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Sigma solutions
Permutations and Combinations
Ex. 8.04
Page 161
Sigma: Page 161Ex 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 161Ex 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 161Ex 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 ? ? ?
9 possibilities for the first digit
Sigma: Page 161Ex 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 161Ex 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 ? ?
8 possibilities for the second digit – one used up
Sigma: Page 161Ex 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 161Ex 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 ?
7 possibilities for the third digit – 2 used up
Sigma: Page 161Ex 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 161Ex 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 6
6 possibilities for the third digit – 3 used up
Sigma: Page 161Ex 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
rn
nPr
= 3024 different PINS are possible.
!5
!94
9 P Same as saying: 9 × 8 × 7 × 6
9 8 7 6
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 161Ex 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 161Ex 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)
Sigma: Page 161Ex 8.04
5 ? ? ?
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 161Ex 8.04
5 ? ? ?
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 161Ex 8.04
5 ? ? ?
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 161Ex 8.04
5 ? ? ?
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 161Ex 8.04
5 ? ? ?
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 161Ex 8.04
5 8 ? ?
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 161Ex 8.04
5 8 ? ?
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 161Ex 8.04
5 8 7 ?
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 161Ex 8.04
5 8 7 ?
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 161Ex 8.04
5 8 7 6
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 161Ex 8.04
5 8 7 6
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
Sigma: Page 161Ex 8.04
5 8 7 6
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
Sigma: Page 161Ex 8.04
5 8 7 6
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.
Sigma: Page 161Ex 8.04
5 8 7 6
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 × 8P3
= 1680
Sigma: Page 161 - Ex 8.04
Total number poss. PIN numbers (from a) = 9P4
= 3024
P(1st digit is odd) = numbers PIN poss.number Total
digit oddan with starting PINS poss.Nbr
3024
1680 =
9
5 = answer
Sigma: Page 161 - Ex 8.042. 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 8.042. 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 8.042. 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 8.042. 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 8.042. 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 8.042. 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 10 ? ?
3 digits from 0 to 9
Sigma: Page 161 - Ex 8.042. 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 10 10 ?
3 digits from 0 to 9
Sigma: Page 161 - Ex 8.042. 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 10 10 10
3 digits from 0 to 9
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.
= 26 000 different ‘numbers’ possible.
Sigma: Page 161 - Ex 8.04
Letter
26 10 10 10
3 digits from 0 to 9
So number of possible ‘numbers’ = 26 × 10 × 10 × 10
or 26 × 103
Sigma: Page 161 - Ex 8.042. 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 10 10 ?
2 digits from 0 to 9 Last digit is either 8 or 9
Sigma: Page 161 - Ex 8.042. 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 10 10 2
2 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 × 102 × 2
= 5 200 different ‘numbers’ possible ending in 8 or 9.
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 × 93
= 18 954
Sigma: Page 161 - Ex 8.04
Total number poss. invoice ‘numbers’ (from a) = 26 × 103
= 26 000
P(does not contain a 1) = numbers invoice poss.number Total
1 a containingnot numbers invoice poss.Nbr
26000
18954 =
= 0.729 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.
There is only one way of selecting both brothers: 2C2 - i.e. 1.
Sigma: Page 161 - Ex 8.04
Total number poss. combinations for the 11 = 14C11
= 364
P(selecting both brothers in the 11) = 11 for the nscombinatio poss. ofnumber Total
12 from brothers-non 9 selecting of waysNbr.
1114
912
C
C =
= or 0.6044 (4sf) answer
This is a selection-only problem (note the word ‘chosen’). Order is not important. So we’re dealing with Combinations.
Number of ways of selecting the other 9 from the other 12 = 12C9 - i.e. 220.
91
55
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: 2C0 - i.e. 1.
Sigma: Page 161 - Ex 8.04
Total number poss. combinations for the 11 = 14C11
= 364
P(selecting neither bro. in the 11) = 11 for the nscombinatio poss. ofnumber Total
12 from brothers-non 11 selecting of waysNbr.
1114
1112
C
C =
= or 0.03297 (4sf) answer
Still is a selection-only problem (note the word ‘chosen’). Order is not important. So we’re dealing with Combinations.
Nbr ways of selecting all 11 from the 12 non-brothers = 12C11 - i.e. 12.
91
3
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: 2C1 - i.e. 2 (could pick either bro).
Sigma: Page 161 - Ex 8.04
Total number poss. combinations for the 11 = 14C11
= 364
P(selecting one bro. in the 11) = 11 for the nscombinatio poss. ofnumber Total
bros-non 10 selecting of Ways 1bro selecting of Ways
1114
1012
12
C
CC =
= or 0.3626 (4sf) answer
Still is a selection-only problem (note the word ‘chosen’). Order is not important. So we’re dealing with Combinations.
Nbr ways of selecting 10 from the 12 non-brothers = 12C10 - i.e. 66.
91
33
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 161 - Ex 8.04
P(selecting 0 red balls) = balls 3 of selections poss. ofnumber Total
yellow 3 selecting of Ways red 0 selecting of Ways
310
34
06
C
CC =
=
Nbr ways of selecting 3 of the 4 yellow balls = 4C3 - i.e. 4.
30
1
Nbr ways of selecting 0 of the 6 red balls = 6C0 - i.e. 1.
Let X: number of red balls selected.
x 0 1 2 3
P(X=x)30
1
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 161 - Ex 8.04
P(selecting 1 red ball) = balls 3 of selections poss. ofnumber Total
yellow 2 selecting of Ways red 1 selecting of Ways
310
24
16
C
CC =
=
Nbr ways of selecting 2 of the 4 yellow balls = 4C2 - i.e. 6.
10
3
Nbr ways of selecting 1 of the 6 red balls = 6C1 - i.e. 6.
Let X: number of red balls selected.
x 0 1 2 3
P(X=x)30
1
30
9
30
9
or
x 0 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)
P(X=2)
Sigma: Page 161 - Ex 8.04
P(selecting 2 red ball) = balls 3 of selections poss. ofnumber Total
yellow 1 selecting of Ways red 2 selecting of Ways
310
14
26
C
CC =
=
Nbr ways of selecting 1 of the 4 yellow balls = 4C1 - i.e. 4.
2
1
Nbr ways of selecting 2 of the 6 red balls = 6C2 - i.e. 15.
Let X: number of red balls selected.
30
1
30
15
30
9
or
30
15
x 0 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)
P(X=3)
Sigma: Page 161 - Ex 8.04
P(selecting 3 red balls) = balls 3 of selections poss. ofnumber Total
yellow 0 selecting of Ways red 3 selecting of Ways
310
04
36
C
CC =
=
Nbr ways of selecting 0 of the 4 yellow balls = 4C0 - i.e. 1.
6
1
Nbr ways of selecting 3 of the 6 red balls = 6C3 - i.e. 20.
Let X: number of red balls selected.
30
1
30
5
30
9
or
30
15
30
5
x 0 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)
P(X= x)
Sigma: Page 161 - Ex 8.04
P(selecting x red balls) = balls 3 of selections poss. ofnumber Total
yellow x)-(3 selecting of Ways red x selecting of Ways
310
)3(46
C
CC xx = answer
Nbr ways of selecting (3- x) of the 4 yellow balls = 4C(3-x)
Nbr ways of selecting x of the 6 red balls = 6Cx
Let X: number of red balls selected.
30
130
9
30
15
30
5
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 161 - Ex 8.04
P(selecting a given pair) = 6 from socks 2any selecting of waysofnumber Total
pair specific a selecting of Ways
26
22
C
C =
=
Nbr ways of selecting a specific pair (selecting 2 socks from 2) = 2C2 - i.e. 1.
15
1
Total number of ways of selecting any 2 socks from the 6 = 6C2 - i.e. 15.
Selection only. Order doesn’t matter (left sock same as right sock). combinations.
Sigma: Page 161Ex 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 161Ex 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:
= 4 × 13C5
Sigma: Page 161Ex 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:
= 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. 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:
= 4 × 13C5 (i.e. 4 suits – each has 13 cards)
= 5148 possible selections of 5 of same suit.
P(getting a flush) = cards 5 ANY of selections possible ofNumber
suit same theof 5 of selections possible ofNumber
552
5134
C
C =
= or 0.001981 (4sf) answer 16660
33
Sigma: Page 161Ex 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 161Ex 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× × × ×
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 161Ex 8.04
? ? ? ? ?
52 51 50 49 48× × × ×
Sigma: Page 161Ex 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:
52 51 50 49 48× × × ×
Sigma: Page 161Ex 8.04
52 ? ? ? ?
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:
52 51 50 49 48× × × ×
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 161Ex 8.04
52 ? ? ? ?
52 51 50 49 48× × × ×
12 possibilities for the second card (the other 12 from the 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?
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 161Ex 8.04
52 12 ? ? ?
52 51 50 49 48× × × ×
12 possibilities for the second card (the other 12 from the 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?
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 161Ex 8.04
52 12 ? ? ?
52 51 50 49 48× × × ×
11 possibilities for the 3rd card (2 used up, so 11 of that suit are remaining)
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 161Ex 8.04
52 12 11 ? ?
52 51 50 49 48× × × ×
11 possibilities for the 3rd card (2 used up, so 11 of that suit are remaining)
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 161Ex 8.04
52 12 11 ? ?
52 51 50 49 48× × × ×
10 possibilities for the 4th card (3 used up, so 10 of that suit are remaining)
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 161Ex 8.04
52 12 11 10 ?
52 51 50 49 48× × × ×
10 possibilities for the 4th card (3 used up, so 10 of that suit are remaining)
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 161Ex 8.04
52 12 11 10 ?
52 51 50 49 48× × × ×
9 possibilities for the 5th card (4 used up, so 9 of that suit are remaining)
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 161Ex 8.04
52 12 11 10 9
52 51 50 49 48× × × ×
9 possibilities for the 5th card (4 used up, so 9 of that suit are remaining)
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 161Ex 8.04
52 12 11 10 9
52 51 50 49 48× × × ×
9 possibilities for the 5th card (4 used up, so 9 of that suit are remaining)
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 161Ex 8.04
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) =
cards 5 of tsarrangemen poss. ofnumber Total
flushes'' possible ofNumber
4849505152
910111252
=
= or 0.001981 (4sf) answer 16660
33