anonymous
  • anonymous
Kay Paso, who is only 3 years old, tears off all 10 of the soup cans on her mother's pantry shelf. Her mother knows that there were 2 cans of tomato soup and 8 cans of vegetabe soup. If Kay's mom selects 4 of the unlabeled cans at random, answer the following: a.)How many possible groups of 4 cans is possible to select? b.)How many groups of cans would have exactly 1 of tomato soup? c.) What is the probability of selecting exactly one tomato soup can? d.) What is the probability that at least 1 of the 4 cans is tomato soup? e.)What is the probability that none of the 4 cans is tomato soup? f.)What relationship exists betwenn the answers to parts d and e?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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KingGeorge
  • KingGeorge
What do you think the answer is for a?
anonymous
  • anonymous
10!/4!6!=1260
anonymous
  • anonymous
right?

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KingGeorge
  • KingGeorge
\(10! \over 4!6!\) is correct, but it's equal to 210. not 1260.
anonymous
  • anonymous
Oh, I see what I did!!! I accidentally put 4 instead of 4! in my work!!!
KingGeorge
  • KingGeorge
Do you think you could also manage b?
anonymous
  • anonymous
Um ,is this like that one with the car with 2 boys and 5 girls? If so, it might be...2C1 times 8C3 =2 times56 =112!!! And c would be 112 divided by 210... but I dont get the rest.
KingGeorge
  • KingGeorge
Perfect. You're doing great.
KingGeorge
  • KingGeorge
d is a little more complicated. First, you have to find the ways you choose only one (this was b, so we've already done it), and then you have to find how many ways to get choose 2 tomato cans. This is given by \(\binom{2}{2} \cdot \binom{8}{2}=28\). Now just add them together to get \(112+28=140\)
KingGeorge
  • KingGeorge
Then of course you have to divide that by 210 to get the probability, but that was the tricky part.
KingGeorge
  • KingGeorge
As for e, it's given by \[\binom{2}{0}\cdot \binom{8}{4}=70\]and divide that by \(210\).
anonymous
  • anonymous
Why add? Is this kind of like the one where you told me if the two people belonged in two categories, you take the perm/com. without them, with one, with both, then add them together?
KingGeorge
  • KingGeorge
Yes, but it's a little tricky to see what the categories are at first. One category would be "One can of tomato soup is chosen" and the other category is "Two cans of tomato soup are chosen." However, if you chose two cans of tomato soup, you also chose one can. So the groups are not exclusive, so instead of multiplying, you have to add.
anonymous
  • anonymous
hh..that somehow makes sense! You are truly miraculous, KingGeorge.
anonymous
  • anonymous
What about f?
KingGeorge
  • KingGeorge
Well, the final value of d is \[{140 \over 210}={2 \over 3}\]and for e\[{70 \over 210}={1 \over 3}\]What do you notice about those two numbers?
KingGeorge
  • KingGeorge
In particular, what happens when you add them together?
anonymous
  • anonymous
they make one!
KingGeorge
  • KingGeorge
Exactly. Since we're using probabilities, and we can only have 0, 1, or 2 can's of tomato soup, it's good that they add to 1.
anonymous
  • anonymous
wai, wah?
KingGeorge
  • KingGeorge
In probability, a probability of 1 means that that thing is guaranteed to happen. You have a 100% chance of it happening. Here, question d asked for the probability of getting 1 or 2 cans of tomato soup, and e asked the probability of getting 0 cans of tomato soup. Since you can't have any other amount of tomato soup, if we sum the probabilities, we should get 1. Since we did get 1, that means we probably did the problem correctly.
anonymous
  • anonymous
OH!!! So since there can only be 2 cans of tomato soup and we got 2 cans of tomato soup plus 0 cans of tomato soup, that should equal that total amount of two, and the probabilities kind of work the same way?
KingGeorge
  • KingGeorge
Basically what we did, was add the probability of getting 1 can to the probability of getting 2 cans in part d. Then, in e, we found the probability of getting 0 cans. By adding them together, we get the probability of getting 0, 1, or 2 cans.
KingGeorge
  • KingGeorge
This stuff probably isn't needed for the question, but it might be good to know anyways.
anonymous
  • anonymous
okay!!! Thank you soooooo much!!! I am forever in your debt!!! Oh, and by the way, I had to refresh my page a ton, so if the thing said I wasn't there, I never really left you, so ... ya. This page is starting to feel like a crowded Facebook page, haha... Oh, and I have a question about a question I saw the teacher do, but really didn t get, so I will put that up as a new question

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