NZ Level 8 (NZC) Level 3 (NCEA) [In development]
Probabilities with Permutations
Lesson

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 of occurring.

Remember this example from our last lesson?

PREVIOUS Example

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

If instead I was asked the question, what is the probability that the $3$3 digit number created was odd, then the answer would

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

Here is another example

$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!

Going one step further!

What if I sent in the first $3$3 people 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.

And even one more step further?

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$1241120=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%

Worked Examples

QUESTION 1

The letters of the word SPACE are to be rearranged.

1. How many different arrangements are possible?

2. What is the probability that the letter E will be the first letter?

3. What is the probability that the letters are arranged in alphabetical order?

QUESTION 2

$8$8 cards have different letters written on them. The letters are $A,R,I,O,S,C,G,U$A,R,I,O,S,C,G,U. The cards are shuffled and laid out on a table with the letters face up next to one another.

1. How many possible arrangements are there?

2. What is the probability that the letters will spell the word GRACIOUS?

Outcomes

M8-3

Use permutations and combinations