can solve the problem without using combinations, just by using the multiplication principle. Total possible number of arrangements of any 5 cards: Number of possible ‘flushes’ – i.e. ways of getting 5 of same suit: 52 × 51 × 50 × 49 × 48 ? ? ? ? ?

58
Sigma: Page 161 Ex 8.04 8. In the card game of Poker, a hand of 5 cards is dealt from a pack of If all 5 cards are from the same suit, the hand is called a ‘flush’. What is the probability of getting a ‘flush’ in the game of Poker? OR can solve the problem without using combinations, just by using the multiplication principle. Total possible number of arrangements of any 5 cards: Number of possible ‘flushes’ – i.e. ways of getting 5 of same suit: 52 × 51 × 50 × 49 × 48 52 possibilities for the first card (could be any card) ? ? ? ? ?

59
Sigma: Page 161 Ex 8.04 8. In the card game of Poker, a hand of 5 cards is dealt from a pack of If all 5 cards are from the same suit, the hand is called a ‘flush’. What is the probability of getting a ‘flush’ in the game of Poker? OR can solve the problem without using combinations, just by using the multiplication principle. Total possible number of arrangements of any 5 cards: Number of possible ‘flushes’ – i.e. ways of getting 5 of same suit: 52 × 51 × 50 × 49 × 48 52 possibilities for the first card (could be any card) 52 ? ? ? ?

60
Sigma: Page 161 Ex 8.04 8. In the card game of Poker, a hand of 5 cards is dealt from a pack of If all 5 cards are from the same suit, the hand is called a ‘flush’. What is the probability of getting a ‘flush’ in the game of Poker? OR can solve the problem without using combinations, just by using the multiplication principle. Total possible number of arrangements of any 5 cards: Number of possible ‘flushes’ – i.e. ways of getting 5 of same suit: 52 × 51 × 50 × 49 × 48 12 possibilities for the second card (the other 12 from the same suit) 52 ? ? ? ?

61
Sigma: Page 161 Ex 8.04 8. In the card game of Poker, a hand of 5 cards is dealt from a pack of If all 5 cards are from the same suit, the hand is called a ‘flush’. What is the probability of getting a ‘flush’ in the game of Poker? OR can solve the problem without using combinations, just by using the multiplication principle. Total possible number of arrangements of any 5 cards: Number of possible ‘flushes’ – i.e. ways of getting 5 of same suit: 52 × 51 × 50 × 49 × 48 12 possibilities for the second card (the other 12 from the same suit) 52 12 ? ? ?

62
Sigma: Page 161 Ex 8.04 8. In the card game of Poker, a hand of 5 cards is dealt from a pack of If all 5 cards are from the same suit, the hand is called a ‘flush’. What is the probability of getting a ‘flush’ in the game of Poker? OR can solve the problem without using combinations, just by using the multiplication principle. Total possible number of arrangements of any 5 cards: Number of possible ‘flushes’ – i.e. ways of getting 5 of same suit: 52 × 51 × 50 × 49 × 48 11 possibilities for the 3rd card (2 used up, so 11 of that suit are remaining) 52 12 ? ? ?

63
Sigma: Page 161 Ex 8.04 8. In the card game of Poker, a hand of 5 cards is dealt from a pack of If all 5 cards are from the same suit, the hand is called a ‘flush’. What is the probability of getting a ‘flush’ in the game of Poker? OR can solve the problem without using combinations, just by using the multiplication principle. Total possible number of arrangements of any 5 cards: Number of possible ‘flushes’ – i.e. ways of getting 5 of same suit: 52 × 51 × 50 × 49 × 48 11 possibilities for the 3rd card (2 used up, so 11 of that suit are remaining) 52 12 11 ? ?

64
Sigma: Page 161 Ex 8.04 8. In the card game of Poker, a hand of 5 cards is dealt from a pack of If all 5 cards are from the same suit, the hand is called a ‘flush’. What is the probability of getting a ‘flush’ in the game of Poker? OR can solve the problem without using combinations, just by using the multiplication principle. Total possible number of arrangements of any 5 cards: Number of possible ‘flushes’ – i.e. ways of getting 5 of same suit: 52 × 51 × 50 × 49 × 48 10 possibilities for the 4th card (3 used up, so 10 of that suit are remaining) 52 12 11 ? ?

65
Sigma: Page 161 Ex 8.04 8. In the card game of Poker, a hand of 5 cards is dealt from a pack of If all 5 cards are from the same suit, the hand is called a ‘flush’. What is the probability of getting a ‘flush’ in the game of Poker? OR can solve the problem without using combinations, just by using the multiplication principle. Total possible number of arrangements of any 5 cards: Number of possible ‘flushes’ – i.e. ways of getting 5 of same suit: 52 × 51 × 50 × 49 × 48 10 possibilities for the 4th card (3 used up, so 10 of that suit are remaining) 52 12 11 10 ?

66
Sigma: Page 161 Ex 8.04 8. In the card game of Poker, a hand of 5 cards is dealt from a pack of If all 5 cards are from the same suit, the hand is called a ‘flush’. What is the probability of getting a ‘flush’ in the game of Poker? OR can solve the problem without using combinations, just by using the multiplication principle. Total possible number of arrangements of any 5 cards: Number of possible ‘flushes’ – i.e. ways of getting 5 of same suit: 52 × 51 × 50 × 49 × 48 9 possibilities for the 5th card (4 used up, so 9 of that suit are remaining) 52 12 11 10 ?

67
Sigma: Page 161 Ex 8.04 8. In the card game of Poker, a hand of 5 cards is dealt from a pack of If all 5 cards are from the same suit, the hand is called a ‘flush’. What is the probability of getting a ‘flush’ in the game of Poker? OR can solve the problem without using combinations, just by using the multiplication principle. Total possible number of arrangements of any 5 cards: Number of possible ‘flushes’ – i.e. ways of getting 5 of same suit: 52 × 51 × 50 × 49 × 48 9 possibilities for the 5th card (4 used up, so 9 of that suit are remaining) 52 12 11 10 9

68
Sigma: Page 161 Ex 8.04 8. In the card game of Poker, a hand of 5 cards is dealt from a pack of If all 5 cards are from the same suit, the hand is called a ‘flush’. What is the probability of getting a ‘flush’ in the game of Poker? OR can solve the problem without using combinations, just by using the multiplication principle. Total possible number of arrangements of any 5 cards: Number of possible ‘flushes’ – i.e. ways of getting 5 of same suit: 52 × 51 × 50 × 49 × 48 9 possibilities for the 5th card (4 used up, so 9 of that suit are remaining) 52 12 11 10 9

69
**× × × × × × × × 52 51 50 49 48 52 12 11 10 9 Sigma: Page 161 Ex 8.04**

8. In the card game of Poker, a hand of 5 cards is dealt from a pack of If all 5 cards are from the same suit, the hand is called a ‘flush’. What is the probability of getting a ‘flush’ in the game of Poker? OR can solve the problem without using combinations, just by using the multiplication principle. Total possible number of arrangements of any 5 cards: Number of possible ‘flushes’ – i.e. ways of getting 5 of same suit: 52 × 51 × 50 × 49 × 48 52 × 12 × 11 × 10 × 9

70
**OR can solve the problem without using combinations, just by using the multiplication principle.**

Total possible number of arrangements of any 5 cards: Number of possible ‘flushes’ – i.e. ways of getting 5 of same suit: 52 × 51 × 50 × 49 × 48 52 × 12 × 11 × 10 × 9 P(getting a flush) = = = or (4sf) answer

Similar presentations

Presentation is loading. Please wait....

OK

Discrete Probability Rosen 5.1.

Discrete Probability Rosen 5.1.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on new seven wonders of the world 2013 Download ppt on transportation in animals Ppt on cse related topics about information Ppt on introduction to bluetooth technology Ppt on essay writing Ppt on different occupations in art Ppt on albert einstein Ppt on online advertising in india Ppt on isobars and isotopes of hydrogen Ppt on layer 3 switching