# 11.7 Probability  (Page 4/18)

 Page 4 / 18

Two number cubes are rolled. Use the Complement Rule to find the probability that the sum is less than 10.

$\text{\hspace{0.17em}}\frac{5}{6}\text{\hspace{0.17em}}$

## Computing probability using counting theory

Many interesting probability problems involve counting principles, permutations, and combinations. In these problems, we will use permutations and combinations to find the number of elements in events and sample spaces. These problems can be complicated, but they can be made easier by breaking them down into smaller counting problems.

Assume, for example, that a store has 8 cellular phones and that 3 of those are defective. We might want to find the probability that a couple purchasing 2 phones receives 2 phones that are not defective. To solve this problem, we need to calculate all of the ways to select 2 phones that are not defective as well as all of the ways to select 2 phones. There are 5 phones that are not defective, so there are $\text{\hspace{0.17em}}C\left(5,2\right)\text{\hspace{0.17em}}$ ways to select 2 phones that are not defective. There are 8 phones, so there are $\text{\hspace{0.17em}}C\left(8,2\right)\text{\hspace{0.17em}}$ ways to select 2 phones. The probability of selecting 2 phones that are not defective is:

## Computing probability using counting theory

A child randomly selects 5 toys from a bin containing 3 bunnies, 5 dogs, and 6 bears.

1. Find the probability that only bears are chosen.
2. Find the probability that 2 bears and 3 dogs are chosen.
3. Find the probability that at least 2 dogs are chosen.
1. We need to count the number of ways to choose only bears and the total number of possible ways to select 5 toys. There are 6 bears, so there are $\text{\hspace{0.17em}}C\left(6,5\right)\text{\hspace{0.17em}}$ ways to choose 5 bears. There are 14 toys, so there are $\text{\hspace{0.17em}}C\left(14,5\right)\text{\hspace{0.17em}}$ ways to choose any 5 toys.
$\text{\hspace{0.17em}}\frac{C\left(6\text{,}5\right)}{C\left(14\text{,}5\right)}=\frac{6}{2\text{,}002}=\frac{3}{1\text{,}001}\text{\hspace{0.17em}}$
2. We need to count the number of ways to choose 2 bears and 3 dogs and the total number of possible ways to select 5 toys. There are 6 bears, so there are $\text{\hspace{0.17em}}C\left(6,2\right)\text{\hspace{0.17em}}$ ways to choose 2 bears. There are 5 dogs, so there are $\text{\hspace{0.17em}}C\left(5,3\right)\text{\hspace{0.17em}}$ ways to choose 3 dogs. Since we are choosing both bears and dogs at the same time, we will use the Multiplication Principle. There are $\text{\hspace{0.17em}}C\left(6,2\right)\cdot C\left(5,3\right)\text{\hspace{0.17em}}$ ways to choose 2 bears and 3 dogs. We can use this result to find the probability.
$\text{\hspace{0.17em}}\frac{C\left(6\text{,}2\right)C\left(5\text{,}3\right)}{C\left(14\text{,}5\right)}=\frac{15\cdot 10}{2\text{,}002}=\frac{75}{1\text{,}001}\text{\hspace{0.17em}}$
3. It is often easiest to solve “at least” problems using the Complement Rule. We will begin by finding the probability that fewer than 2 dogs are chosen. If less than 2 dogs are chosen, then either no dogs could be chosen, or 1 dog could be chosen.

When no dogs are chosen, all 5 toys come from the 9 toys that are not dogs. There are $\text{\hspace{0.17em}}C\left(9,5\right)\text{\hspace{0.17em}}$ ways to choose toys from the 9 toys that are not dogs. Since there are 14 toys, there are $\text{\hspace{0.17em}}C\left(14,5\right)\text{\hspace{0.17em}}$ ways to choose the 5 toys from all of the toys.

$\text{\hspace{0.17em}}\frac{C\left(9\text{,}5\right)}{C\left(14\text{,}5\right)}=\frac{63}{1\text{,}001}\text{\hspace{0.17em}}$

If there is 1 dog chosen, then 4 toys must come from the 9 toys that are not dogs, and 1 must come from the 5 dogs. Since we are choosing both dogs and other toys at the same time, we will use the Multiplication Principle. There are $\text{\hspace{0.17em}}C\left(5,1\right)\cdot C\left(9,4\right)\text{\hspace{0.17em}}$ ways to choose 1 dog and 1 other toy.

$\text{\hspace{0.17em}}\frac{C\left(5\text{,}1\right)C\left(9\text{,}4\right)}{C\left(14\text{,}5\right)}=\frac{5\cdot 126}{2\text{,}002}=\frac{315}{1\text{,}001}\text{\hspace{0.17em}}$

Because these events would not occur together and are therefore mutually exclusive, we add the probabilities to find the probability that fewer than 2 dogs are chosen.

$\text{\hspace{0.17em}}\frac{63}{1\text{,}001}+\frac{315}{1\text{,}001}=\frac{378}{1\text{,}001}\text{\hspace{0.17em}}$

We then subtract that probability from 1 to find the probability that at least 2 dogs are chosen.

$\text{\hspace{0.17em}}1-\frac{378}{1\text{,}001}=\frac{623}{1\text{,}001}\text{\hspace{0.17em}}$

#### Questions & Answers

a colony of bacteria is growing exponentially doubling in size every 100 minutes. how much minutes will it take for the colony of bacteria to triple in size
Divya Reply
I got 300 minutes. is it right?
Patience
no. should be about 150 minutes.
Jason
It should be 158.5 minutes.
Mr
ok, thanks
Patience
what is the importance knowing the graph of circular functions?
Arabella Reply
can get some help basic precalculus
ismail Reply
What do you need help with?
Andrew
how to convert general to standard form with not perfect trinomial
Camalia Reply
can get some help inverse function
ismail
Rectangle coordinate
Asma Reply
how to find for x
Jhon Reply
it depends on the equation
Robert
whats a domain
mike Reply
The domain of a function is the set of all input on which the function is defined. For example all real numbers are the Domain of any Polynomial function.
Spiro
foci (–7,–17) and (–7,17), the absolute value of the differenceof the distances of any point from the foci is 24.
Churlene Reply
difference between calculus and pre calculus?
Asma Reply
give me an example of a problem so that I can practice answering
Jenefa Reply
x³+y³+z³=42
Robert
dont forget the cube in each variable ;)
Robert
of she solves that, well ... then she has a lot of computational force under her command ....
Walter
what is a function?
CJ Reply
I want to learn about the law of exponent
Quera Reply
explain this
Hinderson Reply
what is functions?
Angel Reply
A mathematical relation such that every input has only one out.
Spiro
yes..it is a relationo of orders pairs of sets one or more input that leads to a exactly one output.
Mubita
Is a rule that assigns to each element X in a set A exactly one element, called F(x), in a set B.
RichieRich
If the plane intersects the cone (either above or below) horizontally, what figure will be created?
Feemark Reply

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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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