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### Probability: Marbles, Coins, Cards

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Date: 05/24/97 at 00:00:44
From: Jean-Pierre Berard
Subject: Help wanted

Hello,

I want to write a little program to calculate permutations,
combinations, variations with repetitions, and so on.  I thought that
writing the formulas down would make the reasoning come clear.

Right away, I was confronted with my own shortcomings in this area of
math! In an effort to overcome this, could you please explain the
following problems to me?

1) A bag contains 10 marbles, which are identical except for their
color. There are 4 blue marbles and 6 red marbles. If marbles are
drawn at random from the bag, what is the probability of drawing the
following colored marbles in the order indicated: red, blue, blue,
red.

2) The directors of a company can not decide who should be chairman
and minute secretary. If there are 20 directors.

a) How many different selections are possible?.
b) What is the probability that any two given directors will become
chairman and minute secretary?.
c) What is the probability that any two given directors will become
chairman or minute secretary?.

3) A card player draws 5 cards from a full deck of 52 cards.

a) What is the probability that their hand of cards contains an ace of
spades, a 6 of hearts, a 5 of clubs, a king of hearts and a 9 of
diamonds?
b) What is the probability that their hand of cards contains an ace of
spades, a 6 of hearts, a 5 of clubs, a king of hearts and any
diamond?

4) In a coin tossing experiment, an unbiased coin is flipped 8 times.
What is the probability that 5 heads will result?.

Many thanks,
Jean-Pierre Berard
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Date: 05/24/97 at 07:55:49
From: Doctor Mitteldorf
Subject: Re: Help wanted

Dear Jean-Pierre,

I'm not sure how you want to program these.  One very powerful and
general method is called "Monte Carlo".  You use the computer's number
crunching capability, and you just simulate the entire experiment
perhaps a million times, keeping track of how many times the computer
achieves a successful outcome by whatever criterion you have.  This
method is a bit inelegant and slow, but useful in the many
circumstances when there is no obvious formula or reasoning that will
make the problem simpler.  The answer is approximate, and gets better
the more trials you simulate. You can do a Monte Carlo simulation of
any of the 4 problems you've described, but they are also simple
enough that they can all be done with analytic methods.   I don't
think you need a computer.

1) The probability of drawing the first red is 4/10, which leaves 3
red and 6 blue. The probability of drawing the second blue is 6/9,
which leaves 3 red and 5 blue. The probability of drawing the third
blue is 5/8 which leaves 3 red and 4 blue. The probability of drawing
the fourth red is 3/7 (tt doesn't matter what this leaves.)

The probability of the entire sequence is just the product of the
probability of each individual drawing:

4/10 * 6/9 * 5/8 * 3/7 = 360/5040 = 1/14

2a) There are 20 possibilities for chairman, and having chosen
chairman there are 19 remaining possibilities for secretary.

b) The probability of any given combination of two people for chair
and secretary is 1/380.

c) I'm not sure what you mean by this question.  Do you mean the
probability that either Andrew will be chairman OR Bill will be
secretary?  Out of the 380 possibilities, there are 19 in which Andrew
is chair, and 19 in which Bill is secretary, but these two sets
overlap by 1, in which both conditions are met.  The probability that
either condition is met is thus 37/380.

3a) The probability of drawing these 5 cards in order is 1 in
(52*51*50*49*48). But they don't have to be drawn in that particular
order.  How many different orders are there?  This is the number of
ways to order 5 cards, which is 5! or 5*4*3*2*1.So the probability of
having the given hand is:

(5*4*3*2)/(52*51*50*49*48)

This is just the inverse of the number called "52 choose 5" and
sometimes written as a large pair of parentheses enclosing a 52 and a
5 directly underneath it.  The formula for "n choose r" is

n!/((n-r)!*r!)

3b) This proability is just 13 times as large as (a).

4) If you count as a separate case each sequence (e.g. head-tail-head-
for 8 tosses is 2^8.  This comes from the fact that each separate toss
has 2 outcomes, and there are 8 tosses.

You want to know, of these 2^8, how many have 5 heads and 3 tails?
The answer is "8 choose 5" or (8*7*6) / (3*2*1). This is because there
are 8! ways of arranging 8 coins if they are all different, but
they're not all different:  5 of them are heads and indistinguishable,
so you've overcounted by the orderings of 5 things, a factor of 5!.
Also, 3 of them are tails and indistinguishable, so you've overcounted
by a factor 3!

The probability you want is "8 choose 5" divided by 2^8.

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

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