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Probability Trees


Date: 07/26/98 at 10:12:22
From: charla anderson
Subject: Probabilities

A machine makes bottle caps and packs them automatically into boxes. 
One box in ten has a defective bottle cap.

   a) What is the probability that each of the next four boxes will 
      have a defective bottle cap?

   b) What is the probability that two boxes out of the first four will 
      have a defective bottle cap?

Thanks for your help,
Charla


Date: 07/26/98 at 17:28:32
From: Doctor Ken
Subject: Re: Probabilities

Hi Charla,

These kinds of probability problems can get pretty tricky, but if you 
master them, they can be pretty satisfying. 

One way to approach these problems is to use probability trees. Here's 
how they work:

Let's say the probability of having a sunny day is 0.7, i.e. there's a 
70 percent chance that each day will be sunny (this is a funny weather 
system in which one day's weather has no effect on the next day's 
weather). I can draw a tree like this:

                 start
                 /   \
                /     \
            0.7/       \0.3
              /         \
          sunny        cloudy
           /\             /\
          /  \           /  \
      0.7/    \0.3   0.7/    \0.3
        /      \       /      \
     sunny   cloudy  sunny    cloudy 


Now, what does this diagram mean? Well, we can use it as a tool to find 
probabilities. For instance, what's the probability of having two sunny 
days in a row? We start at the top, choose the sunny path, and choose 
the sunny path again:

                 start
                //   \
               //     \
           0.7//       \0.3
             //         \
          sunny        cloudy
          //\             /\
         //  \           /  \
     0.7//    \0.3   0.7/    \0.3
       //      \       /      \
     sunny   cloudy  sunny    cloudy 

Now we multiply all the numbers we passed along the way: 
0.7 * 0.7 = 0.49. That means there's a 49% chance of ending up at the 
sunny-sunny "leaf" on the tree. If you want to, you can do all the 
leaves' multiplications in advance:

                 start
                 /   \
                /     \
            0.7/       \0.3
              /         \
          sunny        cloudy
           /\             /\
          /  \           /  \
      0.7/    \0.3   0.7/    \0.3
        /      \       /      \
     sunny   cloudy  sunny    cloudy 
     0.49     0.21   0.21      0.09
 
Now, what's the probability of having a sunny day and then a cloudy 
day? Reading the number at the end of the sunny-cloudy path tells us 
that it's 0.21. Piece of cake!

So what's the probability of having at least one sunny day in two days? 
Well, we have to figure out which "leaves" on our tree satisfy the 
condition. Sunny-sunny, sunny-cloudy, and cloudy-sunny do, but cloudy-
cloudy doesn't. So we add up the probabilities of the three leaves: 
0.49 + 0.21 + 0.21 = 0.91, which is the answer.

Now let's see if we can tackle your problem. The probability of any 
box having a defective bottle cap in it is 0.1 (1 divided by 10). So 
we can make a tree. Since your problems ask about sequences of four 
boxes, we'll make it four branches long. I'll use g for good and b 
for bad:

                                start
                               /     \
                           /             \
                    0.9/                     \0.1
                  /                              \
                g                                   b
              /    \                             /    \
         0.9/          \0.1                0.9/          \0.1
         /               \                  /                \
        g                b                 g                 b
       /  \             /  \              /  \              /  \ 
   0.9/    \0.1     0.9/    \0.1      0.9/    \0.1      0.9/    \0.1
     /      \         /      \          /       \         /      \
    g        b       g        b        g         b       g        b
 .9/\0.1 0.9/\0.1 0.9/\0.1 0.9/\0.1 0.9/\0.1 0.9/\0.1 0.9/\0.1 0.9/\.1
  /  \     /  \     /  \     /  \     /  \     /  \     /  \     /  \
 g    b   g    b   g    b   g    b   g    b   g    b   g    b   g    b


You can use the same method as in my example: to find the probability 
that each of the next four boxes will have a defective cap, follow the 
b-b-b-b branch. To find the probability that two of the next four boxes 
will have bad caps, look at all the branches with exactly two g's and 
two b's. I'll let you figure out which branches those are (there are 
six of them). Then add those branches' final numbers together.

Good luck!

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

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