Independent and Dependent Events

Date: 08/16/98 at 19:55:35
From: Kerry
Subject: Independent and dependent events

Dr. Math:

How do I find the probability of an independent and dependent event? 
Can you explain independent and dependent events to me?  

Thanks so much.
Kerry

Date: 08/18/98 at 15:28:35
From: Doctor Margaret
Subject: Re: Independent and dependent events

Hi Kerry,

Thanks for writing to Dr. Math. First let's talk a little about the 
sample space. The sample space is the set of all possible outcomes for 
an event or experiment. For example, the sample space of a die (one of 
a pair of dice) is six: S  = {1, 2, 3, 4, 5, 6}. Each number is what 
you would see on each side of the die. The sample space for a coin is 
two: S = {H, T}, H for heads and T for tails.

Now let's see if we can understand the idea of an independent event 
first. Informally speaking, we say that two events A and B are 
independent if when one of them happens, it doesn't affect the other 
one happening or not. Let's use a real life example.

Let's say that you have a coin and a die (one of a pair of dice). You 
want to find the probability of tossing the coin, getting heads one 
time, and then tossing the die and getting a five one time. We'll call 
the coin toss event A. The plain old probablity of tossing a coin and 
getting heads is 1/2. That is:

       The number of favorable outcomes            1
   A = --------------------------------------  =  ---
       Total possible outcomes (Sample space)      2

The probability of getting a five when you toss the die will be event 
B and that is:

       The number of favorable outcomes              1
   B = --------------------------------------   =   ---
       Total possible outcomes (Sample space)        6

Now for the independent part. Does your chance of getting a five when 
you toss the die have anything to do with whether you get heads or 
tails when you toss the coin? It does not. That's why they are 
independent.

The probability of independent events occurring is found by multiplying 
the probablity of the first event occuring by the probability of the 
second event occurring. Generally, it looks like this:

   P(A,B) = P(A) * P(B)

In our example it looks like this:

   P(H,5) = P(A) * P(B) = 1/2 * 1/6 = 1/12

Now for dependent events. A dependent event is one where the outcome of 
the second event is influenced by the outcome of the first event. For 
example, let's say we have a box with 6 marbles: 3 red, 1 blue, 1 green 
and 1 yellow. What's the probability of picking a yellow marble? We 
know that probablity is 1/6. What's the probability of picking a blue 
marble? Can it be 1/6 also? Well, it could be if we put back the first 
marble we picked. But if we don't put back the first marble, our sample 
space will have changed. We started with six marbles, picked one, and 
now we only have five marbles in the sample space, so the probability 
of picking a blue marble is now 1/5. And in such a case we have 
dependent events, because something about the first one changed the 
second one.

The probability of two dependent events occurring, one right after the 
other, is still found by using the same formula:

   P(A,B)  =  P(A) * P(B)

The big difference is that the individual probabilities won't have the 
same sample spaces. So from our example, what is the probability of 
picking a yellow marble and then a blue marble, without putting the 
first marble back?

   P(Yellow) = 1/6

   P(Blue) = 1/5

   P(Y,B) = 1/6  *  1/5  =  1/30 

This is a very different number from what we would get if the events 
were independent, that is if the sample space remained the same because 
we put the first marble we picked back into the box. Then:

   P(Yellow)  =  1/6

   P(Blue) = 1/6

   P(Y,B) = 1/6 * 1/6 = 1/36

So the trick is to figure out ahead of time if the events are 
independent or dependent, and then use the formula:

   P(A,B)  = P(A) * P(B)

I hope this answers your question. Please write back if you need more 
help.

- Doctor Margaret, The Math Forum
Check out our web site! http://mathforum.org/dr.math/