Permutations and Combinations

Permutations and Combinations
Sigma solutions Permutations and Combinations

Ex. 8.04 Page 161

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. ? ? ? ?

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 ? ? ?

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 ? ? ?

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 ? ?

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 ? ?

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 ?

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 ?

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

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

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? ? ? ? ?

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 ? ? ? ?

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 ? ? ?

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 ? ? ?

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 ? ? ?

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 ? ?

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 ?

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

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

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

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

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

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

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?

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 ? ? ? ?

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 ? ? ?

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 ? ?

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 ?

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

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.

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 ?

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.

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

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

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

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

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)

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)

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)

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)

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)

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) = = =

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:

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

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.

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

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

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 ? ? ? ? ?

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) ? ? ? ? ?

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 ? ? ? ?

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 ? ? ? ?

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 ? ? ?

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 ? ? ?

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 ? ?

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 ? ?

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 ?

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 ?

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

× × × × × × × × 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

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

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