9.7 Probability  (Page 3/18)

 Page 3 / 18

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.
5. Add the probabilities.

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.

12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
sure. what is your question?
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8