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BTG problem 550-700 discussion

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BTG problem 550-700 discussion

by Night reader » Thu Jan 06, 2011 7:39 pm
A box contains 10 pairs of shoes (20 shoes in total). If two shoes are selected at random, what it is the probability that they are matching shoes?

I find it more logical to solve as P(selecting one shoe) 1/20 * P (selecting one matching left shoe) 10/19 OR [+] P(selecting one shoe) 1/20 * P (selecting one matching right shoe) 10/19 = 2*10/(20*19) = 1/19

what's your opinion?

Official explanation follows
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by anshumishra » Thu Jan 06, 2011 7:56 pm
Night reader wrote:A box contains 10 pairs of shoes (20 shoes in total). If two shoes are selected at random, what it is the probability that they are matching shoes?

I find it more logical to solve as P(selecting one shoe) 1/20 * P (selecting one matching left shoe) 10/19 OR [+] P(selecting one shoe) 1/20 * P (selecting one matching right shoe) 10/19 = 2*10/(20*19) = 1/19

what's your opinion?

Official explanation follows
Using Probability theory :
Probability = P(select any shoe)*P(select the matching shoe) = 1*1/19 = 1/19

Using Permutations & combinations :
No. of ways of selecting 2 shoes out of 20 = 20C2
No. of ways of selecting 1 pair from 10 pairs= 10C1

Hence required probability = No. of favorable outcome/Total number of outcomes = 10C1/20C2 = 1/19
Last edited by anshumishra on Thu Jan 06, 2011 8:00 pm, edited 1 time in total.
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by anshumishra » Thu Jan 06, 2011 8:00 pm
Night reader wrote:A box contains 10 pairs of shoes (20 shoes in total). If two shoes are selected at random, what it is the probability that they are matching shoes?

I find it more logical to solve as P(selecting one shoe) 1/20 * P (selecting one matching left shoe) 10/19 OR [+] P(selecting one shoe) 1/20 * P (selecting one matching right shoe) 10/19 = 2*10/(20*19) = 1/19

what's your opinion?
Does this solution assume that any left shoes can be paired with any right shoe ?
Thanks
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by Night reader » Thu Jan 06, 2011 8:07 pm
anshumishra wrote:
Night reader wrote:A box contains 10 pairs of shoes (20 shoes in total). If two shoes are selected at random, what it is the probability that they are matching shoes?

I find it more logical to solve as P(selecting one shoe) 1/20 * P (selecting one matching left shoe) 10/19 OR [+] P(selecting one shoe) 1/20 * P (selecting one matching right shoe) 10/19 = 2*10/(20*19) = 1/19

what's your opinion?
Does this solution assume that any left shoes can be paired with any right shoe ?
partially, the required probability is the sum of two probabilities considering two events i.e. the probability of selecting a pair of right-left shoes and the probability of selecting left-right shoes.
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by anshumishra » Thu Jan 06, 2011 8:16 pm
Night reader wrote:
anshumishra wrote:
Night reader wrote:A box contains 10 pairs of shoes (20 shoes in total). If two shoes are selected at random, what it is the probability that they are matching shoes?

I find it more logical to solve as P(selecting one shoe) 1/20 * P (selecting one matching left shoe) 10/19 OR [+] P(selecting one shoe) 1/20 * P (selecting one matching right shoe) 10/19 = 2*10/(20*19) = 1/19

what's your opinion?
Does this solution assume that any left shoes can be paired with any right shoe ?
partially, the required probability is the sum of two probabilities considering two events i.e. the probability of selecting a pair of right-left shoes and the probability of selecting left-right shoes.
I wouldn't assume that any left shoe can be paired with any right shoe. However, using your approach, here is how I would solve it :

Probability = P(select any left pair of shoe)*P(select the matching right pair) + P(select any right pair of shoe)*P(select the matching left pair) = 10/20*1/19 + 10/20*1/19 = 1/19.
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by Night reader » Thu Jan 06, 2011 8:35 pm
anshumishra wrote:
Night reader wrote:
anshumishra wrote:
Night reader wrote:A box contains 10 pairs of shoes (20 shoes in total). If two shoes are selected at random, what it is the probability that they are matching shoes?

I find it more logical to solve as P(selecting one shoe) 1/20 * P (selecting one matching left shoe) 10/19 OR [+] P(selecting one shoe) 1/20 * P (selecting one matching right shoe) 10/19 = 2*10/(20*19) = 1/19

what's your opinion?
Does this solution assume that any left shoes can be paired with any right shoe ?
partially, the required probability is the sum of two probabilities considering two events i.e. the probability of selecting a pair of right-left shoes and the probability of selecting left-right shoes.
I wouldn't assume that any left shoe can be paired with any right shoe. >>> A box contains 10 pairs of shoes (20 shoes in total)<<< However, using your approach, here is how I would solve it :

Probability = P(select any left pair of shoe)*P(select the matching right pair) + P(select any right pair of shoe)*P(select the matching left pair) = 10/20*1/19 + 10/20*1/19 = 1/19.
oo-o I have to not agree. P(selecting the matching right pair? -may be the matching right item of pair, pair is 2) must be 10/20 with the first selection and 9/19 with the second one... :( where have we gone now
Last edited by Night reader on Thu Jan 06, 2011 8:43 pm, edited 1 time in total.
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by anshumishra » Thu Jan 06, 2011 8:41 pm
Night reader wrote:
anshumishra wrote:
Night reader wrote:
anshumishra wrote:
Night reader wrote:A box contains 10 pairs of shoes (20 shoes in total). If two shoes are selected at random, what it is the probability that they are matching shoes?

I find it more logical to solve as P(selecting one shoe) 1/20 * P (selecting one matching left shoe) 10/19 OR [+] P(selecting one shoe) 1/20 * P (selecting one matching right shoe) 10/19 = 2*10/(20*19) = 1/19

what's your opinion?
Does this solution assume that any left shoes can be paired with any right shoe ?
partially, the required probability is the sum of two probabilities considering two events i.e. the probability of selecting a pair of right-left shoes and the probability of selecting left-right shoes.
I wouldn't assume that any left shoe can be paired with any right shoe. However, using your approach, here is how I would solve it :

Probability = P(select any left pair of shoe)*P(select the matching right pair) + P(select any right pair of shoe)*P(select the matching left pair) = 10/20*1/19 + 10/20*1/19 = 1/19.
that's what I did, may be wording is diff.
Cool ! Nice approach ! Sometimes in hurry it happens (with me it happens a lot). Glad that I learnt a yet another way to solve it.
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by anshumishra » Thu Jan 06, 2011 8:48 pm
Night reader wrote: Probability = P(select any left pair of shoe)*P(select the matching right pair) + P(select any right pair of shoe)*P(select the matching left pair) = 10/20*1/19 + 10/20*1/19 = 1/19.
oo-o I have to not agree. P(selecting the matching right pair? -may be the matching right item of pair, pair is 2) must be 10/20 with the first selection and 9/19 with the second one... :( where have we gone now[/quote]

2 cases :

1st case : 1st draw (left pair) {10/20 10-> left pair , 20->Total} & 2nd draw (right matching pair) {1/19 -> As only one matching right pair is there out of 19 total shoes}
OR
2nd case : 1st draw (right pair) {10/20 10->right pair, 20->Total} & 2nd draw (left matching pair) {1/19 -> As only one matching left pair is there out of 19 total shoes}
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by anshumishra » Thu Jan 06, 2011 8:55 pm
Night reader wrote:
I wouldn't assume that any left shoe can be paired with any right shoe. >>> A box contains 10 pairs of shoes (20 shoes in total)<<< However, using your approach, here is how I would solve it :

Probability = P(select any left pair of shoe)*P(select the matching right pair) + P(select any right pair of shoe)*P(select the matching left pair) = 10/20*1/19 + 10/20*1/19 = 1/19.
oo-o I have to not agree. P(selecting the matching right pair? -may be the matching right item of pair, pair is 2) must be 10/20 with the first selection and 9/19 with the second one... :( where have we gone now


Our solution is similar with one difference. My solution doesn't assume that you can pair any left pair of shoe with any right pair of shoe (you may have one of blck color, the other brown, different brand, style, etc... . Apart from that it is same.
Would be interesting to see how the other people look at it.

Concept-wise we are on same page, I guess.
Thanks
Anshu

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