# 11.7 Probability  (Page 4/18)

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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.

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

can you not take the square root of a negative number
Suppose P= {-3,1,3} Q={-3,-2-1} and R= {-2,2,3}.what is the intersection
can I get some pretty basic questions
In what way does set notation relate to function notation
Ama
is precalculus needed to take caculus
It depends on what you already know. Just test yourself with some precalculus questions. If you find them easy, you're good to go.
Spiro
the solution doesn't seem right for this problem
what is the domain of f(x)=x-4/x^2-2x-15 then
x is different from -5&3
Seid
All real x except 5 and - 3
Spiro
how to prroved cos⁴x-sin⁴x= cos²x-sin²x are equal
Don't think that you can.
Elliott
how do you provided cos⁴x-sin⁴x = cos²x-sin²x are equal
What are the question marks for?
Elliott
Someone should please solve it for me Add 2over ×+3 +y-4 over 5 simplify (×+a)with square root of two -×root 2 all over a multiply 1over ×-y{(×-y)(×+y)} over ×y
For the first question, I got (3y-2)/15 Second one, I got Root 2 Third one, I got 1/(y to the fourth power) I dont if it's right cause I can barely understand the question.
Is under distribute property, inverse function, algebra and addition and multiplication function; so is a combined question
Abena
find the equation of the line if m=3, and b=-2
graph the following linear equation using intercepts method. 2x+y=4
Ashley
how
Wargod
what?
John
ok, one moment
UriEl
how do I post your graph for you?
UriEl
it won't let me send an image?
UriEl
also for the first one... y=mx+b so.... y=3x-2
UriEl
y=mx+b you were already given the 'm' and 'b'. so.. y=3x-2
Tommy
Please were did you get y=mx+b from
Abena
y=mx+b is the formula of a straight line. where m = the slope & b = where the line crosses the y-axis. In this case, being that the "m" and "b", are given, all you have to do is plug them into the formula to complete the equation.
Tommy
thanks Tommy
Nimo
0=3x-2 2=3x x=3/2 then . y=3/2X-2 I think
Given
co ordinates for x x=0,(-2,0) x=1,(1,1) x=2,(2,4)
neil
"7"has an open circle and "10"has a filled in circle who can I have a set builder notation
Where do the rays point?
Spiro
x=-b+_Гb2-(4ac) ______________ 2a
I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
so good
abdikarin
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
strategies to form the general term
carlmark
consider r(a+b) = ra + rb. The a and b are the trig identity.
Mike
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
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Thomas
It is the Â that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas By By  By   By By  By Jordon Humphreys  By Nick Swain