Help with these 2 probability problems

[allison_rivera]

Hello!

I have a question about the answers to these two problems. Obviously, I was provided with an answer and explanation to these questions but I am having trouble with the difference between the two.

Question #1:
There are 4 blue and 6 green marbles in a bag. If you choose 4 marbles, 1 after another and don't replace them, what is the probability that you get 2 blue marbles and 2 green marbles.
---> The answer provided was:
(4C2)*(6C2) / (10C4) = (6*15)/210 = 3/7


Question #2:
You have a class of 24 students. There are 8 seniors and 16 juniors. How many groups of 4 seniors and 2 juniors can you make from all the students in the class?
---> (8C4)*(16C2) = 70 * 120 = 8400

Now, I was inclined to solve #2 in this way:
(8C4)*(16C2) / (24C6)

As I'm writing this, I see that #1 is asking for a probability and #2 is asking for a number (of groups)...but I'm still a little confused.

I thought I was getting the hang of probability problems, but I'm not! I've reviewed the counting methods and probability notes but everything question that I do seems to be a different case. Or, if I do start to figure out a type of question I run into an explanation that does solves the question in a different way!

Could someone provide an explanation for these two problems for clarification?

Thank you in advance,
Alli

[navigalactus]

First one is purely a Probibility problem whereas Second one is PURELY a Combination problem where order of the people in groups formed is not important. So the Second one is simply finding as many ways as there can be to form the groups with the given Junior (m) Senior (n) requirements by the method m X n.

As far as the first problem is concerned, it is first finding out the Required Number of Outcomes that is 2 blue out of 4 and 2 green out of 6, from a total of 10 marbles. So initially all marbles are mixed and you don't know what's gonna come in your hand in 1-2-3-4 time, this is the total number of Outcomes found by 10C4 because marbles taken out are 4 in number.
And as all the 4 events are Exclusive, you have to use Multiplication in 4C2 X 6C2.

[tommywallach]

Hey Guys,

I absolutely HATE "choose two" language for solving combinatorics questions, so I'm not gonna let you. Here's how I would solve both:

However, I need to make sure you understand THE SLOTS METHOD of solving combinatorics:

1. Find # of slots (decision points)
2. Find what goes in each slot
3. Does order matter or not? (If chocolate-vanilla is not the same as vanilla-chocolate, order matters).
4. If Order Matters, multiply all slots; If Order Doesn't Matter, multiply all slots, then divide by the # of slots factorial wherever order doesn't matter

Question #1:
There are 4 blue and 6 green marbles in a bag. If you choose 4 marbles, 1 after another and don't replace them, what is the probability that you get 2 blue marbles and 2 green marbles.

First, let's find the odds of getting BBGG (in that order):

4/10 * 3/9 * 6/8 * 5/7 = 2/5 * 1/3 * 3/4 * 5/7 = 3/42 = 1/14

Now that we have that, we just need to find how many possible orders of BBGG there are. We could use my slots method --> (4 * 3 * 2 * 1)/(2!)(2!) = 6, OR we could just count:

BBGG
BGBG
BGGB
GGBB
GBGB
GBBG

There are six possibilities, so: 6 * 1/14 = 6/14 = 3/7

Question #2:
You have a class of 24 students. There are 8 seniors and 16 juniors. How many groups of 4 seniors and 2 juniors can you make from all the students in the class?

We can do this with pure slots method:

1. Six slots

2. (8 * 7 * 6 * 5) * 16 * 15

3. Order doesn't matter within classes, so...

4. (8 * 7 * 6 * 5)/4! * (16 * 15)/2! = 8400

Hope that helps!

-t