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### Marbles and Dice

```
Date: 08/11/97 at 17:57:50
From: Connie
Subject: Marbles and dice

1. A jar has three black marbles and five red marbles. What is the
probability that one red and one black marble will be picked
without replacing the marbles?

2. Two ordinary dice are thrown. What is the probability that the sum
will be six or at least one of the two faces will be odd?

Thanks,
Connie
```

```
Date: 08/12/97 at 07:06:18
From: Doctor Anthony
Subject: Re: Marbles and dice

Use the nCr notation, where this symbol stands for the number of ways
of selecting a group of r things from n different things.  For example
7C3 is given by:

7!                                             7 x 6 x 5
7C3 =  ------ = 35     It can also be calculated from  -----------
3! 4!                                            1 x 2 x 3

In this problem we must select 1 red from 5 red and 1 black from 3
black.

This is  5C1 x 3C1 = 5 x 3 = 15

However, the number of ways of selecting any 2 marbles from 8 is given
by

8C2 = (8 x 7)/(1 x 2)  =  28

So the probability of 1 red, 1 black =  15/28

An alternative method for this relatively simple problem is to think
that we could draw red, then black with probability 5/8 x 3/7 = 15/56
or, alternatively, we could draw black then red with probability
3/8 x 5/7, which again gives 15/56.

Total probability = 30/56 = 15/28 as before.

>2. Two ordinary dice are thrown. What is the probability that the sum
>will be six or at least one of the two faces will be odd?

Scores        Probability
--------     ----------------
You get sum = 6 in the following ways   5 + 1       1/6 x 1/6 = 1/36
4 + 2           1/36
3 + 3           1/36
2 + 4           1/36
1 + 5           1/36

So the probability that sum = 6 is  5/36

Note that in three of these cases at least one face is odd, so the
probability that sum = 6  AND at least one face is odd is  3/36

The probability, in general, that if we throw two dice at least one
face will be odd can be calculated from

odd, odd = 1/2 x 1/2 = 1/4
odd, even = 1/4
even, odd = 1/4

Giving a total of 3/4. Alternatively, it is 1 - probability both even

=  1 - 1/4
=  3/4

Finally we use the addition rule in the form

P(A or B) = P(A) + P(B) - P(A and B)

P(sum = 6 or at least 1 face odd) = 5/36 + 3/4 - 3/36

= 29/36

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Probability

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