# 2.6 - Independent Events

2.6 - Independent Events

To introduce the idea of independence let's look at a table similar to the one on the previous page. The two-way table below displays the World Campus enrollment predictions for a future semester in terms of level (undergraduate and graduate) and biological sex.

Female Male Total
Total 6600 5400 12000

Choose one student from this future semester, what is the probability that the student is a female?

$P(F)=\dfrac{6600}{12000}=.55$

If it is known that the student is a graduate student, what is the probability that the student is a female?

We now are interested in only graduate students who are female (2750). The total is just the number of graduate students (5000).

$P(F|G)=\dfrac{2750}{5000}=.55$

In this example, the marginal probability is the same as the conditional probability. This will be reflective of events that are independent. In other words knowing the selected student is a graduate student does not give us any additional information about the gender of the student.

We can see that in some situations, the conditional and marginal probabilities are different and for some situations, the two probabilities may be equal. When they are equal, the events are independent events, when they are not equal, they are not independent events.

Unless one is explicitly told that events are independent, one cannot simply assume that they are. However, there are some events for which we naturally assume such independence exists (e.g. a flip of a fair coin or the roll of a fair die). For these examples, one expects that the probability of an outcome of one trial does not affect the probability of subsequent trial outcomes. That is, the probability a tail comes up on the first flip of a coin does not change the probability that a tail comes up on the second, third, fourth, etc. flip of that coin.

We can check for independence of two events by showing that any ONE of the following is true. For any given probabilities for events A and B, the events are independent if:

1. $P(A \cap B)=P(A)\cdot P(B)$
2. $P(A|B)=P(A)$ or,
3. $P(B|A)=P(B)$
Independent Events

Two events, A and B, are considered independent events if the probability of A occurring is not changed based on any knowledge of the outcome of B.

Dependent Events

Two events are not independent, or dependent if the knowledge of the outcome of B changes the probability of A.

## Example 2-7

The probability of getting a tail on any one flip is $\frac{1}{2}$. The probability of getting tails on both flips, (T,T), is $\frac{1}{4}$. Let event A be the outcome of getting a tail on flip one and event B getting a tail on flip two. Find the probability of $P(A \cap B)$

This gives us $P(A)=\frac{1}{2}$ and $P(B)=\frac{1}{2}$.  The probability of getting a tail on both flips can be presented by $P(A \cap B)=\frac{1}{4}$. We can verify that the two events are independent by checking if $P(A \cap B)=P(A)\cdot P(B)$. We see this is true by $P(A)=\frac{1}{4}+\frac{1}{4}$. Therefore the two events, getting a tail on the first flip and getting a tail on the second flip are independent.

## Example 2-8

You apply to grad schools at Harvard (H) and PSU (P).  Probability accepted into Harvard is 0.3. Probability accepted into PSU is 0.6. The probability of being accepted to both is 0.25. Noting the probabilities:

$P(H) = 0.3$,
$P(P) = 0.6$, and
$P(H \cap P)=.25$
Are events getting accepted into Harvard and into Penn State independent events?

Before you jump into an automatic assumption of "of course they are!" be sure you verify one of the above expressions hold true!
$P(H)\cdot P(P)=0.3(0.6)=0.18$ which does not equal 0.25 which is $P(H \cap P)$. Therefore they are dependent!

## Example 2-9

Given the following table, determine whether employment status and gender are independent.

Employment Status and Gender

Employed Unemployed Total
Male 460 40 500
Female 140 260 400
Total 600 300 900

Let A denote the event of being employed and B denote the event of being a male.

$P(A \cap B) =P(employed\ and\ male)=\frac{460}{900}=0.5111$

$P(A) = P(employed) = \frac{600}{900} = \frac{2}{3}$

$P(B) = P(male) = \frac{500}{900}=\frac{5}{9}$

Since $P(A)\cdot P(B) =\frac{2}{3}\cdot \frac{5}{9} = 0.370370370$ is not the same as $P(A \cap B)=0.5111$ Thus, $P(A \cap B)$≠$P(A)\cdot P(B)$ and we can conclude that employment status and gender are not independent for the data provided.

#### Why is this useful?

If the data are not the whole population but represent a random sample from a certain target population, and we want to draw inference about whether the employment status and gender are related for the population, then we know that the sample may not follow the exact relationship $P(A \cap B)=P(A)\cdot P(B)$ even if such a relationship holds for the population due to sampling variability. When we talk about the inference about the independence of a two-way table, we will provide the rationale for the chi-square test for independence which measures how far off the observations are from being independent. This will be discussed in greater detail in a future lesson.

## Independent vs. Mutually Exclusive

Students often confuse independent events and mutually exclusive events. The two terms mean very different things. Recall that if two events are mutually exclusive, they have no elements in common and thus cannot both happen at the same time. In fact, mutually exclusive events are dependent. If A and B are mutually exclusive events, then...

$P(A \cap B)=0 \neq P(A)P(B)$

## Try It! Independent Events

Exxon will open a new gas station at a busy intersection in State College next year and feels that the probability that it will show a profit in its first year is 0.6 taking into consideration that Mobil may or may not open a gas station opposite to it. We know that if Mobil also opens a gas station opposite to it, the probability of a first year profit of the Exxon station will be 0.3 and the probability that Mobil will open the station opposite to it is 0.4.

1. What is the probability that both Exxon shows a profit in the first year and Mobil opens a gas station opposite to it? Let A denote the event that Exxon shows a profit in the first year and B denote the event that Mobil opens a gas station opposite to it.

Given: $P(A) = 0.6, P(B) = 0.4, P(A\mid B)=0.3$

Using the intersection formula for dependent events: $P(A \cap B)=P(B) P(A|B)=0.4\cdot0.3=0.12$

The probability that both Exxon and Mobil will make a profit the first year and Mobil will open the station opposite is .12.

2. What is the probability that either Exxon shows a profit in the first year or Mobil opens a gas station opposite to it?
For an 'or' use the formula for the Union of two dependent events: $P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.6+0.4-0.12=0.88$

1. From an urn with 6 red balls and 4 blue balls, two balls are picked randomly without replacement. Find the probability that both balls picked are red.

Let R be the event of picking a red ball and B for picking a blue ball.

The $P(R)=\dfrac{6}{10}$

The $P(B)=\dfrac{4}{10}$

We are interested in finding $P(both\ balls\ picked \ are\ red) = P(first\ ball\ red)\cap P(second\ ball\ red)$ or $P(R\cap R)$

Since we are choosing without replacement these events are dependent.

$P(R \cap R)=P(R) P(R|R)$

$P(R \cap R)=\dfrac{6}{10}\cdot \dfrac{5}{9}=\frac{1}{3}$

2. Let $A$ and $B$ be the following events:

$A$ = get a job

$B$ = buy a new car

It is a given that $P(A) = 0.9$, $P(B) = 0.7$. What is the probability of double happiness: that you get a job and buy a new car? In other words, we want to find $P(A \cap B)$

There is not yet enough information to answer the question.

First, we will ask whether A, B are independent. In this case, the simplistic approach of saying that the two events are independent is not realistic. Thus, we will think harder and try to assess either $P(A|B)$ or $P(B|A)$? Thinking about it, it is not hard to assess the probability of buying a new car knowing that he/she gets a job. For example, if one thinks that $P(B|A) = 0.75$ (this probability is subjectively chosen and may be different for different individuals), the person happens to think that the chance to buy a new car knowing that he/she gets a job is 75%.

$P(A\cap B)=P(A)P(B|A)=0.9\cdot0.75 = 0.675$

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