There are 5 pairs of shoes out of which 4 shoes are taken one by one. What is the probability that at least...
My Attempt:
It was simple enough to extract from the question the first step that is 1- P( no pairs are present )
But how do we get P( no pairs)
Several solutions dictate that-
From 10 shoes 1 is taken
And then from 8 shoes 1 is taken and so on......
How do we use all this to get the answer?
probability
asked Jan 18 '17 at 15:50
Sidd
616
|
show 1 more comment
My Attempt:
It was simple enough to extract from the question the first step that is 1- P( no pairs are present )
But how do we get P( no pairs)
Several solutions dictate that-
From 10 shoes 1 is taken
And then from 8 shoes 1 is taken and so on......
How do we use all this to get the answer?
probability
asked Jan 18 '17 at 15:50
Sidd
616
2
calculate the probability that no pair is formed - for the first pick this is 1, when you pick the second there is one shoe that forms a pair, and there are 9 left, therefore there is an 8/9 chance of avoiding making a pair - then if the pair has not formed, you have 8 shoes left to choose from, 2 of which will form a pair with 2 of those already chosen, so there is a 6/8 chance of avoiding forming a pair
– Cato
Jan 18 '17 at 15:52
2
another way of doing it is to consider that to get no pair, you will be left with shoes from 4 of 5 pairs, which there are 5C4 = 5 combinations of. Each shoe could be chosen in 2C1 = 2 ways - then it is 5x2^4 / (10 C 4) = 13 /21 ............ the chance of getting at least a pair is 1-13/21 = 8/21
– Cato
Jan 18 '17 at 16:04
I dont get the 5c4 part. What would happen if that were not taken into account
– Sidd
Jan 18 '17 at 16:34
1
in order to 'fail' to choose a pair, you must have one shoe from 4 different pairs - so if the pairs were called A,B,C,D,E then you could have the left shoes from A,B,C, & D - is one way of doing it - but you could also have combinations of left and right shoes, in $2^4=16$ different ways - now that is just for choosing a set of shoes that excludes E, but different selections could have excluded a different shoe. To answer your question, you would not have counted all possible combinations of shoes where no pair was selected.
– Cato
Jan 18 '17 at 16:53
in other words, in order to have no pair, you must have 4 odd shoes which are from 4 different pairs, and none of the shoes from 1 pair - this can happen in 5 combinations of pairs, each combination of pairs can interchange left and right shoes in 16 ways
– Cato
Jan 18 '17 at 16:55
|
show 1 more comment
My Attempt:
It was simple enough to extract from the question the first step that is 1- P( no pairs are present )
But how do we get P( no pairs)
Several solutions dictate that-
From 10 shoes 1 is taken
And then from 8 shoes 1 is taken and so on......
How do we use all this to get the answer?
probability
asked Jan 18 '17 at 15:50
Sidd
616
My Attempt:
It was simple enough to extract from the question the first step that is 1- P( no pairs are present )
But how do we get P( no pairs)
Several solutions dictate that-
From 10 shoes 1 is taken
And then from 8 shoes 1 is taken and so on......
How do we use all this to get the answer?
probability
probability
asked Jan 18 '17 at 15:50
Sidd
616
asked Jan 18 '17 at 15:50
Sidd
616
asked Jan 18 '17 at 15:50
Sidd
616
asked Jan 18 '17 at 15:50
Sidd
616
asked Jan 18 '17 at 15:50
Sidd
616
616
2
calculate the probability that no pair is formed - for the first pick this is 1, when you pick the second there is one shoe that forms a pair, and there are 9 left, therefore there is an 8/9 chance of avoiding making a pair - then if the pair has not formed, you have 8 shoes left to choose from, 2 of which will form a pair with 2 of those already chosen, so there is a 6/8 chance of avoiding forming a pair
– Cato
Jan 18 '17 at 15:52
2
another way of doing it is to consider that to get no pair, you will be left with shoes from 4 of 5 pairs, which there are 5C4 = 5 combinations of. Each shoe could be chosen in 2C1 = 2 ways - then it is 5x2^4 / (10 C 4) = 13 /21 ............ the chance of getting at least a pair is 1-13/21 = 8/21
– Cato
Jan 18 '17 at 16:04
I dont get the 5c4 part. What would happen if that were not taken into account
– Sidd
Jan 18 '17 at 16:34
1
in order to 'fail' to choose a pair, you must have one shoe from 4 different pairs - so if the pairs were called A,B,C,D,E then you could have the left shoes from A,B,C, & D - is one way of doing it - but you could also have combinations of left and right shoes, in $2^4=16$ different ways - now that is just for choosing a set of shoes that excludes E, but different selections could have excluded a different shoe. To answer your question, you would not have counted all possible combinations of shoes where no pair was selected.
– Cato
Jan 18 '17 at 16:53
in other words, in order to have no pair, you must have 4 odd shoes which are from 4 different pairs, and none of the shoes from 1 pair - this can happen in 5 combinations of pairs, each combination of pairs can interchange left and right shoes in 16 ways
– Cato
Jan 18 '17 at 16:55
|
show 1 more comment
2
calculate the probability that no pair is formed - for the first pick this is 1, when you pick the second there is one shoe that forms a pair, and there are 9 left, therefore there is an 8/9 chance of avoiding making a pair - then if the pair has not formed, you have 8 shoes left to choose from, 2 of which will form a pair with 2 of those already chosen, so there is a 6/8 chance of avoiding forming a pair
– Cato
Jan 18 '17 at 15:52
2
another way of doing it is to consider that to get no pair, you will be left with shoes from 4 of 5 pairs, which there are 5C4 = 5 combinations of. Each shoe could be chosen in 2C1 = 2 ways - then it is 5x2^4 / (10 C 4) = 13 /21 ............ the chance of getting at least a pair is 1-13/21 = 8/21
– Cato
Jan 18 '17 at 16:04
I dont get the 5c4 part. What would happen if that were not taken into account
– Sidd
Jan 18 '17 at 16:34
1
in order to 'fail' to choose a pair, you must have one shoe from 4 different pairs - so if the pairs were called A,B,C,D,E then you could have the left shoes from A,B,C, & D - is one way of doing it - but you could also have combinations of left and right shoes, in $2^4=16$ different ways - now that is just for choosing a set of shoes that excludes E, but different selections could have excluded a different shoe. To answer your question, you would not have counted all possible combinations of shoes where no pair was selected.
– Cato
Jan 18 '17 at 16:53
in other words, in order to have no pair, you must have 4 odd shoes which are from 4 different pairs, and none of the shoes from 1 pair - this can happen in 5 combinations of pairs, each combination of pairs can interchange left and right shoes in 16 ways
– Cato
Jan 18 '17 at 16:55
2
2
calculate the probability that no pair is formed - for the first pick this is 1, when you pick the second there is one shoe that forms a pair, and there are 9 left, therefore there is an 8/9 chance of avoiding making a pair - then if the pair has not formed, you have 8 shoes left to choose from, 2 of which will form a pair with 2 of those already chosen, so there is a 6/8 chance of avoiding forming a pair
– Cato
Jan 18 '17 at 15:52
calculate the probability that no pair is formed - for the first pick this is 1, when you pick the second there is one shoe that forms a pair, and there are 9 left, therefore there is an 8/9 chance of avoiding making a pair - then if the pair has not formed, you have 8 shoes left to choose from, 2 of which will form a pair with 2 of those already chosen, so there is a 6/8 chance of avoiding forming a pair
– Cato
Jan 18 '17 at 15:52
2
2
another way of doing it is to consider that to get no pair, you will be left with shoes from 4 of 5 pairs, which there are 5C4 = 5 combinations of. Each shoe could be chosen in 2C1 = 2 ways - then it is 5x2^4 / (10 C 4) = 13 /21 ............ the chance of getting at least a pair is 1-13/21 = 8/21
– Cato
Jan 18 '17 at 16:04
another way of doing it is to consider that to get no pair, you will be left with shoes from 4 of 5 pairs, which there are 5C4 = 5 combinations of. Each shoe could be chosen in 2C1 = 2 ways - then it is 5x2^4 / (10 C 4) = 13 /21 ............ the chance of getting at least a pair is 1-13/21 = 8/21
– Cato
Jan 18 '17 at 16:04
I dont get the 5c4 part. What would happen if that were not taken into account
– Sidd
Jan 18 '17 at 16:34
I dont get the 5c4 part. What would happen if that were not taken into account
– Sidd
Jan 18 '17 at 16:34
1
1
in order to 'fail' to choose a pair, you must have one shoe from 4 different pairs - so if the pairs were called A,B,C,D,E then you could have the left shoes from A,B,C, & D - is one way of doing it - but you could also have combinations of left and right shoes, in $2^4=16$ different ways - now that is just for choosing a set of shoes that excludes E, but different selections could have excluded a different shoe. To answer your question, you would not have counted all possible combinations of shoes where no pair was selected.
– Cato
Jan 18 '17 at 16:53
in order to 'fail' to choose a pair, you must have one shoe from 4 different pairs - so if the pairs were called A,B,C,D,E then you could have the left shoes from A,B,C, & D - is one way of doing it - but you could also have combinations of left and right shoes, in $2^4=16$ different ways - now that is just for choosing a set of shoes that excludes E, but different selections could have excluded a different shoe. To answer your question, you would not have counted all possible combinations of shoes where no pair was selected.
– Cato
Jan 18 '17 at 16:53
in other words, in order to have no pair, you must have 4 odd shoes which are from 4 different pairs, and none of the shoes from 1 pair - this can happen in 5 combinations of pairs, each combination of pairs can interchange left and right shoes in 16 ways
– Cato
Jan 18 '17 at 16:55
in other words, in order to have no pair, you must have 4 odd shoes which are from 4 different pairs, and none of the shoes from 1 pair - this can happen in 5 combinations of pairs, each combination of pairs can interchange left and right shoes in 16 ways
– Cato
Jan 18 '17 at 16:55
|
show 1 more comment
1 Answer
1
active
oldest
votes
Picking 1st shoe from 10 shoes in 10 ways. 2nd shoe in 9 ways. Similarly 3rd in 8 ways. And 4th in 7 ways.
So we have $10cdot9cdot8cdot7$
Now shoes with no pair.
Picking 1st in 10 ways. 2nd in 8 ways. 3rd in 6 ways. 4th in 4 ways.
We have $10cdot8cdot6cdot4$
Probability of picking at least one pair = $1-frac{10cdot8cdot6cdot4}{10cdot9cdot8cdot7}$
answered Jan 18 '17 at 15:58
Kanwaljit Singh
8,5101516
add a comment |
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1 Answer
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Picking 1st shoe from 10 shoes in 10 ways. 2nd shoe in 9 ways. Similarly 3rd in 8 ways. And 4th in 7 ways.
So we have $10cdot9cdot8cdot7$
Now shoes with no pair.
Picking 1st in 10 ways. 2nd in 8 ways. 3rd in 6 ways. 4th in 4 ways.
We have $10cdot8cdot6cdot4$
Probability of picking at least one pair = $1-frac{10cdot8cdot6cdot4}{10cdot9cdot8cdot7}$
answered Jan 18 '17 at 15:58
Kanwaljit Singh
8,5101516
add a comment |
Picking 1st shoe from 10 shoes in 10 ways. 2nd shoe in 9 ways. Similarly 3rd in 8 ways. And 4th in 7 ways.
So we have $10cdot9cdot8cdot7$
Now shoes with no pair.
Picking 1st in 10 ways. 2nd in 8 ways. 3rd in 6 ways. 4th in 4 ways.
We have $10cdot8cdot6cdot4$
Probability of picking at least one pair = $1-frac{10cdot8cdot6cdot4}{10cdot9cdot8cdot7}$
answered Jan 18 '17 at 15:58
Kanwaljit Singh
8,5101516
add a comment |
Picking 1st shoe from 10 shoes in 10 ways. 2nd shoe in 9 ways. Similarly 3rd in 8 ways. And 4th in 7 ways.
So we have $10cdot9cdot8cdot7$
Now shoes with no pair.
Picking 1st in 10 ways. 2nd in 8 ways. 3rd in 6 ways. 4th in 4 ways.
We have $10cdot8cdot6cdot4$
Probability of picking at least one pair = $1-frac{10cdot8cdot6cdot4}{10cdot9cdot8cdot7}$
answered Jan 18 '17 at 15:58
Kanwaljit Singh
8,5101516
Picking 1st shoe from 10 shoes in 10 ways. 2nd shoe in 9 ways. Similarly 3rd in 8 ways. And 4th in 7 ways.
So we have $10cdot9cdot8cdot7$
Now shoes with no pair.
Picking 1st in 10 ways. 2nd in 8 ways. 3rd in 6 ways. 4th in 4 ways.
We have $10cdot8cdot6cdot4$
Probability of picking at least one pair = $1-frac{10cdot8cdot6cdot4}{10cdot9cdot8cdot7}$
answered Jan 18 '17 at 15:58
Kanwaljit Singh
8,5101516
answered Jan 18 '17 at 15:58
Kanwaljit Singh
8,5101516
answered Jan 18 '17 at 15:58
Kanwaljit Singh
8,5101516
answered Jan 18 '17 at 15:58
Kanwaljit Singh
8,5101516
8,5101516
add a comment |
add a comment |
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2
calculate the probability that no pair is formed - for the first pick this is 1, when you pick the second there is one shoe that forms a pair, and there are 9 left, therefore there is an 8/9 chance of avoiding making a pair - then if the pair has not formed, you have 8 shoes left to choose from, 2 of which will form a pair with 2 of those already chosen, so there is a 6/8 chance of avoiding forming a pair
– Cato
Jan 18 '17 at 15:52
2
another way of doing it is to consider that to get no pair, you will be left with shoes from 4 of 5 pairs, which there are 5C4 = 5 combinations of. Each shoe could be chosen in 2C1 = 2 ways - then it is 5x2^4 / (10 C 4) = 13 /21 ............ the chance of getting at least a pair is 1-13/21 = 8/21
– Cato
Jan 18 '17 at 16:04
I dont get the 5c4 part. What would happen if that were not taken into account
– Sidd
Jan 18 '17 at 16:34
1
in order to 'fail' to choose a pair, you must have one shoe from 4 different pairs - so if the pairs were called A,B,C,D,E then you could have the left shoes from A,B,C, & D - is one way of doing it - but you could also have combinations of left and right shoes, in $2^4=16$ different ways - now that is just for choosing a set of shoes that excludes E, but different selections could have excluded a different shoe. To answer your question, you would not have counted all possible combinations of shoes where no pair was selected.
– Cato
Jan 18 '17 at 16:53
in other words, in order to have no pair, you must have 4 odd shoes which are from 4 different pairs, and none of the shoes from 1 pair - this can happen in 5 combinations of pairs, each combination of pairs can interchange left and right shoes in 16 ways
– Cato
Jan 18 '17 at 16:55