# 11.7 Probability  (Page 3/18)

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A card is drawn from a standard deck. Find the probability of drawing a red card or an ace.

$\text{\hspace{0.17em}}\frac{7}{13}\text{\hspace{0.17em}}$

## Computing the probability of mutually exclusive events

Suppose the spinner in [link] is spun again, but this time we are interested in the probability of spinning an orange or a $\text{\hspace{0.17em}}d.\text{\hspace{0.17em}}$ There are no sectors that are both orange and contain a $\text{\hspace{0.17em}}d,\text{\hspace{0.17em}}$ so these two events have no outcomes in common. Events are said to be mutually exclusive events    when they have no outcomes in common. Because there is no overlap, there is nothing to subtract, so the general formula is

$\text{\hspace{0.17em}}P\left(E\cup F\right)=P\left(E\right)+P\left(F\right)\text{\hspace{0.17em}}$

Notice that with mutually exclusive events, the intersection of $\text{\hspace{0.17em}}E\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}F\text{\hspace{0.17em}}$ is the empty set. The probability of spinning an orange is $\text{\hspace{0.17em}}\frac{3}{6}=\frac{1}{2}\text{\hspace{0.17em}}$ and the probability of spinning a $d$ is $\text{\hspace{0.17em}}\frac{1}{6}.\text{\hspace{0.17em}}$ We can find the probability of spinning an orange or a $d$ simply by adding the two probabilities.

The probability of spinning an orange or a $d$ is $\text{\hspace{0.17em}}\frac{2}{3}.$

## Probability of the union of mutually exclusive events

The probability of the union of two mutually exclusive events $\text{\hspace{0.17em}}E\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}F\text{\hspace{0.17em}}$ is given by

$\text{\hspace{0.17em}}P\left(E\cup F\right)=P\left(E\right)+P\left(F\right)\text{\hspace{0.17em}}$

Given a set of events, compute the probability of the union of mutually exclusive events.

1. Determine the total number of outcomes for the first event.
2. Find the probability of the first event.
3. Determine the total number of outcomes for the second event.
4. Find the probability of the second event.

## Computing the probability of the union of mutually exclusive events

A card is drawn from a standard deck. Find the probability of drawing a heart or a spade.

The events “drawing a heart” and “drawing a spade” are mutually exclusive because they cannot occur at the same time. The probability of drawing a heart is $\text{\hspace{0.17em}}\frac{1}{4},\text{\hspace{0.17em}}$ and the probability of drawing a spade is also $\text{\hspace{0.17em}}\frac{1}{4},\text{\hspace{0.17em}}$ so the probability of drawing a heart or a spade is

$\text{\hspace{0.17em}}\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\text{\hspace{0.17em}}$

A card is drawn from a standard deck. Find the probability of drawing an ace or a king.

$\text{\hspace{0.17em}}\frac{2}{13}\text{\hspace{0.17em}}$

## Using the complement rule to compute probabilities

We have discussed how to calculate the probability that an event will happen. Sometimes, we are interested in finding the probability that an event will not happen. The complement of an event $\text{\hspace{0.17em}}E,\text{\hspace{0.17em}}$ denoted $\text{\hspace{0.17em}}{E}^{\prime },\text{\hspace{0.17em}}$ is the set of outcomes in the sample space that are not in $\text{\hspace{0.17em}}E.\text{\hspace{0.17em}}$ For example, suppose we are interested in the probability that a horse will lose a race. If event $\text{\hspace{0.17em}}W\text{\hspace{0.17em}}$ is the horse winning the race, then the complement of event $\text{\hspace{0.17em}}W\text{\hspace{0.17em}}$ is the horse losing the race.

To find the probability that the horse loses the race, we need to use the fact that the sum of all probabilities in a probability model must be 1.

$\text{\hspace{0.17em}}P\left({E}^{\prime }\right)=1-P\left(E\right)\text{\hspace{0.17em}}$

The probability of the horse winning added to the probability of the horse losing must be equal to 1. Therefore, if the probability of the horse winning the race is $\text{\hspace{0.17em}}\frac{1}{9},\text{\hspace{0.17em}}$ the probability of the horse losing the race is simply

$\text{\hspace{0.17em}}1-\frac{1}{9}=\frac{8}{9}\text{\hspace{0.17em}}$

## The complement rule

The probability that the complement of an event    will occur is given by

$\text{\hspace{0.17em}}P\left({E}^{\prime }\right)=1-P\left(E\right)\text{\hspace{0.17em}}$

## Using the complement rule to calculate probabilities

Two six-sided number cubes are rolled.

1. Find the probability that the sum of the numbers rolled is less than or equal to 3.
2. Find the probability that the sum of the numbers rolled is greater than 3.

The first step is to identify the sample space, which consists of all the possible outcomes. There are two number cubes, and each number cube has six possible outcomes. Using the Multiplication Principle, we find that there are $6×6,\text{\hspace{0.17em}}$ or total possible outcomes. So, for example, 1-1 represents a 1 rolled on each number cube.

 $\text{1-1}$ $\text{1-2}$ $\text{1-3}$ $\text{1-4}$ $\text{1-5}$ $\text{1-6}$ $\text{2-1}$ $\text{2-2}$ $\text{2-3}$ $\text{}$ $\text{2-4}$ $\text{2-5}$ $\text{2-6}$ $\text{3-1}$ $\text{3-2}$ $\text{3-3}$ $\text{3-4}$ $\text{3-5}$ $\text{3-6}$ $\text{4-1}$ $\text{4-2}$ $\text{4-3}$ $\text{4-4}$ $\text{4-5}$ $\text{4-6}$ $\text{5-1}$ $\text{5-2}$ $\text{5-3}$ $\text{5-4}$ $\text{5-5}$ $\text{5-6}$ $\text{6-1}$ $\text{6-2}$ $\text{6-3}$ $\text{6-4}$ $\text{6-5}$ $\text{6-6}$
1. We need to count the number of ways to roll a sum of 3 or less. These would include the following outcomes: 1-1, 1-2, and 2-1. So there are only three ways to roll a sum of 3 or less. The probability is
$\text{\hspace{0.17em}}\frac{3}{36}=\frac{1}{12}\text{\hspace{0.17em}}$
2. Rather than listing all the possibilities, we can use the Complement Rule. Because we have already found the probability of the complement of this event, we can simply subtract that probability from 1 to find the probability that the sum of the numbers rolled is greater than 3.

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
Â
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