Transcript Sigma Ex. 8.04 solutions - perms & combos

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
C112 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
=
413C5
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