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If we have 10 books labeled from a to j. what is the probability that a,b and c are...

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svjr | Student, Undergraduate | Honors

Posted January 25, 2012 at 3:56 PM via web

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If we have 10 books labeled from a to j. what is the probability that a,b and c are chosed at random?

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txmedteach | High School Teacher | (Level 3) Associate Educator

Posted January 25, 2012 at 4:40 PM (Answer #1)

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I just want to clarify that we get four different answers based on whether or not the order of selection of the books matters and whether or not the books are replaced after being chosen.

Let's suppose that the order does not matter and that the books are not replaced.

We are now looking at the probability that three things happen in sequence:

1) One of the 3 books (a,b,c) is chosen from the 10 books on the shelf P = 3/10

2) one of the other 2 books is chosen from the 9 remaining books on the shelf. P = 2/9

3) the last book is chosen from the 8 remaining books on the shelf. P = 1/8

This set of events requires that we multiply each probability to find the final probability (let's call `P_1` because this is our first case:

`P_(1) = 3/10*2/9*1/8 = 1/120`

So, our probability given the first situation is 1/120.

Let's suppose now, that order does matter, and we need to select A, then B, then C. Now we're looking at 3 other events:

1) A is chosen out of the first 10 books. P = 1/10

2) B is chosen out of the remaining 9 books. P = 1/9

3) C is chosen out of the remaining 8 books. P = 1/8

Our total probability (`P_2`) for this case is then:

`P_2 = 1/10*1/9*1/8 = 1/720`

Notice how much lower this is than in our first case! Incidentally, if we multiply by the number of ways we can order the 3 books (3*2*1 = 6 ways of arranging A,B, and C) then we get the probability `P_1` of the case where we didn't care about order.

Now, let's move on to the third case, where we replace books, but it doesn't matter in which order they are selected. Now, we look at a new set of sequential probabilities:

1) One of the 3 books (a,b,c) is selected out of 10. P = 3/10

2) Book replaced, and one of the remaining 2 books is selected from the 10. P = 2/10

3) Book replaced and the last remaining book is selected from the 10. P = 1/10

Now, we do the same as we did before. We multiply the probabilities to see what happens in this case (We'll call this probability `P_3`:

`P_3 = 3/10*2/10*1/10 = 3/500`

Like the 2nd overall case, this probability is lower than `P_1` because now, after each step, we are putting what is effectively a new "wrong" book into the shelf. As expected, this makes it harder to pick the 3 books randomly.

Finally, let's calculate the probability of the case in which the order of the books matters AND we replace the books after each step. Let's write out the sequence of events:

1) A is chosen out of 10 books. P = 1/10

2) A is replaced and B is chosen out of 10 books. P = 1/10

3) B is replaced and C is chosen out of 10 books. P = 1/10

As you might expect by now, we'll call the overall probability for this case `P_4` and we'll calculate it by multiplying the individual probabilities of each step in the sequence:

`P_4 = 1/10*1/10*1/10 = 1/1000`

It's clear to see that this is the lowest probability of the 4 overall proabilities we calculated. Again, this makes sense because we are restricting our "correct" choices AND we are including more "wrong" choices in the 2nd and 3rd step.

The way the question is asked, though, leads me to think that it is looking for the `P_1` that we calculated. So, we'll say the probability of the mentioned event occuring is 1/120.

Hope that helps!

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