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Sigma solutions Permutations and Combinations

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Ex Page 161

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Sigma: Page 161 Ex 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. ????

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Sigma: Page 161 Ex 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. ????

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Sigma: Page 161 Ex 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??? 9 possibilities for the first digit

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Sigma: Page 161 Ex 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???

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Sigma: Page 161 Ex 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. 98?? 8 possibilities for the second digit – one used up

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Sigma: Page 161 Ex 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. 98??

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Sigma: Page 161 Ex 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. 987? 7 possibilities for the third digit – 2 used up

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Sigma: Page 161 Ex 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. 987?

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Sigma: Page 161 Ex 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 possibilities for the third digit – 3 used up

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Sigma: Page 161 Ex 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. = 3024 different PINS are possible. Same as saying: 9 × 8 × 7 ×

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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 ????

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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 ????

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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) Sigma: Page 161 Ex ???

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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) Sigma: Page 161 Ex ???

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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. Sigma: Page 161 Ex ???

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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. Sigma: Page 161 Ex ???

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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. Sigma: Page 161 Ex ???

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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. Sigma: Page 161 Ex ??

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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. Sigma: Page 161 Ex ??

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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. Sigma: Page 161 Ex ?

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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. Sigma: Page 161 Ex ?

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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. Sigma: Page 161 Ex

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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. Sigma: Page 161 Ex

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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. 8 P 3 Nbr poss. PINS starting with an odd digit = 5 × 8 P 3 Sigma: Page 161 Ex

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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. 8 P 3 Nbr poss. PINS starting with an odd digit = 5 × 8 P 3 or 5 × 8 × 7 × 6 Sigma: Page 161 Ex

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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. 8 P 3 Nbr poss. PINS starting with an odd digit = 5 × 8 P 3 or 5 × 8 × 7 × 6 = 1680 PINS. Sigma: Page 161 Ex

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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 Nbr poss. PINS starting with an odd digit (from b)= 5 × 8 P 3 = 1680 Sigma: Page Ex 8.04 Total number poss. PIN numbers (from a)= 9 P 4 = 3024 P(1 st digit is odd) = = = answer

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Sigma: Page Ex 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?

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Sigma: Page Ex 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

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Sigma: Page Ex 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

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Sigma: Page Ex 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

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Sigma: Page Ex 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

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Sigma: Page Ex 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 2610?? 3 digits from 0 to 9

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Sigma: Page Ex 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 2610 ? 3 digits from 0 to 9

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Sigma: Page Ex 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 digits from 0 to 9

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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. = different numbers possible. Sigma: Page Ex 8.04 Letter digits from 0 to 9 So number of possible numbers = 26 × 10 × 10 × 10 or 26 × 10 3

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Sigma: Page Ex 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 2610 ? 2 digits from 0 to 9 Last digit is either 8 or 9

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Sigma: Page Ex 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 digits from 0 to 9 Last digit is either 8 or 9 2 possibilities So number of possible numbers = 26 × 10 × 10 × 2 or 26 × 10 2 × 2 = different numbers possible ending in 8 or 9.

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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). Nbr poss. invoice numbers not containing a 1= 26 × 9 3 = Sigma: Page Ex 8.04 Total number poss. invoice numbers (from a)= 26 × 10 3 = P(does not contain a 1) = = = answer

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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. There is only one way of selecting both brothers: 2 C 2 - i.e. 1. Sigma: Page Ex 8.04 Total number poss. combinations for the 11= 14 C 11 = 364 P(selecting both brothers in the 11) = = = or (4sf) answer This is a selection-only problem (note the word chosen). Order is not important. So were dealing with Combinations. Number of ways of selecting the other 9 from the other 12 = 12 C 9 - i.e. 220.

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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. There is only one way of selecting neither brother: 2 C 0 - i.e. 1. Sigma: Page Ex 8.04 Total number poss. combinations for the 11= 14 C 11 = 364 P(selecting neither bro. in the 11) = = = or (4sf) answer Still is a selection-only problem (note the word chosen). Order is not important. So were dealing with Combinations. Nbr ways of selecting all 11 from the 12 non-brothers = 12 C 11 - i.e. 12.

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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) Nbr ways of selecting 1 brother from 2: 2 C 1 - i.e. 2 (could pick either bro). Sigma: Page Ex 8.04 Total number poss. combinations for the 11= 14 C 11 = 364 P(selecting one bro. in the 11) = = = or (4sf) answer Still is a selection-only problem (note the word chosen). Order is not important. So were dealing with Combinations. Nbr ways of selecting 10 from the 12 non-brothers = 12 C 10 - i.e. 66.

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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) P(X=0) Sigma: Page Ex 8.04 P(selecting 0 red balls) = = = Nbr ways of selecting 3 of the 4 yellow balls = 4 C 3 - i.e. 4. Nbr ways of selecting 0 of the 6 red balls = 6 C 0 - i.e. 1. Let X: number of red balls selected. x0123 P(X=x)

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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) P(X=1) Sigma: Page Ex 8.04 P(selecting 1 red ball) = = = Nbr ways of selecting 2 of the 4 yellow balls = 4 C 2 - i.e. 6. Nbr ways of selecting 1 of the 6 red balls = 6 C 1 - i.e. 6. Let X: number of red balls selected. x0123 P(X=x) or

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x0123 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) P(X=2) Sigma: Page Ex 8.04 P(selecting 2 red ball) = = = Nbr ways of selecting 1 of the 4 yellow balls = 4 C 1 - i.e. 4. Nbr ways of selecting 2 of the 6 red balls = 6 C 2 - i.e. 15. Let X: number of red balls selected. or

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x0123 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) P(X=3) Sigma: Page Ex 8.04 P(selecting 3 red balls) = = = Nbr ways of selecting 0 of the 4 yellow balls = 4 C 0 - i.e. 1. Nbr ways of selecting 3 of the 6 red balls = 6 C 3 - i.e. 20. Let X: number of red balls selected. or

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x0123 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) P(X= x ) Sigma: Page Ex 8.04 P(selecting x red balls) = = answer Nbr ways of selecting (3- x ) of the 4 yellow balls = 4 C (3- x ) Nbr ways of selecting x of the 6 red balls = 6 C x Let X: number of red balls selected.

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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? Sigma: Page Ex 8.04 P(selecting a given pair) = = = Nbr ways of selecting a specific pair (selecting 2 socks from 2) = 2 C 2 - i.e. 1. Total number of ways of selecting any 2 socks from the 6 = 6 C 2 - i.e. 15. Selection only. Order doesnt matter (left sock same as right sock). combinations.

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Sigma: Page 161 Ex 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 doesnt matter. combinations. Total nbr of possible selections of ANY 5 cards = 52 C 5 = Number of possible flushes – i.e. selections of 5 of same suit:

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Sigma: Page 161 Ex 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 doesnt matter. combinations. Total nbr of possible selections of ANY 5 cards = 52 C 5 = Number of possible flushes – i.e. selections of 5 of same suit: = 4 × 13 C 5

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Sigma: Page 161 Ex 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 doesnt matter. combinations. Total nbr of possible selections of ANY 5 cards = 52 C 5 = Number of possible flushes – i.e. selections of 5 of same suit: = 4 × 13 C 5 (i.e. 4 suits – each has 13 cards and were choosing 5 of them) = 5148 possible selections of 5 of same suit.

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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 doesnt matter. combinations. Total nbr of possible selections of ANY 5 cards = 52 C 5 = Number of possible flushes – i.e. selections of 5 of same suit: = 4 × 13 C 5 (i.e. 4 suits – each has 13 cards) = 5148 possible selections of 5 of same suit. P(getting a flush) = = = or (4sf) answer

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Sigma: Page 161 Ex 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: ××××

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Sigma: Page 161 Ex 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: ××××

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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: Number of possible flushes – i.e. ways of getting 5 of same suit: Sigma: Page 161 Ex 8.04 ????? ××××

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Sigma: Page 161 Ex 8.04 ????? 52 possibilities for the first card (could be any card) 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: Number of possible flushes – i.e. ways of getting 5 of same suit: ××××

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Sigma: Page 161 Ex ???? 52 possibilities for the first card (could be any card) 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: Number of possible flushes – i.e. ways of getting 5 of same suit: ××××

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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: Number of possible flushes – i.e. ways of getting 5 of same suit: Sigma: Page 161 Ex ???? ×××× 12 possibilities for the second card (the other 12 from the same suit)

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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: Number of possible flushes – i.e. ways of getting 5 of same suit: Sigma: Page 161 Ex ??? ×××× 12 possibilities for the second card (the other 12 from the same suit)

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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: Number of possible flushes – i.e. ways of getting 5 of same suit: Sigma: Page 161 Ex ??? ×××× 11 possibilities for the 3 rd card (2 used up, so 11 of that suit are remaining)

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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: Number of possible flushes – i.e. ways of getting 5 of same suit: Sigma: Page 161 Ex ?? ×××× 11 possibilities for the 3 rd card (2 used up, so 11 of that suit are remaining)

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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: Number of possible flushes – i.e. ways of getting 5 of same suit: Sigma: Page 161 Ex ?? ×××× 10 possibilities for the 4 th card (3 used up, so 10 of that suit are remaining)

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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: Number of possible flushes – i.e. ways of getting 5 of same suit: Sigma: Page 161 Ex ? ×××× 10 possibilities for the 4 th card (3 used up, so 10 of that suit are remaining)

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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: Number of possible flushes – i.e. ways of getting 5 of same suit: Sigma: Page 161 Ex ? ×××× 9 possibilities for the 5 th card (4 used up, so 9 of that suit are remaining)

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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: Number of possible flushes – i.e. ways of getting 5 of same suit: Sigma: Page 161 Ex ×××× 9 possibilities for the 5 th card (4 used up, so 9 of that suit are remaining)

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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: Number of possible flushes – i.e. ways of getting 5 of same suit: Sigma: Page 161 Ex ×××× 9 possibilities for the 5 th card (4 used up, so 9 of that suit are remaining)

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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: Number of possible flushes – i.e. ways of getting 5 of same suit: Sigma: Page 161 Ex ×××× ××××

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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: ×××× ×××× P(getting a flush) = = = or (4sf) answer

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