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### Colored and Numbered Discs

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Date: 05/25/2001 at 17:28:24
From: Gerard Morley
Subject: Probability

Hi, I hope you can help me. I am doing my Leaving Cert (state exam in
Ireland).

Question:

Six red discs, numbered from 1 to 6, and 4 green discs, numbered from
7 to 10, are placed in box A.

Ten blue discs, numbered from 1 to 10, are placed in box B.

Two discs are drawn from Box A and two discs are drawn from Box B. The
four discs are drawn at random and without replacement.

Find the probability that the discs drawn are:

(1) one red disc, one green disc and two blue discs with all four
discs odd numbered.

(2) one red disc, one green disc and two blue discs with the total on
the red and green discs equal to 10 and the total on the blue discs
also equal to 10.

Part (1)

P = (3/10 x 2/9 x 5/10 x 4/9) = 2/135

The real solution is 2/135 x 2 = 4/135

But I can't see why they multiply by 2.

Part (2)

P = (1/10 x 1/9)x3 x (1/10 x 1/9)x8 = 2/675

The real solution is (1/10 x 1/9)x3x2 x (1/10 x 1/9)x4x2 = 4/675

Why is the first part of real solution above multiplied by 2, i.e.
(1/10 x 1/9)x3x2?
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Date: 05/25/2001 at 20:15:37
From: Doctor Schwa
Subject: Re: Probability

Hi Gerard,

Thanks for stating the problem so clearly, and making it so easy to
help you by showing us exactly where you get confused.

>(1)
>   P = (3/10 x 2/9 x 5/10 x 4/9) = 2/135
>The real solution is 2/135 x 2 = 4/135
>But I can't see why they multiply by 2.

The reason they multiply by 2 is that your multiplication here can be
read as "and then."

That is, you can choose red odd and then green odd and then blue odd
and then blue odd, *or* green odd and then red odd and then blue odd
and then blue odd.

There are two different ways to do it, each of which has the same
probability.

If that explanation is confusing, think of it this way: there aren't
only three possible choices for the first disc you draw out of the
red/green bin, there are really 5. Any of the odd ones will do. Then
the second disc has to be an odd one of the opposite color.

So, all together, the possibilities are:

1,7
1,9
3,7
3,9
5,7
5,9

7,1
7,3
7,5
9,1
9,3
9,5

You'll see that they split into two groups of 6. Your calculation had
the probability at 6/90 for the first bin, but the real probability is
12/90 when you take into account the order.

Yet a third way of thinking about it is that there are six possible
combinations as you found, but since order doesn't matter there are
only (10 choose 2) = 45 choices which should be in the denominator.
That is, you can find the probability as 6 combinations out of 45, or
as 12 permutations out of 90.

>(2)
>   P = (1/10 x 1/9)x3 x (1/10 x 1/9)x8 = 2/675
>The real solution is (1/10 x 1/9)x3x2 x (1/10 x 1/9)x4x2 = 4/675
>Why is the first part of real solution above multiplied by 2, i.e.
>(1/10 x 1/9)x3x2?

The arguments are exactly the same as above: you can have
red-then-green or green-then-red, which doubles it; or, in other
words, you can have:

1,9
2,8
3,7
7,3
8,2
9,1

which is six choices, not three.

I hope that helps clear things up! These sorts of problems are
conceptually very tricky.

- Doctor Schwa, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 05/26/2001 at 13:18:59
From: Gerard Morley
Subject: Probability

Hi, Gerard here again. Thanks very much for your help on the last
question - it was much appreciated. I have another question:

On an unbiased die, the numbers 1, 3, and 4 are colored red and the
numbers 2, 5, and 6 are colored black.

Q1. The die is thrown three times with the following outcome: the
second throw shows a red number and the sum of the numbers on the
first and second throws is equal to the number on the third throw.

Find the probability of this outcome.

I have no idea how to figure this out! Hope you can help.
```

```
Date: 05/26/2001 at 16:25:58
From: Doctor Schwa
Subject: Re: Probability

Hi Gerard,

I think in this case, with such a strange rule, it would be simplest
to just list all the possible outcomes.

There are 216 possible ways to throw the die three times (6*6*6), and
of those, we have to have a 1, 3, or 4 on the second throw, and the
first plus second has to equal the third, so the winning ones are:

1,1,2
2,1,3
3,1,4
4,1,5
5,1,6

1,3,4
2,3,5
3,3,6

1,4,5
2,4,6

Probably it would have been easier to realize that with a 1 on the
second throw there are 5 choices; with a 3 there are 3 choices; and
with a 4 there are 2 choices. That would have saved actually having to
list all of them.

Thanks for the fun question!

- Doctor Schwa, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Probability
High School Probability

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