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?
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 |
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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 |
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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).
$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!
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 |
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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.
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 |
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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%
The letters of the word SPACE are to be rearranged.
How many different arrangements are possible?
What is the probability that the letter E will be the first letter?
What is the probability that the letters are arranged in alphabetical order?
$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.
How many possible arrangements are there?
What is the probability that the letters will spell the word GRACIOUS?