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Probabilities with Combinations

To work out the probability of something occurring we consider the two values:  the number of ways the thing you want can happen and the total number of ways it could happen

Probabilities can be represented as a fraction

$\text{Probability of Something Occurring}=\frac{\text{What you want}}{\text{Total possible}}$Probability of Something Occurring=What you wantTotal possible

Or as a percentage

$\text{Probability of Something Occurring}=\frac{\text{What you want}}{\text{Total possible}}\times100%$Probability of Something Occurring=What you wantTotal possible×100%


We still approach the calculations of the combinations the same, but then we have the added step of calculating the probabilities. 




A magazine editor is deciding which of $5$5 articles to print as the cover story on the front page.  

If only $2$2 stories can be chosen, how many different selections are possible?

First, we confirm that it is a combination, not a permutation in that the order DOESN'T matter.  So we know now we need to use the combination formula.  We check the question looking for restrictions, but there isn't any.  


Where $n$n is the total number of articles $n=5$n=5 and $r$r is the number we are choosing, $r=2$r=2






So there are $10$10 combinations.


If you were the writers of one of the stories, what is the probability that your article is chosen for the front page?

If your article is chosen, then it is definitely one of the two. So there are now 4C1 possible combinations of your article (and one other) being on the front cover.  

Total combinations of your article appearing = $C(4,1)=4$C(4,1)=4

Thus the probability becomes, 

$\text{Probability of Your Article}=\frac{\text{What you want}}{\text{Total possible}}\times100%$Probability of Your Article=What you wantTotal possible×100%

$\text{Probability of Your Article}=\frac{4}{10}\times100%=40%$Probability of Your Article=410×100%=40%

Or you could write it as $\frac{2}{5}$25 to leave the probability as a fraction

Worked Examples


A box contains 6 pens of different colours: red, green, blue, yellow, black and white. Two pens are drawn at random without replacement.

  1. How many possible selections are there?

  2. What is the probability of drawing the green and black pens?


A menu has three entrées ($E_1,E_2,E_3$E1,E2,E3), four mains ($M_1,M_2,M_3,M_4$M1,M2,M3,M4) and two desserts ($D_1,D_2$D1,D2). A meal is made up of one of each.

  1. How many different meals are possible?

  2. What is the probability of selecting $E_1$E1 , $M_3$M3 and $D_2$D2?

  3. How many different meals are possible given that $E_1$E1 is the entrée?


$5$5 people are to be selected from a larger group of $10$10 candidates. If Amelia is among the candidates, what is the probability that she will be among those selected?



Use permutations and combinations

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