Guide 4, 4th Edition, Chapter 5, Problem 9-11

[adm45]

Can you explain the underlying concept tested in these types of problems? I am always confused when it seems the process is to find the combinations and 1) times or put over the number of combinations or 2) probability of something else? THANKS


Guide 4, 4th Edition, Chapter 5, Problem 9

Question: In a bag of marbles, there are 3 red, 2 white, and 5 blue. If Bob takes 2 marbles out of the bag, what is the probability that he will have one white and one blue marble?

Guide 4, 4th Edition, Chapter 5, Problem 10

Question: A florist has 2 Azaleas, 3 Buttercups, and 4 Petunias. She put two flowers together at random in a bouquet. However a customer does not want two of the same flower. What is the probability that the florist does not have to change the bouquet ?

Guide 4, 4th Edition, Chapter 5, Problem 11

Question: Five A-list actresses are vying for the three leading roles in the new film, "Catfight in Denmark." The actresses are Julia Robards, Merly Strep, Sally Fieldstone, Lauren Bake-all, and Hallie Strawberry. Assuming that no actresses has any advantage in getting any role, what is the probability that Julia and Hallie will star in the film together?

[jlucero]

That's a pretty broad question here. I think you'll need to be a bit more detailed if you're trying to get a more specific explanation to any of these. The similarity is that all of them involve finding two combinations: the combinations of something specific happening divided by the combinations of anything possible happening.

[adm45]

For question number 10, how is it different than question 9, why can't the "List the winning scenarios" methods be used like in Q9. Why can't I have (2/10)(1/9) + (3/10)(2/9) + (4/10)(3/9) = 2/9?

Question 10: Also for the counting method, is there another method to find the desired outcome of two same flowers than the one shown listing all flower 1 and 2 combinations?

For question number 11, what is this method called and why is it used: (prob of J then H then other)( number of combinations of JHother) (1/5)(1/4)(3/3)= 1/20 X 6 equals 3/10.


I have a better understanding that although these problems asks for probability we can use counting methods to find probability of desired divided total possibilities. I am confused as to why each of these problems can use alternative methods besides counting method. thanks

[tim]

Sorry, our rules indicate that in order for us to help you with a question you need to post the full text of the question so other students can benefit from the explanations as well.

[adm45]

Guide 4, 4th Edition, Chapter 5, Problem 9

Question: In a bag of marbles, there are 3 red, 2 white, and 5 blue. If Bob takes 2 marbles out of the bag, what is the probability that he will have one white and one blue marble?

Guide 4, 4th Edition, Chapter 5, Problem 10

Question: A florist has 2 Azaleas, 3 Buttercups, and 4 Petunias. She put two flowers together at random in a bouquet. However a customer does not want two of the same flower. What is the probability that the florist does not have to change the bouquet ?

Guide 4, 4th Edition, Chapter 5, Problem 11

Question: Five A-list actresses are vying for the three leading roles in the new film, "Catfight in Denmark." The actresses are Julia Robards, Merly Strep, Sally Fieldstone, Lauren Bake-all, and Hallie Strawberry. Assuming that no actresses has any advantage in getting any role, what is the probability that Julia and Hallie will star in the film together?


For question number 10, how is it different than question 9, why can't the "List the winning scenarios" methods be used like in Q9. Why can't I have (2/10)(1/9) + (3/10)(2/9) + (4/10)(3/9) = 2/9?

Question 10: Also for the counting method, is there another method to find the desired outcome of two same flowers than the one shown listing all flower 1 and 2 combinations?

For question number 11, what is this method called and why is it used: (prob of J then H then other)( number of combinations of JHother) (1/5)(1/4)(3/3)= 1/20 X 6 equals 3/10.


I have a better understanding that although these problems asks for probability we can use counting methods to find probability of desired divided total possibilities. I am confused as to why each of these problems can use alternative methods besides counting method. thanks

[tim]

First, I noticed that the full text of the problems appears in the first post. If those were there all along, I apologize for asking you for the text again; if you added them later, thank you for doing so.

On question 10, you ask a "why can't I do X?" question. Please note that it is inappropriate to ask such a question unless you have actually been told that you can't do X. In this case, you most certainly CAN do what you are asking to do! The only problem is that you have performed your calculations incorrectly; there are 9 flowers, not 10.

You ask whether there is another way to approach 10. Rest assured that in a combinatorics or probability problem the answer is always YES - there is ALWAYS another way to do the problem; it just might not be the most efficient way. :)

I'm going to punt on your questions regarding 11, because I see them as irrelevant. It doesn't matter what the method is called as long as it works. And the reason we use any given strategy that works is because it works. :)

[jk21]

I am actually confused about the exact same set of problems here and the solutions as explained in MGMAT:

Marbles: probability of selecting a White and Blue marble is explained in the solution as 5/10 (Blue) X 2/9 (White) = 1/9 and then multiplied X 2 for the reverse situation (W then B). Order matters, and probability of both orderings count.

Flowers: the probability of pairs was 5/18 as follows.
Azalea Azalea = 2/9 X 1/8 = 2/72
Buttercup Buttercup = 3/9 X 2/8 = 6/72
Petunia Petunia = 4/9 X 3/8 = 12/72

In the flower solution, notice we did not multiply each matching pair probability X 2 (e.g. we did not account for match orders e.g. Azalea#1 - Azalea#2 AND Azalea#2 - Azalea#1). We only counted each match once.

Looking at it from the counting method, we also see a discrepancy: with marbles, we counted BXW marble pairs as 5X2 = 10. Why wouldn't we count the azalea pairs as 2X1 = 2, buttercups as 3X2 = 6, and Petunias as 4X3 =12 for a total of 20 combo's of bouquet pairs, rather than just 10 that do not care about order in solution.

Isn't selecting 2 marbles out of a bag the same as selecting a pair of flowers for a bouquet (you don't replace the flowers in the exact same way that you don't replace the marbles)? I'm totally bewildered here. Azalea 1 could be in spot #1 or #2.

[RonPurewal]

the quick and dirty answer to your question is this:
"black, THEN white" is different from "white, THEN black".
"azalea, THEN another azalea" is not any different from "azalea, THEN another azalea".

if you still don't see why this is a thing, then consider 2 dice (more familiar to most students). imagine 2 situations:
* if you want to roll 1 odd number and 1 even number, then there are (3)(3) + (3)(3) ways to do that: "odd then even", and "even then odd".
* on the other hand, if you just want 2 odd numbers, that's just (3)(3). this is just "odd then odd"; there's no second case left untreated.
if you still have doubts, just draw out a 6x6 grid with all the possible outcomes of the dice throw. at exactly 18 positions in the grid, there will be an even number on one die and an odd number on the other one. at only 9 positions will there be two odd numbers.