The Professor's Response

Hi Professor!

I keep getting this question wrong please help. A bag contains 15 yellow marbles, 10 red marbles, and 5 white marbles. The marbles are identical except for their colors. What is the probability that when three marbles are selected without replacement that one of them is white. How could I know that just one is going to come out to be white????

See the professor's answer below.

Hi Katherine,

What you need to do to solve this problem is to first write out a probability statement that lists all of the ways this can happen. Here is the exact statement from the problem:

P(that when three marbles are selected without replacement that one of them is white) =

The statement above is fine, but we really need to use it now to think about all of the different ways you can get just one white marble out of three. For example, here is one way that this occurs: we could end up taking a yellow marble, then a red marble, and then a white one. That would be YRW. That is one way that this event can happen. If we could list all of these scenarios that have just one white marble, we could then calculate the probability of each of those and add them all together to find the probability we are looking for.

The only problem with that plan is that it would take a very long time. Maybe we can simplify that approach somehow. For example, instead of listing the outcome with the specific colors that resulted, maybe we can use **not white** and **white** only. Then YRW becomes NNW. This will greatly reduce the number of cases to consider. Now, the only outcomes are: NNW, NWN, and WNN. These are considered different because the white marble is drawn from the bag in a different place for each of them (in other words, it's either drawn first, second, or last.)

Finally, we can rewrite the probability statement:

P(that when three marbles are selected without replacement that one of them is white) =P(NNW or NWN or WNN) =

Now, there are two ways to solve this from here. The first way involves simple addition, the second way involves the combination formula... Watch the video for the rest:

Professor McGuckian