Thread: Probability Q - Help please!
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Feb 7th, 2009 11:41 PM
#1
Probability Q - Help please!
Cant' seem to figure this one out. My friend asked for some help and its bugging me.
Q1) A box has 8 pairs of shoes. 4 Shoes are randomly selected, what is the probability that there will be exactly one complete pair?
Q2) From a team of 6 men and 9 women, what is the probability that 3 men and 2 women are chosen? (Choose 5)
Thanks, so I can go to bed.
Last edited by CSR; Feb 7th, 2009 at 11:59 PM.
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Feb 7th, 2009 11:51 PM
#2

Originally Posted by
CSR
Cant' seem to figure this one out. My friend asked for some help and its bugging me.
Q1) A box has 8 pairs of shoes. 4 socks are randomly selected, what is the probability that there will be exactly one complete pair?
0% since the box had 8 pairs of shoes, not socks.
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Feb 7th, 2009 11:59 PM
#3

Originally Posted by
Hellfire
0% since the box had 8 pairs of shoes, not socks.

Sorry, typo, shoes only.
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Feb 8th, 2009 08:57 AM
#4
Q1) A box has 8 pairs of shoes. 4 Shoes are randomly selected, what is the probability that there will be exactly one complete pair?
(1st, 2nd) + (1st, 3rd) + (1st, 4th) + (2nd, 3rd) + (2nd, 4th) + (3rd, 4th)
(1/8*1/7) + (1/8*1/6) + (1/8*1/5) + (1/7*1/6) + (1/7*1/5)+(1/6*1/5)
?
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Feb 8th, 2009 09:32 AM
#5
Whats the probability that after you graduate, you will wonder why you studied probabilities instead of a skill that would get you a job?
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"Giving money to government is like giving whiskey and car keys to teenage boys."
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Feb 8th, 2009 10:01 AM
#6

Originally Posted by
CSR
Cant' seem to figure this one out. My friend asked for some help and its bugging me.
Q1) A box has 8 pairs of shoes. 4 Shoes are randomly selected, what is the probability that there will be exactly one complete pair?
Q2) From a team of 6 men and 9 women, what is the probability that 3 men and 2 women are chosen? (Choose 5)
Thanks, so I can go to bed.
Q1
1 * (1/15) * (1) * (12/13) = 12/195 or 1 in 16.25
Odds of exactly one completed pair? I dont know for sure Im not very good at math :p
(First shoe can be anything) * (one shoe out of 15 remaining makes the pair) * (third shoe can be any of the remaining 14) * (shoe can be any of the 13 remaining EXCEPT the one that completes the second pair)
Q2
24 / 1001
Probably wrong on Q2
Last edited by EmperorOfCanada; Feb 8th, 2009 at 10:19 AM.
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Feb 8th, 2009 10:04 AM
#7
The male and female should be
(3/15) * (2/15) * (1/15) = Answer
But I was never good at this.
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Feb 8th, 2009 10:14 AM
#8
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Feb 8th, 2009 10:26 AM
#9

Originally Posted by
Jay Hova
Q1) A box has 8 pairs of shoes. 4 Shoes are randomly selected, what is the probability that there will be exactly one complete pair?
(1st, 2nd) + (1st, 3rd) + (1st, 4th) + (2nd, 3rd) + (2nd, 4th) + (3rd, 4th)
(1/8*1/7) + (1/8*1/6) + (1/8*1/5) + (1/7*1/6) + (1/7*1/5)+(1/6*1/5)
?
crap
change it to
(((1 / 16) * 1) / 15) + (((1 / 16) * 1) / 14) + (((1 / 16) * 1) / 13) + (((1 / 15) * 1) / 14) + (((1 / 15) * 1) / 13) + (((1 / 14) * 1) / 13) = 0.0288232601
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Feb 8th, 2009 10:27 AM
#10
actually forget it
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Feb 8th, 2009 10:45 AM
#11
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Feb 8th, 2009 10:48 AM
#12

Originally Posted by
Jon Lai
Use combinatorics.
hahah that was funny. I did some lulz.
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Feb 8th, 2009 10:51 AM
#13

Originally Posted by
Jay Hova
crap
change it to
(((1 / 16) * 1) / 15) + (((1 / 16) * 1) / 14) + (((1 / 16) * 1) / 13) + (((1 / 15) * 1) / 14) + (((1 / 15) * 1) / 13) + (((1 / 14) * 1) / 13) = 0.0288232601
actually, i think this should be right.
picking 4 shoes, we can either have 1-2, 1-3, 1-4, 2-3, 2-4, 3-4.
1-2: 1st shoe is "right" AND (hence the *) 2nd shoe (1/15) is "right"
OR (+)
1-3: 1st shoe is "right" AND (*) the 3rd shoe (1/14)
so on and so forth.
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Feb 8th, 2009 10:57 AM
#14
And if you really want to know....
you could write a VB or any other app..that could solve this.
if you run it for 10,000 to 100,000 times...it'll eventually converge to it's theoretical probability. 
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Feb 8th, 2009 11:20 AM
#15

Originally Posted by
Jay Hova
actually, i think this should be right.
picking 4 shoes, we can either have 1-2, 1-3, 1-4, 2-3, 2-4, 3-4.
1-2: 1st shoe is "right" AND (hence the *) 2nd shoe (1/15) is "right"
OR (+)
1-3: 1st shoe is "right" AND (*) the 3rd shoe (1/14)
so on and so forth.
I really honestly have no idea what you are talking about... Explain why yours is right and mine is wrong?
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