This is a question that I am having problems coming to the answer with: A bag co

Question

This is a question that I am having problems coming to the answer with:
A bag contains eight red marbles and 3 blue marbles. If you remove five random marbles what it the probability that one marble is blue and the other four are red?

I have the answer but don't know how to arrive to it. The answer is:
# of possible outcomes: 11 x 10 x 9 x 8 x 7 = 55 440 (but where do these numbers come from?)
# of successful outcomes 3 x 8 x 7 x 6 x 5 = 5040 (and where do these numbers come from?)
Therefore : p(x) = 5040/55 440 = 1/11

Explanation / Answer

These probabilities stem from the multiplication rule. Essentially you have a total of 11 marbles (8 red +3 blue). From these 11 we want to chose 5. Think of this as we have 5 slots to fill. I.e. __ x __ x __ x __ x __ Because we are not allowing repetition, once an item is selected, we remove it from the bag and throw it away, thus, for our first selection, there are 11 marbles we can choose from. Ther for we now have the following diagram: 11 x __ x __ x __ x __ For our next selection we only have 10 marbles to choose from (because we have already selected one and we cannot use it again) Thus our diagram is now 11 x 10 x __ x __ x __ We will continue doing this, until we have selected 5 marbles and our diagram will be 11 x 10 x 9 x 8 x 7 = 55,440 (this represents the total permutations of marbles we can select. For the # of success full outcomes it is similar. we start with the 5 slots to fill: __ x __ x __ x __ x __ However, we are constrained by what we are looking for. We are looking for the probability that only 1 marble is blue. For simplicity, we will place this value into the first slot. Because we only have 3 blue marbles, our # of choices for the first slot is 3. Our diagram looks like this: 3 x __ x __ x __ x __ Now for the second selection we must pay attention to the fact that we have selected one blue marble (it is now removed from the bag), and we only have 2 blue marbles and 8 red marbles remaining. Because we only want to select one blue marble, we must now remove the remaining blue marbles from our possible permutations. This is we take the total # remaining (10) and subtract the remaining blue marbles = 10-2 = 8. This shows us that we have 8 possible selections for the second step, thus our diagram looks like this 3 x 8 x __ x __ x __ As we have selected this marble, we remove it from the bag and we only have 7 marbles remaining, thus we only have 7 possibilities for our third step, and our diagram looks like this: 3x8x7x_x_. We continue this process of selecting a marble and removing it from the bag and we find that our total choices given the contraint of only one blue marble is 3 x 8 x 7 x 6 x 5 = 5040. If you have any other questions, please let me know

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