## Using tree diagrams to find conditional probabilities

Those problems that ask you to find the probability of a series of events “without replacement” can be scary because the probabilities of each event keep changing. (These are known as conditional probability problems.) If the number of possible outcomes isn’t too large, you can tame these problems by using a tree diagram to simplify your calculations.

1. For the first event, draw a tree branch for each possible outcome.
2. At the end of each branch, draw a tree branch for each possible outcome of the second event.
3. Continue until you have a column for every event.
4. For every branch on the tree, write down the probability of that event occurring at that location.
5. Then multiply all the branches from first event to last event to find the probability of any one outcome.
6. Add various events together to get the probability of any compound outcome.

Here’s a simple example that shows how this process works. Let’s say you have a candy dish with 10 red candies, 15 green candies and 20 blue candies. You want to know the probability that you draw at least two red candies or at least two blue candies. There are a lot of different possibilities here, but a tree diagram simplifies everything greatly. Start by drawing a tree with every possible outcome (R, G and B in this example). Then from each outcome, draw another tree representing each outcome for the second draw. Repeat for the third draw. Your tree will look like this:

(You can see that this process will get pretty unwieldy if there are too many outcomes or too many events.)

Next, label each branch with the probability for that outcome. Note that the probabilities change depending on which outcomes have already occurred. For our example, the tree would now look like this:

Finally, for each of the branch ends at the right, multiply together all the probabilities leading to that endpoint. For example, the very top branch, which represents RRR, you would multiply 10/45*9/44*8/43 to find the probability of getting a red candy on all three draws. The final table looks like this (to make the table easier to read, we have calculated only those branches that represent at least two reds or at least two blues):

The probability of our desired event is then the sum of all of listed probabilities: 0.5352.

## Calculating the probability of an event by its complement

You will often be asked to calculate the probability of a compound event; that is, an event that contains two or more simple outcomes. For example, if you flip five coins, what is the probability that you get either four or five heads? To solve this problem, you calculate the probability of getting exactly four heads and the probability of getting exactly five heads, then add the numbers together:

$P(x=4 \text{ or } x=5)=P(x=4)+P(x=5)=$

$\displaystyle \binom{5}{4} \! \left ( \dfrac{1}{2} \right )^4 \!\! \left ( \dfrac{1}{2} \right)^1+ \binom{5}{5} \! \left (\dfrac{1}{2} \right )^5 \!\! \left ( \dfrac{1}{2} \right )^0= \dfrac{5}{32}+ \dfrac{1}{32}= \dfrac{6}{32}= \dfrac{3}{16}$

But what if a compound event contains a lot of simple events? For example, if you roll ten dice, what is the probability you get at least two sixes? To solve this the way we did the previous example, we would need to find the probability of getting exactly two sixes, the probability of getting exactly three sixes, and so on, up to the probability of getting exactly 10 sixes. Although the calculations are not difficult, it is very tedious to find nine different probabilities in order to add them all together. For this problem, it is much simpler to calculate the complement of the given event. The complement of “at least two sixes” is “at most one six”. So calculate the probability of getting at most one six when you roll ten dice:

$P(x=0 \text{ or } x=1)=P(x=0)+P(x=1)=$

$\displaystyle \binom{10}{0} \! \left ( \frac{1}{6} \right )^0 \!\! \left ( \frac{5}{6} \right )^{10}+ \binom{10}{1} \! \left ( \frac{1}{6} \right )^1 \!\! \left ( \frac{5}{6} \right )^9=0.1615+0.3230=0.4845$

Because the probability of an event A and its complement (not A) add up to 1, the probability of getting at least two sixes is then $1-0.4845=0.5155.$

Calculating a compound probability by finding the probability of the complement instead is the basis of many problems in statistics. It’s up to you to determine when this is the best approach to solving a problem.

## The addition rule and multiplication rule in probability

Two of the many formulas you learn in statistics are referred to as the Multiplication Rule and the Addition Rule. How do you figure out when to use each one? This is actually a pretty easy one. The multiplication rule is used to determine the probability of two events both happening, one after the other. The key word to look for in a problem is ‘and’.

Example 1: You draw a single card from a standard deck, replace it and draw a second card. What is the probability you draw a seven first AND a heart second? Because you want the probability of both events occurring, you use the multiplication rule.

The addition rule is used to find the probability that either one of two events occurs. The key word here is “or’.

Example 2: You draw a single card from a standard deck, replace it and draw a second card. What is the probability you draw a seven on the first draw OR a heart on the second draw? Because you want the probability of either event occurring, you use the addition rule.

How do the rules work? The multiplication rule comes in two forms. If the two events are independent, that is, if the probability of the second event does not change when the first event occurs, then the formula is simple:

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

If the probability of the second event depends on the first event occurring, the formula is modified slightly to show this:

$P(A \cap B) = P(A) \cdot P(B|A)$

You can see from either formula why this is called the multiplication rule. The addition rule is just slightly more complicated.

$P(A \cup B)=P(A)+P(B)-P(A \cap B)$

You can see from this formula why it is called the addition rule. Let’s solve both problems.

Example 1: The two events are independent, so we use the first version of the rule:

$P(A \cap B) = P(A) \cdot P(B)=\dfrac{4}{52} \cdot \dfrac{13}{52}= \dfrac{1}{52}$

Example 2: (Note that we use the result from Example 1 in solving this problem.)

$P(A \cup B)=P(A)+P(B)-P(A \cap B)=\dfrac{4}{52}+ \dfrac{13}{52}- \dfrac{1}{52}= \dfrac{16}{52}= \dfrac{4}{13}$

## Calculating probabilities as a fraction

The first day you are asked to find probabilities, the questions are relatively simple:

• What is the probability of rolling a 2 or a 3 on a standard die? [Ans: 2/6 = 1/3]
• What is the probability of drawing a five from a standard deck of cards? [Ans: 4/52 = 1/13]

But very quickly, the problems get a lot more complicated:

• What is the probability of flipping a coin ten times and getting exactly 8 heads?
• What is the probability of being dealt a full house (three of a kind plus a pair) in a five card poker hand?
• A class has 14 boys and 12 girls. If the teacher randomly creates a new seating chart, what is the probability that the six students in the front row comprise exactly four boys and two girls?

How do you attack these problems? The best way is to think of probability as a fraction. The numerator counts all the ways the specified event can occur, and the denominator counts all the ways any event can occur: