next up previous contents index
Next: Independence Up: Probability Previous: Relative Frequency

Determination of Probabilities 1: Tree Diagrams

We will discuss three ways of determining probabilities: enumeration (listing of the sample space), tree diagrams, and resampling. Resampling will be discussed in the next chapter and enumeration is what we have been doing. For instance, we solved the probability of a "7" or "11" in the game of craps by listing the sample space and just observing the number of times we get "7" or "11". This gets tedious very quickly and for many problems is quite unrealistic. For example consider the following problem:

Urn Problem : Suppose we have an urn with 30 blue balls and 50 red balls in it and that these balls are identical except for color. Suppose further the balls are well mixed and that we draw 3 balls, without replacement. Determine the probability that the balls are all of the same color.

Even for this simple problem there are 82160 elements in the sample space. But note that this problem is sequential in nature; i.e, there are 3 steps (draw Ball 1, draw Ball 2, and draw Ball 3). In such cases a simple tree diagram  can often solve the problem or even if unfinished lead to the answer. We will draw our trees horizontally. So for this problem, begin by putting a dot on the center left side of your paper. Then for the first ball, it is either blue or red. Beginning at the dot, trace a branch up for blue, putting the probability of the first ball being blue, 30/80, on it and a "B" at the end. Likewise, trace a branch down for red with 50/80 on it and an R at the end. Hence at the end of the first step your tree looks like Figure 2.1.


  
Figure 2.1: Tree diagram : Step 1
\begin{figure}
\begin{center}
\epsfig{file=fig13.ps, height=5in, width=5in, angle= -90}\end{center}\end{figure}

Next do the second step at each of the ends of the first step. The second ball is either blue or red. That is, at the "B", draw one branch up for second ball blue with the probability of 29/79 on it (on this branch, you have already drawn one of the blue balls, so there are 29 blue balls left out of 79 balls), and end it with a "B". Next draw one branch down for second ball red with the probability 50/79 (a blue ball was drawn on the first step so there are 50 red balls left out of 79 balls) and end it with an "R".

Now you try the second step at the "R" of the first step. If you have done it right, your tree diagram at the end of the second step should look like Figure 2.2.


  
Figure 2.2: Tree diagram : Step 2
\begin{figure}
\begin{center}
\epsfig{file=fig14.ps, height=5in, width=5in, angle= -90}\end{center}\end{figure}

Hey this is easy stuff! Now you try the third step. The ball can be blue or red so there will be two branches at the end of the four second step branches. If you have done it right, your tree diagram at the end of the third and last step should look like:


  
Figure 2.3: Tree diagram : Step 3
\begin{figure}
\begin{center}
\epsfig{file=fig15.ps, height=5in, width=5in, angle= -90}\end{center}\end{figure}

Look at the node of the final "B", "B", "B" branch. This means blue ball on first step, blue ball on second step, and blue ball on third step. What's the probability of this? It's easy. 30 out of 80 times you go up to the first "B", and of those times 29 out of 79 times you go up to the second "B", and of those times 28 out of 78 times you go up to the third "B". The key word, here, is of. That is 28/78 of 29/79 of 30/80 times you get to the "B", "B", "B" node. Hence the probability of three blue balls is:

\begin{displaymath}\frac{28}{78} \times \frac{29}{79} \times \frac{30}{80} = .0494.
\end{displaymath}

Likewise the probability of three reds is (follow the bottom most branch to its final node) is

\begin{displaymath}\frac{48}{78} \times \frac{49}{79} \times \frac{40}{80} = .2386.
\end{displaymath}

Finally, the probability that the balls are the same color is the probability of either 3 reds or 3 blues. These events cannot happen at the same time (in fact all last 8 nodes are disjoint events); hence, the probability that the balls are the same color is .0494 + .2386 = .2880. That is, almost 30% of the time you will draw three balls out of the urn which are the same color.


Exercise 3.5.1  
1.
In the urn example, find the probability of getting 2 redballs and one blue ball.
2.
In the urn example, find the probability of getting the first two balls red.
3.
In the urn example, suppose we darw 4 balls without replacement. Now find the probability that all 4 balls are of the same color.
4.
Use a tree diagram to determine the probability of getting three red cards, when three cards are dealt (without replacement) from a well shuffled standard deck of 52 cards.
5.
Six cards with the numbers 1 through 6 on them are well shuffled and two cards are dealt (without replacment). Use a tree diagram to determine the probability of that the sum of the numbers on the two cards is 7.
6.
In the urn example, find the probability of getting 2 redballs and one blue ball.


next up previous contents index
Next: Independence Up: Probability Previous: Relative Frequency

2001-01-01