S1 permutations and combinations help
Hi
I would appreciate help on the following questions:
1. The 6 letters of the word LONDON are each written on a card and the 6 letters are shuffled and lined up, face down. The 1st 2 letters are turned over and they spell LO.
Find the probability that when all of the cards are turned over they spell LONDON
and
2. There are 8 competitors in the final of a 100m race. The first 3 competitors to finish the course will receive an accolade. 2 of the competitors are from the 'Spears Athletic club', and the other 6 represent different clubs.
Find the probability that 'Spears' will at least 1 accolade.
I'm not looking for the answers, just some guidelines on how to approach these questions.
Much appreciated.
I would appreciate help on the following questions:
1. The 6 letters of the word LONDON are each written on a card and the 6 letters are shuffled and lined up, face down. The 1st 2 letters are turned over and they spell LO.
Find the probability that when all of the cards are turned over they spell LONDON
and
2. There are 8 competitors in the final of a 100m race. The first 3 competitors to finish the course will receive an accolade. 2 of the competitors are from the 'Spears Athletic club', and the other 6 represent different clubs.
Find the probability that 'Spears' will at least 1 accolade.
I'm not looking for the answers, just some guidelines on how to approach these questions.
Much appreciated.
(edited 12 years ago)
1.
Well, since you don't want the answers exactly I'll use a similar example question with an example solution since I don't really know the theory behind it, I just know 'how to do it'
If you have n symbols there are n! possible rearrangements if the symbols were to be shuffled, where n in the amount of symbols.
For example 5 symbols a b c d e would be 5! (5x4x3x2x1 = 120) = 120 combinations. If you wanted to know the probability of one rearrangement of this it would be 1/120.
Apply the same principle to 4 symbols N D O N and that's your answer
2.
Let's drop all the fancy wording and think of this in a different way. This is just like saying there are 8 balls in a bag numbered 1-8 and 3 are taken out, and the answer is the probability of either (or both) of 2 chosen balls being picked (ball 1 and 2 for example). It's also easier to think of this by 1 being picked at a time, 3 times, instead of all at once. The chance of ball 1 or 2 being picked on the first pick is 2/8. Then the second pick it would be 2/7, and the final pick (if neither have still been picked out), 2/6. The demoninator is decreased by 1 each time because the sample is being reduced by 1 per pick. Just add all the 3 probabilities together and that would be your answer.
Well, since you don't want the answers exactly I'll use a similar example question with an example solution since I don't really know the theory behind it, I just know 'how to do it'
If you have n symbols there are n! possible rearrangements if the symbols were to be shuffled, where n in the amount of symbols.
For example 5 symbols a b c d e would be 5! (5x4x3x2x1 = 120) = 120 combinations. If you wanted to know the probability of one rearrangement of this it would be 1/120.
Apply the same principle to 4 symbols N D O N and that's your answer
2.
Let's drop all the fancy wording and think of this in a different way. This is just like saying there are 8 balls in a bag numbered 1-8 and 3 are taken out, and the answer is the probability of either (or both) of 2 chosen balls being picked (ball 1 and 2 for example). It's also easier to think of this by 1 being picked at a time, 3 times, instead of all at once. The chance of ball 1 or 2 being picked on the first pick is 2/8. Then the second pick it would be 2/7, and the final pick (if neither have still been picked out), 2/6. The demoninator is decreased by 1 each time because the sample is being reduced by 1 per pick. Just add all the 3 probabilities together and that would be your answer.
(edited 12 years ago)
Original post by Lewk
1.
Well, since you don't want the answers exactly I'll use a similar example question with an example solution since I don't really know the theory behind it, I just know 'how to do it'
If you have n symbols there are n! possible rearrangements if the symbols were to be shuffled, where n in the amount of symbols.
For example 5 symbols a b c d e would be 5! (5x4x3x2x1 = 120) = 120 combinations. If you wanted to know the probability of one rearrangement of this it would be 1/120.
Apply the same principle to 4 symbols N D O N and that's your answer
2.
Let's drop all the fancy wording and think of this in a different way. This is just like saying there are 8 balls in a bag numbered 1-8 and 3 are taken out, and the answer is the probability of either (or both) of 2 chosen balls being picked (ball 1 and 2 for example). It's also easier to think of this by 1 being picked at a time, 3 times, instead of all at once. The chance of ball 1 or 2 being picked on the first pick is 2/8. Then the second pick it would be 2/7, and the final pick (if neither have still been picked out), 2/6. The demoninator is decreased by 1 each time because the sample is being reduced by 1 per pick. Just add all the 3 probabilities together and that would be your answer.
Well, since you don't want the answers exactly I'll use a similar example question with an example solution since I don't really know the theory behind it, I just know 'how to do it'
If you have n symbols there are n! possible rearrangements if the symbols were to be shuffled, where n in the amount of symbols.
For example 5 symbols a b c d e would be 5! (5x4x3x2x1 = 120) = 120 combinations. If you wanted to know the probability of one rearrangement of this it would be 1/120.
Apply the same principle to 4 symbols N D O N and that's your answer
2.
Let's drop all the fancy wording and think of this in a different way. This is just like saying there are 8 balls in a bag numbered 1-8 and 3 are taken out, and the answer is the probability of either (or both) of 2 chosen balls being picked (ball 1 and 2 for example). It's also easier to think of this by 1 being picked at a time, 3 times, instead of all at once. The chance of ball 1 or 2 being picked on the first pick is 2/8. Then the second pick it would be 2/7, and the final pick (if neither have still been picked out), 2/6. The demoninator is decreased by 1 each time because the sample is being reduced by 1 per pick. Just add all the 3 probabilities together and that would be your answer.
thanks for your help, but i still don't understand
i guess i will need to know how to do the actual question and how to get the answers
..sorry for being thick lolOriginal post by hatnan
thanks for your help, but i still don't understand
i guess i will need to know how to do the actual question and how to get the answers ..sorry for being thick lol
i guess i will need to know how to do the actual question and how to get the answers
..sorry for being thick lolWell my explanation for 2. isn't very good but i'm no good at explaining things so it's the best i can do for that question :/
for 1. though, do you understand factorials, i.e theres 3x2x1=6 possible rearrangements for 3 symbols, 4x3x2x1=24 for 4, and so on and since you're only looking for 1 specific rearrangements of those 4 symbols (NDON) the probability would be 1/(4x3x2x1) (1 in 24)
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