probability question

[AFF2009]

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In how many ways can a group of 2 basketballs and 1 tennis ball be formed from a total of 8 basketballs and 2 tennis balls?

Asnwer is 30.

But I kinda of disgaree in adding the possibilites I thought we had to multiply them. I think it's 28 * 2= 56 rather than 28+2= 30

does anybody agree with me???

Thanks

 

[vvvv]

In how many ways can a group of 2 basketballs and 1 tennis ball be formed from a total of 8 basketballs and 2 tennis balls?

Asnwer is 30.

But I kinda of disgaree in adding the possibilites I thought we had to multiply them. I think it's 28 * 2= 56 rather than 28+2= 30

does anybody agree with me???

Thanks

I agree

 

[thebozz1975]

No, 30 is the correct answer. Just think through it logically. Don't rely on formulas. I know that has caused vvvv to get at least 2 prob questions wrong in 2 days.

 

[AFF2009]

sorry thebozz1975, your arguement doesnt convince me. no offense....

I still think of it as 28 possibilites AND 2 possibilites thus they are multiplied

Like if i told u we have a box that has 3 rows of apples and 2 columns of apples how many apples total: 3*2= 6

I hope I took u inside my head, so u know what I see...😳

 

[vvvv]

No, 30 is the correct answer. Just think through it logically. Don't rely on formulas. I know that has caused vvvv to get at least 2 prob questions wrong in 2 days.

hey boy, be cool. if you think it should be 28+2 so why then.....

 

[thebozz1975]

My goal is not to "convince you" or teach basic probability. I'm just saying that 30 IS the correct answer. No question about it. This is a very easy problem. Hint: it is not a permutation problem, which is the way you are approaching it.

 

[AFF2009]

ok thanks for your help!🙄

 

[AFF2009]

Is there anybody else that would like to share their opinion on this. It's not a difficult problem, maybe I am thinking too much.

Thanks

 

[Streetwolf]

My goal is not to "convince you" or teach basic probability. I'm just saying that 30 IS the correct answer. No question about it. This is a very easy problem. Hint: it is not a permutation problem, which is the way you are approaching it.

First I'd like to dedicate this 1,000th post to all the little people out there.

But seriously, the answer is 56. List the possibilities if you doubt it.

 

[jay47]

Can someone explain this whole problem to me? Maybe I don't understand what it is actually asking.

I thought that you could choose one of 8 basketballs, then one of 7. That would be 8*7=56. Then you could choose one of 2 tennis balls, 56 * 2= 112.

Obviously, something about this isn't right. Please help.

 

[thebozz1975]

.

 

[vvvv]

.

C8 ^2 . C2 ^1 = 8!/2! (8-2!) * 2!/1! = 28 * 2 = 56. No body need you to teach. we're here for info

 

[Streetwolf]

Can someone explain this whole problem to me? Maybe I don't understand what it is actually asking.

I thought that you could choose one of 8 basketballs, then one of 7. That would be 8*7=56. Then you could choose one of 2 tennis balls, 56 * 2= 112.

Obviously, something about this isn't right. Please help.

You're on the right track, but don't forget that it doesn't matter which basketball you choose first. By doing 8*7 you've selected two basketballs, but you're also allowing the person to choose basketball A then basketball B, *OR* basketball B then basketball A. In both cases you get the same two basketballs, so you're counting everything twice. That's why you take 56 and divide by 2. Then of course you multiply by 2 because you could choose either of 2 tennis balls.

 

[jay47]

You're on the right track, but don't forget that it doesn't matter which basketball you choose first. By doing 8*7 you've selected two basketballs, but you're also allowing the person to choose basketball A then basketball B, *OR* basketball B then basketball A. In both cases you get the same two basketballs, so you're counting everything twice. That's why you take 56 and divide by 2. Then of course you multiply by 2 because you could choose either of 2 tennis balls.

Thanks a ton Streetwolf, and congrats on the 1000th post. Your explanation makes perfect sense. Its been 2 years since I took any math classes at all, but I still enjoy math problems every once in a while.

Oh, and another thing; could you simply consider this a combination type problem? And would it be safe to assume that the way I was doing it was called a permutation? Just checking up on the old math definitions.

 

[Streetwolf]

Yes that was a permutation you did. We want a combination in this problem. Order does not matter in a combination.

 

[allstardentist]

know probability well. I had like four of those on my test.