Ok... so... here's my nice little grid..
So there's 20/210 ways to get 2 chocolate chips. So, 2/21 .. ok, that works
The same thing.
a) Put hand in, pull out cookie. Put hand in again, pull out cookie.
b) Put one hand in, pull out cookie. Put second hand in, pull out cookie. Same thing as a).
c) Put one hand in, grab cookie. Put second hand in, grab cookie. Pull both hands out, both with cookies in. Same thing as b).
d) Put both hands in at same time, grab cookie in each hand. Pull both hands out, both with cookies in. Same thing as c).
e) Put one hand in, pull out two cookies. Same thing as d).
Therefore, pulling 2 cookies out at the same time *is the same thing* as pulling them out one at a time. EXCEPT you can't distinguish between which is the "first" and which is the "second".
Very smart.
Now you're home and dry.
To determine the probability of any combination, divide the number of occurrences of that combination by the total number of boxes. So, for example, the probability of peanut then choc is the number of boxes with PC in them divided by
which works out as (3 x 5) / (15 x 14)
which is the same as what was got further up the thread with that (3 / 15) x (5 / 14).
Does it start to make sense yet?
Yes and no .. it was determined that (3 / 15) x (5 / 14) wasn't a possible answer which is what led me to be confused in the first place
6. What is the probability that the first is peanut butter and the second is chocolate chip?
a. 1/9
b. 2/21
c. 1/5
d. 2/7
e. none of these
7. What is the probability that the second is chocolate chip given that the first is oatmeal?
a. 1/9
b. 1/3
c. 1/5
d. 5/14
e. none of these
I got 1/14 for #6 and 1/6 for #7.
I highly doubt both are "none" ... "none" has never been the answer.
I'm still not convinced it's the samething, but I could be wrong... Need to think about it more. But in the interest of not cluttering up OP's thread, let's try to help him on the original question.
I think the grid he drew up is the correct grid for the given problem.
I'm afraid the answer does appear to be 1/14 which (as you say) does not appear in the list of answers. So my answer would be "none of the above" after all.
Anyone else care to have a go? Am I doing this all wrong? I hope not
Come back to us and tell us what the answer is when you get the results back.
In the interest of saving my sanity, I submitted the exercise. It was just a practice thing, so it wouldn't have hurt my grade regardless.
Turns out, the answer was "none" for
What is the probability that the first is peanut butter and the second is chocolate chip?
So, that's good..
But for
What is the probability that the second is chocolate chip given that the first is oatmeal?
I got 1/6, and it turns out it was 5/14 .. did I count wrong?
I wonder what the difference between this question and a question that is worded the same style as #6, which would be
What is the probability the first is O and second is C? If there was no difference in these two questions then the answer for #7 should also be (e). But the answer is not (e) so there must be a difference in the way #7 is worded from the way #6 is worded.
You're given that the first is O.
So all the lines where it's not O are out of the equation.
You're left with one line (the one corresponding to the one with the cookie you picked, it doesn't matter precisely which one) which has 6 possibilities for the 2nd one to be O, 3 for it to be P and 5 for it to be C.
So that's 5 out of 5 + 6 + 3 which is 5/14.
Sensible?