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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. ? ? ? ? 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. n! n Pr ( n r )! 9 9! P4 5! Same as saying: 9 × 8 × 7 × 6 = 3024 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) 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 ? ? ? 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 ? ? 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 ? 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 5 8 7 6 = 1680 PINS. 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. … (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) = Nbr poss. PINS starting with an odd digit Total number poss. PIN numbers 1680 = 3024 5 = answer 9 Sigma: Page 161 - 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 161 - 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 161 - 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 161 - 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 26 3 digits from 0 to 9 ? ? ? Sigma: Page 161 - 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 26 3 digits from 0 to 9 ? ? ? Sigma: Page 161 - 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 26 3 digits from 0 to 9 10 ? ? Sigma: Page 161 - 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 26 3 digits from 0 to 9 10 10 ? Sigma: Page 161 - 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 26 3 digits from 0 to 9 10 10 10 Sigma: Page 161 - 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 26 3 digits from 0 to 9 10 10 10 So number of possible ‘numbers’ = 26 × 10 × 10 × 10 or 26 × 103 = 26 000 different ‘numbers’ possible. Sigma: Page 161 - 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 26 2 digits from 0 to 9 10 10 Last digit is either 8 or 9 ? Sigma: Page 161 - 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 26 2 digits from 0 to 9 10 10 Last digit is either 8 or 9 2 2 possibilities So number of possible ‘numbers’ = 26 × 10 × 10 × 2 or 26 × 102 × 2 = 5 200 different ‘numbers’ possible ending in 8 or 9. Sigma: Page 161 - 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 = 26 000 Nbr poss. invoice numbers not containing a 1 = 26 × 93 = 18 954 P(does not contain a 1) = Nbr poss. invoice numbers not containing a 1 Total number poss. invoice numbers = = 18954 26000 0.729 answer Sigma: Page 161 - Ex 8.04 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: 2C 2 - 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) = Nbr. ways of selecting 9 non - brothers from 12 Total number of poss. combinatio ns for the 11 12 C9 = 14C 11 = 55 or 0.6044 (4sf) 91 answer Sigma: Page 161 - Ex 8.04 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: 2C 0 - i.e. 1. Nbr ways of selecting all 11 from the 12 non-brothers = 12C11 - i.e. 12. Nbr. ways of selecting 11 non - brothers from 12 Total number of poss. combinatio ns for the 11 P(selecting neither bro. in the 11) = 12 C11 = 14C 11 = 3 or 0.03297 (4sf) 91 answer Sigma: Page 161 - Ex 8.04 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 Nbr ways of selecting 1 brother from 2: 2C 1 = 14C11 = 364 - 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) = Ways of selecting 1bro Ways of selecting 10 non - bros Total number of poss. combinatio ns for the 11 C112 C10 = 14C 11 2 = 33 91 or 0.3626 (4sf) answer Sigma: Page 161 - Ex 8.04 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 = 6C 0 - i.e. 1. Nbr ways of selecting 3 of the 4 yellow balls = 4C3 - i.e. 4. P(selecting 0 red balls) = Ways of selecting 0 red Ways of selecting 3 yellow Total number of poss. selections of 3 balls 4 C C3 0 = 10 C3 1 = 30 0 1 6 x P(X=x) 1 30 2 3 Sigma: Page 161 - Ex 8.04 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 = 6C 1 - i.e. 6. Nbr ways of selecting 2 of the 4 yellow balls = 4C2 - i.e. 6. P(selecting 1 red ball) = Ways of selecting 1 red Ways of selecting 2 yellow Total number of poss. selections of 3 balls 4 C C2 1 = 10 C3 3 9 = 10 or 30 0 1 6 x P(X=x) 1 30 9 30 2 3 Sigma: Page 161 - Ex 8.04 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 = 6C 2 - i.e. 15. Nbr ways of selecting 1 of the 4 yellow balls = 4C1 - i.e. 4. Ways of selecting 2 red Ways of selecting 1 yellow Total number of poss. selections of 3 balls P(selecting 2 red ball) = = = x 0 P(X=x) 1 30 C2 4 C1 10 C3 1 15 2 or 30 1 6 9 30 2 15 30 3 Sigma: Page 161 - Ex 8.04 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 = 6C 3 - i.e. 20. Nbr ways of selecting 0 of the 4 yellow balls = 4C0 - i.e. 1. P(selecting 3 red balls) = = = x 0 P(X=x) 1 30 Ways of selecting 3 red Ways of selecting 0 yellow Total number of poss. selections of 3 balls C3 4 C 0 10 C3 1 5 6 or 30 1 6 9 30 2 3 15 30 5 30 Sigma: Page 161 - Ex 8.04 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) = Ways of selecting x red Ways of selecting (3 - x) yellow Total number of poss. selections of 3 balls 6 = C x 4 C ( 3 x ) 10 answer C3 x 0 1 2 3 P(X=x) 1 30 9 30 15 30 5 30 Sigma: Page 161 - 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 = 6C 2 - i.e. 15. Nbr ways of selecting a specific pair (selecting 2 socks from 2) = 2C2 - i.e. 1. P(selecting a given pair) = = = Ways of selecting a specific pair Total number of ways of selecting any 2 socks from 6 2 C2 6 C2 1 15 Sigma: Page 161 Ex 8.04 8. In the card game of Poker, a hand of 5 cards is dealt from a pack of 52. 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 = 2 598 960 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 52. 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. \ combinations. Selecting only so order doesn’t matter. Total nbr of possible selections of ANY 5 cards = 52C5 = 2 598 960 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 52. 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. \ combinations. Selecting only so order doesn’t matter. Total nbr of possible selections of ANY 5 cards = 52C5 = 2 598 960 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) = 5148 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 52. 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. \ combinations. Selecting only so order doesn’t matter. Total nbr of possible selections of ANY 5 cards = 52C5 = 2 598 960 Number of possible ‘flushes’ – i.e. selections of 5 of same suit: = 4 × 13C5 (i.e. 4 suits – each has 13 cards) = 5148 possible selections of 5 of same suit. P(getting a flush) = Number of possible selections of 5 of the same suit Number of possible selections of ANY 5 cards = 413C5 52 C5 33 = 16660 or 0.001981 (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 52. 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 52. 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 52. 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 Number of possible ‘flushes’ – i.e. ways of getting 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 52. 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 Number of possible ‘flushes’ – i.e. ways of getting 5 of same suit: 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 52. 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 Number of possible ‘flushes’ – i.e. ways of getting 5 of same suit: 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 52. 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 Number of possible ‘flushes’ – i.e. ways of getting 5 of same suit: 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 52. 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 Number of possible ‘flushes’ – i.e. ways of getting 5 of same suit: 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 52. 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 Number of possible ‘flushes’ – i.e. ways of getting 5 of same suit: 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 52. 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 Number of possible ‘flushes’ – i.e. ways of getting 5 of same suit: 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 52. 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 Number of possible ‘flushes’ – i.e. ways of getting 5 of same suit: 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 52. 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 Number of possible ‘flushes’ – i.e. ways of getting 5 of same suit: 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 52. 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 Number of possible ‘flushes’ – i.e. ways of getting 5 of same suit: 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 52. 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 Number of possible ‘flushes’ – i.e. ways of getting 5 of same suit: 9 possibilities for the 5th card (4 used up, so 9 of that suit are remaining) 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 52. 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 Number of possible ‘flushes’ – i.e. ways of getting 5 of same suit: 9 possibilities for the 5th card (4 used up, so 9 of that suit are remaining) 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 52. 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 Number of possible ‘flushes’ – i.e. ways of getting 5 of same suit: 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: 52 × 51 × 50 × 49 × 48 Number of possible ‘flushes’ – i.e. ways of getting 5 of same suit: 52 × 12 P(getting a flush) = = = × 11 × 10 × 9 Number of possible ' flushes' Total number of poss. arrangemen ts of 5 cards 52 12 11 10 9 52 51 50 49 48 33 16660 or 0.001981 (4sf) answer