8.06 Probabilities with permutations (Enrichment)

Now that we know how to count the number of arrangements (permutations) possible, we can work out the probability of a particular (or set of particular) arrangements occurring.  

Worked example

example 1 

Suppose I have the digits $1,3,4,7,8,9$1,3,4,7,8,9.  

(a) How many $3$3 digit odd numbers can be made?

I'll start with $3$3 boxes, and write a $4$4 in the final position because there are only 4 possibilities for this to make the number an odd number.  

		$4$4

 

After fulfilling this requirement I have $5$5 options for the first position and $4$4 for the second.

$5$5	$4$4	$4$4

 

This means there are $80$80 possible arrangements.  Without the restriction of the number needing to be odd, there would have been $120(P(6,3)=6\times5\times4)$120(P(6,3)=6×5×4) arrangements.  

(b) What is the probability that the $3$3 digit number created was odd?

Do:

$\frac{\text{number of desired outcomes}}{\text{number of total outcomes}}=\frac{80}{120}=66.67%$number of desired outcomesnumber of total outcomes​=80120​=66.67% (or $\frac{2}{3}$23​ if you need to leave it as a fraction). 

 

example 2

$8$8 people enter a room and randomly stand in a line along the back wall.  What is the probability that they stand from tallest to shortest left to right?

Probability requires us to know $2$2 things -> $\frac{\text{number of desired outcomes}}{\text{number of total outcomes}}$number of desired outcomesnumber of total outcomes​

So we need to know 

Number of desired outcomes - which is how many ways can they line up from tallest to shortest, left to right.  Well the answer to this is just $1$1 way.  There is only going to be one arrangement of the tallest to shortest left to right.

Number of total outcomes - is the total possible arrangements that $8$8 people can stand in a line

$P(8,8)=8!=40320$P(8,8)=8!=40320

This means that: 

$\frac{\text{number of desired outcomes}}{\text{number of total outcomes}}=\frac{1}{40320}=0.00248%$number of desired outcomesnumber of total outcomes​=140320​=0.00248%

Now that's a very slim chance of that happening!

 

example 3

If the first $3$3 people are in the right order, what is the probability then that the remaining people randomly stand in height order?

My diagramatic calculation looks like this, 

$1$1	$1$1	$1$1	$5$5	$4$4	$3$3	$2$2	$1$1

 

The first $3$3 spots are decided, only $1$1 person is the tallest, only $1$1 person is the second tallest and only $1$1 is the third tallest.  I have chosen them and sent them in.  This means that there are $5!$5! possible arrangements that the remaining $5$5 people stand in order.  $5!=120$5!=120.

So the probability of all $8$8 now being in height order is 

$\frac{\text{number of desired outcomes}}{\text{number of total outcomes}}=\frac{1}{120}=0.83%$number of desired outcomesnumber of total outcomes​=1120​=0.83%.  This is over $300$300 times more likely.  But still, less than $1%$1% chance of it occurring.  

 

example 4

How many times more likely will this arrangement of tallest to shortest, left to right occur if I send in the $4$4th tallest person (in addition to the first $3$3)?

So this means that I send in the first 4, and we are calculating the odds that the final four randomly assign themselves into the right order.

$1$1	$1$1	$1$1	$1$1	$4$4	$3$3	$2$2	$1$1

 

So total possible arrangements of the final four standing in order are $4!=24$4!=24 ways

This makes the probability of them doing that randomly

$\frac{\text{number of desired outcomes}}{\text{number of total outcomes}}=\frac{1}{24}$number of desired outcomesnumber of total outcomes​=124​

We are asked by what value has this chance increase?

$\frac{\frac{1}{24}}{\frac{1}{120}}=5$124​1120​​=5 So it is $5$5 times more likely, that I get the correct arrangement if I send in $4$4 of the $8$8 people.  The percentage probability has increased from $0.83%$0.83% to $4.17%$4.17%

Practice questions

QUESTION 1

QUESTION 2