Session 4: Probability - Module D: Conditional Probability

1. Discussion: Probability of Multiple Events

How to calculate the probability of multiple events, or a sequence of events, happening?
What are dependent and independent events?

2. Independent Probabilities

Now that we have seen the probability of one event or another, what happens when we look at the probability of one event occurring with another?

First we'll start with independent events.
Two events are independent if the outcome of one does not affect the other.
For example, the outcome of one dice roll does not affect the outcome of the next dice roll; nor does the flip of a coin.

Lets calculate the probability of rolling an even number and then an odd number:
Because the two events (rolling an even number, rolling an odd number) are independent we can can calculate their probabilities as we normally would.
In this case they both have the probability of 1/2.

With that done, we can calculate the probability of both.
We do this by multiplying the two probabilities together (1/2) * (1/2) = 1/4.
Thus the probability of rolling an even number and an odd number is 1/4.

Now you try! Calculate the following probabilities:

3. Dependent Probability: Socks

Now let's try probabilities of dependent events. Two events are dependent if the outcome of one affects the other. Let's consider a bag of socks with 3 green socks and 2 blue socks.

We can see that if I were to take a sock out of the bag, there is a 3/5 probability that I will pull a green sock, and 2/5 that I will pull a blue sock. But what's the probability of pulling a green sock, then pulling a blue sock? Like we said above, the probability of pulling a green sock is 3/5.

After pulling the green sock, now what's the probability of pulling a blue sock? Remember, now there are only 2 green socks and 2 blue socks, making it 4 socks total. Since there are 2 blue socks, and 4 total socks, the probability of pulling a blue sock is 1/2 (reduced from 2/4).

We now know the probabilities of both our events. To calculate them, we do as we did with independent events and multiply the two probabilities to get 3/5 * 1/2 = 3/10.

4. Dependent Probability: Cards

Another example is drawing cards from a deck and not replacing them. If we pull a card out of the deck and keep it, that affects the probability of drawing the rest of the cards, because we have reduced the number of possible outcomes on the next draw! Without replacement, what is the probability of drawing a king, then drawing another king?

First let's calculate the probability of pulling a king. There are 4 kings, and 52 cards, so the probability of drawing the first king is 4/52.

Now we need to calculate the probability of drawing another king. We didn't put the first king back into the deck, so we are down to 51 cards. Additionally, because we have removed the first king, we are also down to only 3 kings. So the probability of pulling the second king is 3/51.

Now that we know the probability of drawing our two cards, the final step is to multiply the probabilities together: 4/52 * 3/51 = 12/2652 = 1/221.

Try out what you've learned on the following events:

5. Conditional Probability: Hobbies

Conditional probability refers to the probability of an event A given that an event B is true. An example of this is what is the probability a car is red, given that it is a race car. It may be that a higher percent of race cars are red than other types of cars.

Lets say there are 100 cars and they are broken up as follows:

. Race Car Mini-Van SUV Truck Total
Red 8 7 0 9 24
Silver 3 3 7 7 20
Black 2 8 5 12 27
Green 0 16 4 9 29
Total 13 34 16 37 100

Now to find the chance of A given B we write:

P(A given B) = P(A and B) / P(B)

In words that is the probability of A given B is equal to the probability of A and B divided by the probability of B.

For a more concrete example lets say we wanted to find the probability of a car being red, if we know it is a race car. Using the table above we see that the probability of a red race car is 8/100 and the probability of a race car is 13/100. So we need:

P(red given race-car) = P(red and race-car) / P(race-car) = 8 / 13 = .615 = 61.5%

Now you try with the following events: