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# Conditional Probability

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Describe the two ways that you can use probabilities to check whether or not two events are independent.

#### Watch This

http://www.youtube.com/watch?v=u03NipAbyYg James Sousa: Conditional Probability

#### Guidance

Consider a high school with 400 students. The two-way table below shows the number of students in each grade who earned different midterm letter grades.

 9th Grade 10th Grade 11th Grade 12th Grade A 20 18 25 18 B 30 35 38 55 C 30 17 20 20 D 15 15 12 4 F 5 15 5 3

Suppose you choose a student at random from the school. Let  $G$ be the event that the student is from ninth grade. Let  $A$ be the event that the student got an $A$ . You can calculate the probabilities of each of these events using the information in the table.

$P(G) &= \frac{20+30+30+15+5}{400}=\frac{100}{400}=\frac{1}{4}=0.25\\P(A) &= \frac{20+18+25+18}{400} =\frac{81}{400}=0.2025$

Now suppose you choose a student at random and they tell you they are in 9th grade. What's the probability that they got an $A$ ? Now, your sample space is only the 100 students in 9th grade. $P(9th \ grade \ student \ got \ an \ A)=\frac{20}{100}=0.20$ . When you have additional information that causes you to restrict the sample space you are considering, you are working with conditional probability.

The conditional probability of event  $A$ given event  $B$ is the probability of event  $A$   occurring given event  $B$ occurred . The notation is $P(A|B)$ , which is read as “the probability of  $A$ given $B$ ”.

With the students example, you were calculating $P(A|G)$  (“the probability that a student got an  $A$ given that they are in 9th grade”). To calculate this probability, you found the number of ninth grade students with an  $A(A \cap G)$ and divided by the number of ninth grade students $(G)$ . You would have gotten the same result had you found $P(A \cap G)$  and divided by $P(G)$ , because the 400's cancel each other out.

$P(A|G)=\frac{20}{100}=\frac{\frac{20}{400}}{\frac{100}{400}}=\frac{P(A \cap G)}{P(G)}$

To calculate a conditional probability, you can always use this formula, generalized below:

$P(A|B)=\frac{P(A \cap B)}{P(B)}$

Example A

Consider the experiment of tossing three coins and recording the sequence of heads and tails. Let  $A$ be the event of getting at least two heads. Let  $B$ be the event of getting three heads.

a) Find $P(A|B)$ .

b) Find $P(B|A)$ .

c) Does $P(A|B)=P(B|A)$ ?

Solution: The sample space for tossing three coins is $S=\{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT\}$ .

a)  $P(A|B)$ is the probability of getting at least two heads given that you have gotten three heads. If you KNOW that you got three heads, then you automatically have gotten at least two heads.  $P(A|B)$ should be 1. Using the formula:

$P(A|B)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{8}}{\frac{1}{8}}=1$

b)  $P(B|A)$ is the probability of getting three heads given that you have gotten at least two heads. Since you know you have gotten at least two heads, the new sample space is $\{HHT,HTH,THH,HHH\}$ . The probability of getting three heads is $\frac{1}{4}$ . Using the formula:

$P(B|A)=\frac{P(B \cap A)}{P(A)}=\frac{\frac{1}{8}}{\frac{4}{8}}=\frac{1}{4}$

c) $P(A|B) \neq P(B|A)$ The order of the letters within the probability statement matters!

Example B

Consider two independent events  $C$ and $D$ . What is $P(C|D)$  in terms of $P(C)$  and $P(D)$ ? What is $P(D|C)$  in terms of $P(C)$  and $P(D)$ ?

Solution: Since the two events are independent,  $P(C \cap D)=P(C) P(D)$ .

$P(C|D) &= \frac{P(C \cap D)}{P(D)}=\frac{P(C)P(D)}{P(D)}=P(C)\\P(D|C) &= \frac{P(D \cap C)}{P(C)}=\frac{P(D)P(C)}{P(C)}=P(D)$

If  $C$ and  $D$ are independent, then whether or not  $D$ has occurred has no effect on the probability of  $C$ occurring and vice versa. This should make sense given the definition of independent events. One way to test if two events  $C$ and  $D$ are independent is to verify that $P(C|D)=P(C)$  and $P(D|C)=P(D)$ .

Example C

You have two coins, one regular coin and one special coin with heads on both sides. You put the two coins in a bag and choose one at random. Let  $S$ be the event that the coin is the special coin with heads on both sides. Let  $H$ be the event that when the coin is tossed it comes up heads.

a) What is $P(S)$ ?

b) What is $P(S|H)$ ?

Solution: a)  $P(S)$ is the probability that you have chosen the special coin. There are two coins in the bag, one of which is the special coin.  $P(S)=\frac{1}{2}$

b) The experiment of choosing a coin and tossing it has four outcomes in its sample space:

1. Regular Coin, Tails
2. Regular Coin, Heads
3. Special Coin, Heads 1st Side
4. Special Coin, Heads 2nd Side

$P(S|H)$  is the probability that you have chosen the special coin given that when you tossed it, it came up heads. In order to compute this probability, you need to know $P(S \cap H)$  and $P(H)$ . There are two outcomes that are special coins and heads, so $P(S \cap H)=\frac{2}{4}=\frac{1}{2}$ . There are three outcomes that are heads, so $P(H)=\frac{3}{4}$ . Now you can compute the conditional probability:

$P(S|H)=\frac{P(S \cap H)}{P(H)}=\frac{\frac{1}{2}}{\frac{3}{4}}=\frac{2}{3}$

Compare the answers to parts a and b. In each case you are calculating the probability that the coin is the special coin; however, in part b you have additional information that supports that it is the special coin . Because you have additional information, the probability that it is the special coin goes up. Note that because $P(S) \neq P(S|H)$ , the two events  $S$ and  $H$ are NOT independent.

Concept Problem Revisited

Two events  $A$ and  $B$ are independent if and only if:

1) $P(A \cap B)=P(A) P(B)$

2) $P(A|B)=P(A)$  and $P(B|A)=P(B)$

#### Vocabulary

An experiment is an occurrence with a result that can be observed.

An outcome of an experiment is one possible result of the experiment.

The sample space for an experiment is the set of all possible outcomes of the experiment.

An event for an experiment is a subset of the sample space containing outcomes that you are interested in (sometimes called favorable outcomes ).

The complement of an event is the event that includes all outcomes in the sample space not in the original event. The symbol for complement is ′.

The union of two events is the event that includes all outcomes that are in either or both of the original events. The symbol for union is $\cup$ .

The intersection of two events is the event that includes all outcomes that are in both of the original events . The symbol for intersection is $\cap$ .

A Venn diagram is a way to visualize sample spaces, events, and outcomes.

The probability of an event is the chance of the event occurring.

Two events are independent if one event occurring does not change the probability of the second event occurring. $P(A \cap B)=P(A) P(B)$ if and only if  $A$ and  $B$ are indendent events. Also, $P(A|B)=P(A)$ and  $P(B|A)=P(B)$ if and only if  $A$ and  $B$ are independent events.

Two events are dependent if one event occurring causes the probability of the second event to go up or down.

Two events are disjoint ( mutually exclusive ) if they do not have any outcomes in common.

The conditional probability of event  $A$ given event  $B$ is the probability of event  $A$ occurring given event  $B$ occurred. The notation is $P(A|B)$ , which is read as “the probability of  $A$ given $B$ ”.

#### Guided Practice

Consider the experiment of rolling a pair of dice. Let  $A$ be the event that the sum of the numbers on the dice is an 8. Let  $B$ be the event that the two numbers on the dice are a 3 and a 5.

1. What is $P(A)$ ? What is $P(B)$ ?
2. What is $P(B|A)$ ?
3. Are events  $A$ and  $B$ independent?

The sample space for this experiment has 36 outcomes:

$S &= \{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),\\& (3,1),(3,2),(3,3),(3,4),(3,5),(3,6), (4,1),(4,2),(4,3),(4,4),(4,5),(4,6), \\& (5,1),(5,2),(5,3),(5,4),(5,5),(5,6), (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$

1. There are 5 pairs of numbers that have a sum of 8, so $P(A)=\frac{5}{36}$ . There are 2 pairs of numbers that are a 3 and a 5, so $P(B)=\frac{2}{36}=\frac{1}{18}$ .
2. $P(B|A)=\frac{P(B \cap A)}{P(A)}=\frac{\frac{2}{36}}{\frac{5}{36}}=\frac{2}{5}$ . The sample space has been restricted to the five outcomes with a sum of 8.
3. The two events are independent if $P(B|A)=P(B)$ . $P(B)=\frac{1}{18}$ , but $P(B|A)=\frac{2}{5}. \ P(B) \neq P(B|A)$ . so the events are not independent.

#### Practice

Consider the experiment of drawing a card from a standard deck. Let  $A$ be the event that the card is a diamond. Let  $B$ be the event that the card is a red card. Let  $D$ be the event that the card is a four.

1. Find $P(A), P(B), P(C), P(D)$ .

2. Find $P(A|B)$  and $P(B|A)$ .

3. Are events  $A$ and  $B$ independent?

4. Find  $P(D|B)$ and $P(B|D)$ .

5. Are events  $B$ and  $D$ independent?

Consider the experiment of flipping three coins and recording the sequence of heads and tails. Let  $A$ be the event that all the coins are the same. Let  $B$ be the event that there is at least one heads. Let  $C$ be the event that the third coin is a tails. Let  $D$ be the event that the first coin is a heads.

6. Find $P(A), P(B), P(C), P(D)$ .

7. Find $P(A|B)$  and $P(B|A)$ .

8. Are events  $A$ and  $B$ independent?

9. Find $P(C|D)$  and $P(D|C)$ .

10. Are events  $C$ and  $D$ independent?

11. Find  $P(A|D)$ and $P(D|A)$ .

12. Are events  $A$ and  $D$ independent?

13. Explain what conditional probability is in your own words.

14. Explain two ways to test whether or not two events are independent. When does it make sense to use one method over the other?

15. Explain where the conditional probability formula $P(A|B)=\frac{P(A \cap B)}{P(B)}$  comes from.