~Here is an example: A container contains 6 blue & 4 green marbles. All marbles are identical in size. Also, the colors of the blue & green marbles are identical. You are to seclect one marble at a time (without looking at them) & do not replace it until you have 5. Find the probability that you have selected 3 blue & 2 green marbles.

~When forming a tree diagram for this problem, start at the top with S or with the total number of each item. Then draw two branches coming down, one for B & the other for G (stage one) (your first selection could be a blue or a green marble).

~From each of these selections, draw two more branches with a B or G again (stage two)(your second selection). Continue in this manner until all 5 stages are complete. Just make sure that each individual path to the end has exactly 3 blue & 2 green.

~I've indicated the 10 paths with their corresponding probabilities along each section.

~This is what it would look like:

          6B & 4G
                        | 6/10                            | 4/10
                B                       G
              | 5/9                   | 4/9                            | 6/9             | 3/9
        B             G                 B         G
       |4/8       |4/8       |5/8      |3/8             | 5/8     | 3/8     |6/8
      B        G         B        G              B       G        B
      |4/7    |4/7 |3/7   |4/7   |3/7    |5/7            |3/7 |4/7    |5/7        |5/7
       G       B     G      B     G       B               G    B       B           B
       |3/6    |3/6   |4/6 |3/6 |4/6     |4/6           |4/6 |3/6   |4/6       |4/6
      G        G     B     G     B        B               B     G      B          B

~NOTE: This is a very long way to do this problem...notice the 10 mutually exclusive paths, each of which have a probability of 1/21 (multiply the probabilities along each path). Adding the 10 results, we get our final answer of 10/21 or .4762

~NOTE: The best way (much quicker & easier) to do this problem is by finding the # of ways you can select 3 blues & 2 green marbles, then divide by the total # of ways of selecting any 5 marbles. This is the basic definition of probability...you would get the following which can be easily found on your TI-83..

              (6C3)(4C2) divided by 10C5 = .4762

~This technique is particularly helpful when solving problems involving winning so many games out of a total number (i.e., the world series in baseball, 2 out of 3 sets in tennis, 3 out of 5 games of chess, etc.). However, in these real life situations, the probabilities are statistical & must be estimated, where as, in the above problem, the probabilities are known in advance (A Priori).

~In a men's tennis match, the probability that player A loses a set is 7/11. Find the odds that player A wins the match (3 out of 5 sets).

The probability of person A losing a set is 7/11, so the probability of winning a set=4/11. This is what we use since the problem calls for the odds person A wins 3 out of 5 sets.

Here is the shorter approach.

case 2) A wins in 4 sets: WWLW, WLWW, LWWW (3 ways)

case 3) A wins in 5 sets: WLLWW, LLWWW, LWLWW,
WWLLW, WLWLW, LWWLW (6 ways)

Now, figure the probabilities in each case (use the multiplication rule)

case 2) {(4/11)^3] (7/11) times 3 = 1344/14641

case 3) [(4/11)^3][(7/11)^2] times 6 = 18816/161051

Now, add them. Need all denominators the same, so, multiply first by (11)^2/(11)^2 to get 7744/161051

now, add the tops to get 41344/161051. This is the probability A wins 3 out of 5 sets (wins the match).

since this is less than 1/2, this is unlikely, so the odds are against.

to find them, find the number of times person A loses to the number of times A wins in 161051 (bottom of the probability fraction).

we get, (161051-41344) to 41344 or 119707 to 41344 against.
(a little better than 3 to 1, approximately)

~In most basic problems, the odds come out much nicer or can be reduced to much smaller numbers.