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# I found this problem on the internet while trying to find

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Manager
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I found this problem on the internet while trying to find [#permalink]

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26 Oct 2007, 11:02
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

I found this problem on the internet while trying to find some practice material:

Tom has 30 apples, 15 of them are bad.Steve has 30 oranges and 6 of them are bad. Both Tom and Steve have to put two apples and two oranges in a basket at random. What is the probability that the basket would have exactly two bad apples and two bad oranges?

I tried to solve it but looks like something is missing..........can someone help.
Manager
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26 Oct 2007, 11:06

P = (2/15)*(2/6) = 4/90 = 2/45
Manager
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26 Oct 2007, 11:36
But don't you have to consider the probability of picking BAD one's only from the entire lot ?
Current Student
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26 Oct 2007, 11:42
i'm getting 4/29, but I fear I'm wrong.
VP
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26 Oct 2007, 12:13
15/30*14/29 + 6/30*5/29 = 7/29 + 1/29 = 8/29

this is wrong - see followup posts

Last edited by KillerSquirrel on 26 Oct 2007, 15:22, edited 1 time in total.
Manager
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26 Oct 2007, 13:15
I got 8/29 too like KS
Manager
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26 Oct 2007, 13:25
Why do we add the probability of these two events? These are two mutually exclusive events, and we are looking for the probability of 1 event AND the other event.
VP
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26 Oct 2007, 13:28
yuefei wrote:
Why do we add the probability of these two events? These are two mutually exclusive events, and we are looking for the probability of 1 event AND the other event.

AND ---> multiply

Last edited by KillerSquirrel on 26 Oct 2007, 15:21, edited 1 time in total.
Manager
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26 Oct 2007, 13:32
Well if you throw a dice twice, what is the probability that you throw six both times?
P = (1/6)*(1/6) = 1/36

What if the question had asked what is the probability of selecting 2 bad apples or 2 bad oranges? The result would have two scenarios, which would provide for p = p(1) + p(2). Thoughts?
Current Student
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26 Oct 2007, 13:36
yuefei wrote:
Well if you throw a dice twice, what is the probability that you throw six both times?
P = (1/6)*(1/6) = 1/36

What if the question had asked what is the probability of selecting 2 bad apples or 2 bad oranges? The result would have two scenarios, which would provide for p = p(1) + p(2). Thoughts?

this is also tripping me up!
VP
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26 Oct 2007, 13:41
yuefei wrote:
Well if you throw a dice twice, what is the probability that you throw six both times?

P = (1/6)*(1/6) = 1/36

What if the question had asked what is the probability of selecting 2 bad apples or 2 bad oranges? The result would have two scenarios, which would provide for p = p(1) + p(2). Thoughts?

total

21/60*14/59+21/60*5/59 =

Manager
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26 Oct 2007, 13:46
KillerS, you're a machine!!
VP
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26 Oct 2007, 13:48
Does anyone has an OA here ? Or is this an academic debate ?

Director
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26 Oct 2007, 14:40
OA here. I was thinking the answer is 7/841 (7/29*1/29)
SVP
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26 Oct 2007, 14:42
To me

o p(2 bad apples) = 15*14 / (30*29) = 7/29
o p(2 bad oranges) = 6*5 / (30*29) = 1/29

Then,
Director
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26 Oct 2007, 15:01
nitinneha wrote:
I found this problem on the internet while trying to find some practice material:

Tom has 30 apples, 15 of them are bad.Steve has 30 oranges and 6 of them are bad. Both Tom and Steve have to put two apples and two oranges in a basket at random. What is the probability that the basket would have exactly two bad apples and two bad oranges?

I tried to solve it but looks like something is missing..........can someone help.

agree with calculations 7/29 x 1/29 = 7/29^2, but the result seems too low...
SVP
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26 Oct 2007, 15:16
IrinaOK wrote:
nitinneha wrote:
I found this problem on the internet while trying to find some practice material:

Tom has 30 apples, 15 of them are bad.Steve has 30 oranges and 6 of them are bad. Both Tom and Steve have to put two apples and two oranges in a basket at random. What is the probability that the basket would have exactly two bad apples and two bad oranges?

I tried to solve it but looks like something is missing..........can someone help.

agree with calculations 7/29 x 1/29 = 7/29^2, but the result seems too low...

Well... it's 1 chance on 120 to have this specific set ... I think it's fine, the number of bad orange is low too

If the number of bad oranges were also equal to the good one, a set like this would have 1 chance on 17 to be...
VP
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26 Oct 2007, 15:21
Fig wrote:
To me

o p(2 bad apples) = 15*14 / (30*29) = 7/29
o p(2 bad oranges) = 6*5 / (30*29) = 1/29

Then,

Yes - perfect as always - the answer should be 7/(29^2)

SVP
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26 Oct 2007, 15:27
KillerSquirrel wrote:
Fig wrote:
To me

o p(2 bad apples) = 15*14 / (30*29) = 7/29
o p(2 bad oranges) = 6*5 / (30*29) = 1/29

Then,

Yes - perfect as always - the answer should be 7/(29^2)

Thanks ... U are not that bad too .... Did u fix a date for your test?

[I know... I asked u it one time already... Perhaps, this time I will have answer ... Never giving up ! ]
VP
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26 Oct 2007, 15:29
Fig wrote:
KillerSquirrel wrote:
Fig wrote:
To me

o p(2 bad apples) = 15*14 / (30*29) = 7/29
o p(2 bad oranges) = 6*5 / (30*29) = 1/29

Then,

Yes - perfect as always - the answer should be 7/(29^2)

Thanks ... U are not that bad too .... Did u fix a date for your test?

[I know... I asked u it one time already... Perhaps, this time I will have answer ... Never giving up ! ]

My test will be ~ December

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