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/