Conditional+Probability

=Conditional Probability= //How to handle **Dependent Events**// Life is full of random events! You need to get a "feel" for them to be a smart and successful person.

Independent Events
Events can be "[|Independent]", meaning each event is **not affected** by any other events.

Example: Tossing a coin.
Each toss of a coin is a perfect isolated thing. What it did in the past will not affect the current toss. The chance is simply 1-in-2, or 50%, just like ANY toss of the coin. So each toss is an **Independent Event**.

Dependent Events
But events can also be "dependent" ... which means they **can be affected by previous events** ... ===Example: Marbles in a Bag=== 2 blue and 3 red marbles are in a bag. What are the chances of getting a blue marble? The chance is **2 in 5** So the next time: See how the chances change each time? Each event **depends on** what happened in the previous event, and is called **dependent**. That is the kind of thing we will be looking at here.
 * But after taking one out** you change the chances!
 * if you got a **red** marble before, then the chance of a blue marble next is **2 in 4**
 * if you got a **blue** marble before, then the chance of a blue marble next is **1 in 4**

"Replacement" Note: if you had **replaced** the marbles in the bag each time, then the chances would **not** have changed and the events would be [|independent]:
 * **With** Replacement: the events are **Independent** (the chances don't change)
 * **Without** Replacement: the events are **Dependent** (the chances change)

Tree Diagram
A [|Tree Diagram]: is a wonderful way to picture what is going on, so let's build one for our marbles example. There is a 2/5 chance of pulling out a Blue marble, and a 3/5 chance for Red: We can even go one step further and see what happens when we select a second marble: If a blue marble was selected first there is now a 1/4 chance of getting a blue marble and a 3/4 chance of getting a red marble. If a red marble was selected first there is now a 2/4 chance of getting a blue marble and a 2/4 chance of getting a red marble. Now we can answer questions like **"What are the chances of drawing 2 blue marbles?"** Answer: it is a **2/5 chance** followed by a **1/4 chance**: Did you see how we multiplied the chances? And got 1/10 as a result.
 * The chances of drawing 2 blue marbles is 1/10**

Notation
P(A) means "Probability Of Event A" In our marbles example Event A is "get a Blue Marble first" with a probability of 2/5: And Event B is "get a Blue Marble second" ... but for that we have 2 choices: So we have to say **which one we want**, and use the symbol "|" to mean "given": P(B|A) means "Event B **given** Event A" In other words, event A has already happened, now what is the chance of event B? P(B|A) is also called the "Conditional Probability" of B given A. And in our case: So the probability of getting **2 blue marbles** is: And we write it as //"Probability of **event A and event B** equals//
 * We love notation in mathematics!** It means we can then use the power of algebra to play around with the ideas. So here is the notation for probability:
 * P(A) = 2/5**
 * If we got a **Blue Marble first** the chance is now **1/4**
 * If we got a **Red Marble first** the chance is now **2/4**
 * P(B|A) = 1/4**

//the probability of **event A** times the probability of **event B given event A**"//

Let's do the next example using only notation:

Example: Drawing 2 Kings from a Deck
Event A is drawing a King first, and Event B is drawing a King second. For the first card the chance of drawing a King is 4 out of 52 P(A) = 4/52 But after removing a King from the deck the probability of the 2nd card drawn is **less** likely to be a King (only 3 of the 51 cards left are Kings): P(B|A) = 3/51 And so: =(4/52) x (3/51)= 12/2652 = **1/221** So the chance of getting 2 Kings is 1 in 221, or about 0.5%
 * P(A and B) = P(A) x P(B|A)**

Finding Hidden Data
Using Algebra we can also "change the subject" of the formula, like this: And we have another useful formula: //"The probability of **event B given event A** equals//
 * Start with: ||  || P(A and B) = P(A) x P(B|A) ||
 * Swap sides: ||  || P(A) x P(B|A) = P(A and B) ||
 * Divide by P(A): ||  || P(B|A) = P(A and B) / P(A) ||

//the probability of **event A and event B** divided by the probability of **event A**//

Example: Ice Cream
70% of your friends like Chocolate, and 35% like Chocolate AND like Strawberry. What percent of those who like Chocolate also like Strawberry? P(Strawberry|Chocolate) = P(Chocolate and Strawberry) / P(Chocolate) 0.35 / 0.7 = 50% 50% of your friends who like Chocolate also like Strawberry

Big Example: Soccer Game
You are off to soccer, and want to be the Goalkeeper, but that depends who is the Coach today: Sam is Coach more often ... about 6 out of every 10 games (a probability of **0.6**). So, what is the probability you will be a Goalkeeper today?
 * with Coach Sam the probability of being Goalkeeper is **0.5**
 * with Coach Alex the probability of being Goalkeeper is **0.3**

Let's build a [|tree diagram]. First we show the two possible coaches: Sam or Alex: The probability of getting Sam is 0.6, so the probability of Alex must be 0.4 (together the probability is 1) Now, if you get Sam, there is 0.5 probability of being Goalie (and 0.5 of not being Goalie): If you get Alex, there is 0.3 probability of being Goalie (and 0.7 not): The tree diagram is complete, now let's calculate the overall probabilities. Remember that: P(A and B) = P(A) x P(B|A) Here is how to do it for the "Sam, Yes" branch: (When we take the 0.6 chance of Sam being coach and include the 0.5 chance that Sam will let you be Goalkeeper we end up with an 0.3 chance.) But we are not done yet! We haven't included Alex as Coach: An 0.4 chance of Alex as Coach, followed by an 0.3 chance gives 0.12 0.3 + 0.12 = **0.42 probability** of being a Goalkeeper today (That is a 42% chance)
 * And the two "Yes" branches of the tree together make:**

Check
One final step: complete the calculations and make sure they add to 1: 0.3 + 0.3 + 0.12 + 0.28 = 1 Yes, they add to **1**, so that looks right.

Friends and Random Numbers
Here is another quite different example of Conditional Probability. Let's add our friends one at a time ...
 * 4 friends** (Alex, Blake, Chris and Dusty) each choose a random number between 1 and 5. What is the chance that any of them chose the same number?

First, what is the chance that Alex and Blake have the same number?
Blake compares his number to Alex's number. There is a 1 in 5 chance of a match. As a [|tree diagram]: Note: "Yes" and "No" together makes 1

(1/5 + 4/5 = 5/5 = 1)

Now, let's include Chris ...
But there are now two cases to consider: And we get this: For the top line (Alex and Billy **did** match) we already have a match (a chance of 1/5). But for the "Alex and Billy **did not** match" there is now a **2/5** chance of Chris matching (because Chris gets to match his number against both Alex and Billy). And we can work out the combined chance by **multiplying the chances** it took to get there: Following the "No, Yes" path ... there is a 4/5 chance of No, followed by a 2/5 chance of Yes: (4/5) × (2/5) = 8/25 Following the "No, No" path ... there is a 4/5 chance of No, followed by a 3/5 chance of No: (4/5) × (3/5) = 12/25 Also notice that when you add all chances together you still get 1 (a good check that we haven't made a mistake): (5/25) + (8/25) + (12/25) = 25/25 = 1
 * If Alex and Billy **did** match, then Chris has only **one number** to compare to.
 * But if Alex and Billy **did not** match then Chris has **two numbers** to compare to.

Now what happens when we include Dusty?
It is the same idea, just more of it: OK, that is all 4 friends, and the "Yes" chances together make 101/125: Answer: **101/125**

But notice something interesting ... if we had followed the "No" path we could have **skipped all the other calculations** and made our life easier: The chances of **not matching** are: (4/5) × (3/5) × (2/5) = **24/125** So the chances of **matching** are: 1 - (24/125) = **101/125** (And we didn't really need a tree diagram for that!)

And that is a popular trick in probability: It is often easier to work out the "No" case (This idea is shown in more detail at [|Shared Birthdays] .)

=Reference=

@http://www.mathsisfun.com