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Math Help - Two bags with 100 balls each, probability of sum >80

  1. #1
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    Two bags with 100 balls each, probability of sum >80

    I got a probability problem we got in homework and I could use some help with it.

    It goes like this:
    One bag contains red balls numbered from 1 to 100 and another bag contains blue balls numbered from 1 to 100. You take one ball from each of the bags.
    What is the probability that the sum exceed 80?

    The only way I can come up with a solution would be to write down every possible result that exceeds 80 but hey that's A LOT of work,, there must be another way, right? Help is appreciated!
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    Quote Originally Posted by Fjolliver View Post
    I got a probability problem we got in homework and I could use some help with it.

    It goes like this:
    One bag contains red balls numbered from 1 to 100 and another bag contains blue balls numbered from 1 to 100. You take one ball from each of the bags.
    What is the probability that the sum exceed 80?

    The only way I can come up with a solution would be to write down every possible result that exceeds 80 but hey that's A LOT of work,, there must be another way, right? Help is appreciated!
    That is a lot of work! This is a common problem in Probability!

    There are numerous ways we can come to an answer, but I like this method. \mu_x = 50.5. Standard deviation is \sigma_x=29.01. The set x is same as the set y...

    P(X+Y>80) = P(Z>\frac{80-101}{58.02})

    P(Z > -.362) = .6413
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    Thanks for the fast answer!

    But I really didn't get how you did all the way and the answer is wrong though it sounded very smart at least according to my key but it have had wrong before so I'm not sure. If you'd like the answer in the key, let me know..

    <br />
\mu_x = 50.5<br />
    <br />
\sigma_x=29.01<br />
    The two mentioned above, where did you get 'em from and what do they mean and come from and how to use 'em? Also how did you get 58.02? Sorry I've never been in contact with 'em before. If I got you wrong somewhere it could be because math is not a everyday thing I do in English...

    Edit: Noticed I might posted in the wrong section.. How old are you when you go to high school? Elementary? Middle?
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    Quote Originally Posted by Fjolliver View Post
    Thanks for the fast answer!

    But I really didn't get how you did all the way and the answer is wrong though it sounded very smart at least according to my key but it have had wrong before so I'm not sure. If you'd like the answer in the key, let me know..

    <br />
\mu_x = 50.5<br />
    <br />
\sigma_x=29.01<br />
    The two mentioned above, where did you get 'em from and what do they mean and come from and how to use 'em? Also how did you get 58.02? Sorry I've never been in contact with 'em before. If I got you wrong somewhere it could be because math is not a everyday thing I do in English...
    The mean is \mu and that is the sum of 1 to 100 divided by 100. The standard deviation is \sigma, and there is no easy way to calculate that. You need a calculator for large samples. There is no way around that.
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    Okey, I see..But you're pretty sure your result is the right and the key is lieing?

    I figured another way to make it easier would be to use 10 instead of 100 and it would generate the same answer and then the collection of succeding ways would be lesser. I think =) though still it's not the "right" way to do it...

    If the key is lieing wouldn't this be possible(Right thinking and not only a coincident): 8/10 * 8/10 = 0.64
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    Quote Originally Posted by Fjolliver View Post
    Okey, I see..But you're pretty sure your result is the right and the key is lieing?

    I figured another way to make it easier would be to use 10 instead of 100 and it would generate the same answer and then the collection of succeding ways would be lesser. I think =) though still it's not the "right" way to do it...

    If the key is lieing wouldn't this be possible(Right thinking and not only a coincident): 8/10 * 8/10 = 0.64
    I never said the key is false. What is the solution; I am having one of those proverbial brain farts!
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    The key says 0.68.

    Well 0.64 is very close to that at least Maybe just a miss on one or two possible results, that you have to add after the calculation?
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    The was thinking of this and wasn't sure if I was on the right track or not.

    The sums can range from 2 to 200. If the first ball is an 81 or greater, then the second doesn't matter. That's a 20% chance.

    If the first ball is less than or equal to 80 that is a 80% chance and a 60% chance the second ball will cause the sum to be greater than 80.
    Because 120/200=.6

    .20+(.8)(.6)=.68
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    Quote Originally Posted by galactus View Post
    The was thinking of this and wasn't sure if I was on the right track or not.

    The sums can range from 2 to 200. If the first ball is an 81 or greater, then the second doesn't matter. That's a 20% chance.

    If the first ball is less than or equal to 80 that is a 80% chance and a 60% chance the second ball will cause the sum to be greater than 80.
    Because 120/200=.6

    .20+(.8)(.6)=.68
    Wow, I knew it was a lot easier than I was thinking... sometimes I over complicate these problems!
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    WOW Thanks man! After this horrible day that was a perfect ending at least! It sounds exactly right! I guess top grade is awaiting tomorrow with that explanation

    First thread in this forum but I surely will come back here next time, thanks to both of you who helped me =)
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    Quote Originally Posted by colby2152 View Post
    That is a lot of work! This is a common problem in Probability!

    There are numerous ways we can come to an answer, but I like this method. \mu_x = 50.5. Standard deviation is \sigma_x=29.01. The set x is same as the set y...

    P(X+Y>80) = P(Z>\frac{80-101}{58.02})

    P(Z > -.362) = .6413
    Err.. hummm.. Why are you using a normal table? Why are you using the normal table in this way?

    RonL
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    Quote Originally Posted by CaptainBlack View Post
    Err.. hummm.. Why are you using a normal table? Why are you using the normal table in this way?

    RonL
    Honestly, I'm not sure... major brain fart, didn't have lunch yet!

    First thing I thought of was using the normal approximation in this manner. Close, but no cigar!
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    Quote Originally Posted by colby2152 View Post
    Honestly, I'm not sure... major brain fart, didn't have lunch yet!

    First thing I thought of was using the normal approximation in this manner. Close, but no cigar!
    This should be a discrete triangular distribution.

    RonL
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    Hello, Fjolliver!

    One bag contains red balls numbered from 1 to 100
    and another bag contains blue balls numbered from 1 to 100.
    You take one ball from each of the bags.
    What is the probability that the sum exceed 80?

    Imagine rolling two 100-sided dice and visualize the table of outcomes.

    Write 1-to-100 across the top and 1-to-100 down the left side.
    . . It will be a 100 x 100 "addition table" with 100^2 = 100,000 entries.


    Where are the entries that do not exceed 80?
    They lie to the left of the diagonal line that start at (1,79) at the top
    . . and ends at (79,1) at the lower-left.
    There are: . 1 + 2 + 3 + \cdots + 79 \:=\:3160 of them.

    Therefore . . .

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    Yes you're totally right Soroban. That's if you want the exactly right answer which you probably want Thanks!
    ..but..
    ..I think this formula was the one to use to calculate it the "easy" way (no need to write a lot of numbers):
    Sn=n(a1+an/2)

    I think it's standard equation for series or what to call it. However instead of "n" you put the amount of numbers in the series and instead of "a1" you take the first number in the series and instead of "an" the last number in the series.

    In this case:
    Sn=79(1+79/2)
    Sn=3160

    Now if anyone would trip over this one he/she got multiple choices to solve it at!
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