# Thread: [SOLVED] Simple Probability has me stumped

1. ## [SOLVED] Simple Probability has me stumped

There's this exercise question that's completely stumping me.
Ok, so there's 3 types of cookies:
5 Chocolate Chip - (C)
7 Oatmeal - (O)
And 3 Peanut Butter - (P)

Two cookies are taken (without replacing them)

Now, I would say the set of possible solutions is
{CC, CO, CP, OC, OO, OP, PC, PO, PP}
So, 9 solutions

1) What's the chance both are chocolate chip?
Since this is only 1 of the 9 solutions, I would say 1/9

2) What's the chance one is oatmeal and the other is peanut butter?
There are 2 chances. OP and PO. So, I would say 2/9 .. except 2/9 isn't an answer! (It's multi choice)

3) Whats the chance the first is peanut butter and the second is chocolate chip?
Again, I'd say 1/9

4) What's the chance the first is oatmeal and the second is chocolate chip?
... Again, 1/9

But these have to be wrong. Firstly, #2 doesn't follow because 2/9 isn't an acceptable solution. And the rest just FEEL wrong.

So I started wondering if I had to look at the NUMBER of cookies.
If I do, there is about a 33% chance of getting chocolate chip, 46% chance of getting oatmeal, and a 20% chance of getting peanut butter.

But after you get 1 cookie, there's 3 different ways it could go! If it's a chocolate chip, suddenly you have less chance of getting a chocolate chip and more of the others! Should this even be considered? The way it's worded, it seems like both are taken at the same time.

Any help is appreciated

EDIT:

So, for the first question:

33% chance of chocolate chip
46% chance of oatmeal
20% chance of peanut butter

If a chocolate chip is taken, then

29% chance of chocolate chip
50% chance of oatmeal
21% chance of peanut butter

So what is the chance that you will get both chocolate chip?

2. I'm running out of time on this, no one can take a stab?

3. I'm not sure how you are coming up with the set
{CC, CO, CP, OC, OO, OP, PC, PO, PP}, but that is not how I would approach the problem.

You are picking one cookie at a time, as opposed to reaching in once and grabbing two.

So, I'll do the first question and from that you should be able to follow the same logic and do the rest... assuming my method is correct.

P(both are chocolate chip)
= P(1st pick is chocolate chip and 2nd pick is chocolate chip)
= P(1st pick is chocolate chip) * P(2nd pick is chocolate chip)
= (5/15) * (4/14)
= .0952

I'm pretty sure this is the way to approach it. By the way if you use this method on the 2nd question, do you get an answer that is one of the multiple choices?

4. Heck, let me do #2 because I see that this isn't quite like the 1st one.

P(Oatmeal and Peanut Butter)
= P(1st pick is Oatmeal and 2nd pick is Peanut Butter) +
P(1st pick is Peanut Butter and 2nd pick is Oatmeal)
= P(1st pick is Oatmeal) * P(2nd pick is Peanut Butter) +
P(1st pick is Peanut Butter) * P(2nd pick is Oatmeal)
= ((7/15)*(3/14)) + ((3/15)*(7/14))
= 0.2

Is this one of the choices for question 2?

5. Oh wow, I was looking at this completely wrong. I'll try the second and edit this post with results.

Oh, I see you've already done it.

I was ... over-thinking this, I think.

Thanks a ton

Attempt at number 3 using your method

P(Peanut butter then chocolate chip)
= P(1st pick is Peanut Butter and 2nd pick is Chocolate chip)
= P(1st pick is Peanut Butter) * P(2nd pick is Chocolate chip)
= ((3/15)*(5/14))
= 0.0714

hmm.. thats not right..

6. As an aside, I wonder if the solytion method will be different if the question said, you are reaching in ONCE and grabbing two AT THE SAMETIME.

7. I don't think you can assume this. Imagine 2 people reaching in simultaneously and grabbing a cookie at random. This is fine unless they grab the same cookie, which is possible.

I think you *have* to assume they're taken one at a time.

Also, the problem says they are "Taken without replacement", so this implies that we are looking to decrease the total number of cookies.

Any thoughts on 3 and 4? I ended up getting answers that didn't work ... somehow..

8. Question 3

P(1st is P and 2nd is C)
= (3/15)*(5/14)
= .0714

Question 4

P(1st is O and 2nd is C)
= (7/15)*(5/14)
= 0.1667

EDIT: note that in both of these, the question DOES NOT say "... the probability that BOTH are ... ".
Rather, it says "... the probability that 1st is this and the 2nd is that ... "

By the way, what I was saying on the previous post, was a WHAT IF they were asking that. I know the question at hand is describing a scenario where the person is picking one at a time, but WHAT IF the question said, reach in once and grab two?

9. Possible answers for 3 and 4 are

3)
A. 1/9
B. 2/21
C. 1/5
D. 2/7

4)
A. 1/9
B. 1/3
C. 1/5
D. 5/14

I came up with the same answers you did, neither of which are possible.
Well, letter "E" for both is "None of these", but that's never the answer.

Yes, I see you meant "What if"

But, this can go one of 2 ways

1) The person closes their eyes, reaches in and grabs two at the exact same time. This is fine, but the person may accidentally grab the same cookie with both hands, which ruins the problem.

2) The person has their eyes open and, before grabbing them, chooses the cookies that they will take. However, they will be choosing in their head 1 at a time. So even though they haven't grabbed any yet, they've already picked cookie 1 of the 15, then cookie 2 of the 14, which is the same as physically grabbing 1 at a time.

10. One way of working it out is to work out the total number of possible pairs of cookies, which is $(5 + 7 + 3)^2$ and enumerating each pair, possibly by drawing a 15x15 grid and writing "OO", "PC", CP", OP" etc in each box. Then count up each combination.

Then you can see what the difference is between counting them together (in which case you would total all the "PC" and "CP" and call them one possibility) and counting them separately (in which case "PC" would mean "peanut then choc" and "CP" would mean "choc then peanut".

Okay, so it's long-winded and tedious, but remarkably therapeutic, and should give you a more comprehensive grasp of what's going on. Seriously, I recommend it as an exercise - try it out. Then you'll know better what you're doing and you'll fly through this sort of exercise.

11. Originally Posted by Matt Westwood
One way of working it out is to work out the total number of possible pairs of cookies, which is $(5 + 7 + 3)^2$ and enumerating each pair, possibly by drawing a 15x15 grid and writing "OO", "PC", CP", OP" etc in each box. Then count up each combination.

Then you can see what the difference is between counting them together (in which case you would total all the "PC" and "CP" and call them one possibility) and counting them separately (in which case "PC" would mean "peanut then choc" and "CP" would mean "choc then peanut".

Okay, so it's long-winded and tedious, but remarkably therapeutic, and should give you a more comprehensive grasp of what's going on. Seriously, I recommend it as an exercise - try it out. Then you'll know better what you're doing and you'll fly through this sort of exercise.
I am up for doing that, but I don't really see how. The combinations without regard to amount are just
{C,C} {C,O} {C,P} {O,C} {O,O} {O,P} {P,C} {P,O} {P,P}

Since I am taking these 2 at a time without regard for the other situations, how do I incorporate the amount into it?

12. Originally Posted by Matt Westwood
One way of working it out is to work out the total number of possible pairs of cookies, which is $(5 + 7 + 3)^2$ and enumerating each pair, possibly by drawing a 15x15 grid and writing "OO", "PC", CP", OP" etc in each box. Then count up each combination.

Then you can see what the difference is between counting them together (in which case you would total all the "PC" and "CP" and call them one possibility) and counting them separately (in which case "PC" would mean "peanut then choc" and "CP" would mean "choc then peanut".

Okay, so it's long-winded and tedious, but remarkably therapeutic, and should give you a more comprehensive grasp of what's going on. Seriously, I recommend it as an exercise - try it out. Then you'll know better what you're doing and you'll fly through this sort of exercise.
I don't want to clutter up the OP's thread, but the method you are describing would apply to a case where the cookies are chosen two at a time with one reach into the jar correct? As opposed to what the question is asking, which is reach in twice, grabbing one at a time.

13. Oh jeez, now I'm getting confused.

14. There's 15 cookies (5+7+3) so across the top you put:

C C C C C O O O O O O O P P P

... and down the left hand side you put the same.

Then you've got a 15x15 grid which you fill up each square with 2 letters, the first one matching the letter at the left hand side of the row, and the second one matching the letter at the top of the column.

You will find the top left hand 5x5 square is all "CC" so that's the 25 out of $15^2$ possible combinations which is all chocolate chip.

There will be 5 x 7 where the first is C and second is O, and 7 x 5 where the first is O and the second is C. And so on.

Yes, you've got the 9 different combinations, what this exercise does is tell you exactly how many possible ways of drawing each of those combinations is.

Hope this helps.

15. Originally Posted by Matt Westwood
There's 15 cookies (5+7+3) so across the top you put:

C C C C C O O O O O O O P P P

... and down the left hand side you put the same.

Then you've got a 15x15 grid which you fill up each square with 2 letters, the first one matching the letter at the left hand side of the row, and the second one matching the letter at the top of the column.

You will find the top left hand 5x5 square is all "CC" so that's the 25 out of $15^2$ possible combinations which is all chocolate chip.

There will be 5 x 7 where the first is C and second is O, and 7 x 5 where the first is O and the second is C. And so on.

Yes, you've got the 9 different combinations, what this exercise does is tell you exactly how many possible ways of drawing each of those combinations is.

Hope this helps.
I've just noticed this is all wrong. This is for the "Replacement" scenario, in which you replace what you've just picked!

I think that you can get round this by blanking out the main diagonal because that's where the same cookie is picked twice.

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