Probability in Flipping Coins

Date: 2/25/96 at 19:16:27
From: Dana Kaye
Subject: Probability Question

I am looking for some help with a couple of probability questions.
On Friday we had a substitute teacher who explained probability
to us and gave us some questions.  Since we share textbooks at 
school we can't take the textbooks home, and my parents don't 
have a clue.

I am not looking for you to do my homework for me (Though that 
would be nice), but I would like some instruction on how to go 
about solving these problems.  Here are the problems:

12. Six pennies are flipped.  What is the probability of getting

    a. two heads and four tails?
    b. four heads and two tails?
    c. at least three heads?
    d. at most two tails?

13. A bag contains one red marble and one white marble.  Marbles
    are drawn and replaced in a bag before the next draw.

    a. What is the probability that a red marble will be drawn two
       times in a row?
    b. In ten draws, what is the probability that a red marble will
       be drawn at least five times.

Thanks for your help!
Sincerely,
Alexia Kaye

Date: 2/26/96 at 15:22:46
From: Doctor Syd
Subject: Re: Probability Question

Dear Alexia,

Hello!  I'm glad you wrote for help -- that is what we are here 
for!  

Probability can be pretty cool once you understand the basic 
concepts, so let's figure out how to approach your problems and 
see where we can go from there.  

Okay, so if you are flipping 6 coins, and you want to know the 
probability of a  certain event, the first and most important thing 
you need to know is what the probability that the coin will land 
on heads is, and what the probability that the coin will land on 
tails is.  This is information that should be given to you in the 
problem - otherwise you don't have enough information.  

Usually these problems deal with a "fair coin," that is a coin in 
which there is an equal probability that it will land on head or tail.  
Thus, in the fair coin, the probability that the coin lands on heads 
is 1/2 because on the average, one out of every two times the coin 
lands on heads.  Likewise the probability that the coin will land 
on tails is also 1/2.   

It is important to note that in the real world there really isn't an 
actual coin that is "fair."  Most coins have probabilities that are 
nearly equal to 1/2, but are more like, for instance,  .4899 
probability for heads and .5101 probability for tails.

Let's tackle that first problem.  Suppose you want to know the 
probability that when you flip your six coins you will get the 
sequence:HHHHTT where H=heads and T=tails.  Well, the total 
number of sequences is going to be 2^6 (that's 2 to the sixth power).  
This may be something you have seen before; if not, feel free to 
write back and I'll explain more.  

So, the probability of getting this one sequence is going to be 1/2^6, 
right?  Well, now let's use the same general strategy to solve your 
problem.  How many sequences have 4 heads and 2 tails in them?  
Certainly the sequence above does, and there are others:  HHTTHH 
is an example.  So how can we count them?  Well, it turns out that 
there are "6 choose 2" such sequences.  I'm not sure how much 
combinatorics (combinatorics is a fancy word for the subject of 
math that has to do with counting things like the number of sequences 
of H's and T's such that there are 4 H's and 2 T's!) you have had, 
so if you don't understand some of what I write, please write back 
so I can clear it up.  

Basically, we can think of the place of the tails as a distinguishing 
feature of the sequences with 4 heads and 2 tails.  In other words, 
once we know where the tails are, since we have just 4 other slots 
left, those slots must be filled by heads, and we are okay.  So, in how 
many ways can we put 2 T's into 6 different slots?  The combinatorial 
answer is "6 choose 2" = 6!/ (4!)(2!).  Thus, the number of sequences 
with 4 heads and 2 tails is  6!/ (4!)(2!).  Thus the probability that 
we'll get such a sequence is:  [6!/ (4!)(2!)]/2^6 = 6!/(4!)(2!)(2^6).

Phew!  That was a long answer to a seemingly innocent and simple 
question, eh?!

Your second question is quite similar to the first, so I leave that to 
you. 

The third and fourth questions can be figured out using methods 
similar to the first two problems, but by extending just a bit.  
Remember that it is sometimes easier to figure out the probability 
that a certain event will NOT happen, and then once you know that, 
you can use the fact the probability that an event will happen + the 
probability that an event won't happen  = 1  to figure out the desired 
probability.  

For instance, in problem c, it might be easier to figure out the 
probability that you will not get at least 3 heads.  If you don't have 
at least 3 heads, what does that mean?  It means  you must have 
0, 1, or 2 heads, right?  So, what is the probability that you have 
0 heads?  The probability that you have 1 head?  The probability 
that you have 2 heads?  (These are things you can solve using exactly 
the same method we used to solve your first two problems).  Since the 
events are mutually exclusive (that is, if you have a sequence with 
exactly 1 head, then it is impossible for it to have exactly 2 heads.  
Whoa!  What a thought!  Sometimes that goofy math terminology is 
expressing the simplest of notions!), the probability that you have 
0, 1, or 2 heads is the probability of having exactly 0 heads + 
probability of having exactly 1 head + the probability of having 
exactly 2 heads.  So, from here you can probably get an answer.  

I'm going to leave the marble questions to you, for now.  See if 
you can figure out what to do with them, having done the coin 
problems.  If I've said anything that is unclear, or you have any 
more questions, feel free to write back.

-Doctor Syd,  The Math Forum