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Mathematics 66 Online
OpenStudy (anonymous):

a cooler contains 7 cans of soda 3 colas, 3 oranges, and 1 cherry. Two cans are selected at random without replacement. Find the prob that at least one can is cola

OpenStudy (anonymous):

3/7. There

OpenStudy (anonymous):

Oops tyop

OpenStudy (yuki):

to find the probability of your case, you need to compute the prob P(exactly one can is cola) + P(exactly two cans are cola) + P(all three cans are cola) but tat will take a while so we use another technique

OpenStudy (anonymous):

3/7 right?

OpenStudy (yuki):

sorry, not need for all three part

OpenStudy (anonymous):

No that was the first part but my ipad is spazzing. I'll answer if yuki cant

OpenStudy (yuki):

anyway, the complement of (at least one cola) is (no cola at all) and the probability of a complement is p(complement) = 1-p(statement) so 1-p(no cola) would be easier to calculate

OpenStudy (yuki):

can you find the probability that the two cans you pulled out contains no cola can ?

OpenStudy (anonymous):

how would u do that

OpenStudy (yuki):

okay so since you are pulling out two cans from a container at random without any replacement, the the way you pull out two cans will have these following steps 1), pull out a can from 7 cans 2), pull out another from the "remaining 6" so, try to answer step 1). what is the probability of pulling out a non-cola can out of 7 ?

OpenStudy (anonymous):

6/7

OpenStudy (yuki):

that is not right you have 4 cans that ar non-cola out of 7 so ...

OpenStudy (anonymous):

4/7?

OpenStudy (yuki):

very good

OpenStudy (yuki):

now, how many non-cola cans are there in the container left ?

OpenStudy (radar):

First draw 3/7 probability for a cola second draw 3/6 probality for a cola probability from both draws getting at least on cola is 92% Doesn't seem right.

OpenStudy (anonymous):

2

OpenStudy (anonymous):

NO ONE GIVE CHICKA THE ANSWER!!!!

OpenStudy (yuki):

nope initially you had 4 non-cola cans right ? are you having trouble identifying what a non-cola can is ? Cherry and Orange are the non-cola cans. we had 1 cherry and 3 oranges , 4 total. since we pulled out one , so how many should be left ?

OpenStudy (anonymous):

3

OpenStudy (yuki):

exactly, so now you are ready to answer step 2) what is the probability that you can pull out the 3 non-cola cans from the remaining 6 ?

OpenStudy (anonymous):

Hahaha having trouble identifying what a non-cola is...

OpenStudy (yuki):

well, there is cherry cola and lime cola right ?

OpenStudy (anonymous):

is it .71

OpenStudy (anonymous):

I like Lime Cola better :)

OpenStudy (anonymous):

Go with that one.

OpenStudy (anonymous):

CAN SOMEONE HELP ME

OpenStudy (yuki):

you are only two steps before the answer ok? so what was the prob of 3 cans out of 6 ?

OpenStudy (anonymous):

3/6

OpenStudy (yuki):

very good

OpenStudy (yuki):

the last step is to multiply those two probabilities, 4/7 * 3/6 = ?

OpenStudy (anonymous):

20.5714

OpenStudy (yuki):

haha you can just use fractions you know, 2/7 anyways, remember that we were looking for the complement ? all you have to do now is to subtract this from 1. so the answer is 1 - 2/7 = ?

OpenStudy (anonymous):

so 5/7

OpenStudy (yuki):

very good

OpenStudy (anonymous):

thank u

OpenStudy (yuki):

Now if you really seek help, I recommend you to hire a private tutor or go to a tutoring center or an after school program. the type of help you need is not easy to obtain online and as you noticed, it takes a bit while to solve just one problem. I will be more than happy to help you but you need to relax and get help efficiently, ok? Now good luck. :)

OpenStudy (mathmagician):

P(there is no cola)=7/10*6/9=21/45 P(there is a least 1 cola)=1-P(there is no cola)=1-21/45=24/45

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