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Old 08-02-2007   #1
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Default Probability question about permutations:?

Here's the setup:5 hats are filled with slips of paper. Each hat has a slip of paper with the number 20, a slip with 21, consecutively by integer through 35. This gives each hat 16 numbers.Question:If I draw out a number from each hat at random, what is the chance that the sum of the slips will equal 100 or more by the 3rd hat? Not 3rd, but 4th? Not 4th, but 5th?So far, I have:3 hats, total number of possibilities: 16^3=40964 hats: 16^4=655365 hats: 16^5=1048576I can't figure out how to get the number of permutations (if not combinations).Bonus: What if it was 101, 6 hats? 124, 7 hats?I am looking for a formula to find the answer, not just an add-em-up strategy, as this becomes prohibitive after the 3 hats question, out of the sheer possibilities.
Old 08-02-2007   #2
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Let's start off with reaching 100 by the third hat. You need to enumerate all the ways you could do that. The first number needs to be at least 30.If the first number is 30, there is only one way to get 100. (1)If the first number is 31, there are two ways to get to 100 plus one (above) to get over 100. (3)If the first number is 32, there are three ways to get to 100 plus 3 to get over. (6)If the first number is 33, there are four ways to get to 100 plus 6 to get over. (10)If the first number is 34, there are five ways to get to 100 plus 10 to get over. (15)If the first number is 35, there are six ways to get to 100 plus 15 to get over. (21)In all, there are 56 ways to equal or exceed 100 by adding three numbers between 20 and 35. Your chance of doing it is then 56/4096. That also tells you there are 4040 ways of NOT getting to 100. You can take advantage of that when working on the four hats and so forth on the five hats. I hope this helps.
Old 08-02-2007   #3
death in a blink
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I am not 100% sure, but this should be the answer.Firstly, set the first number being picked. The omitted possibility is 20 to 29, 10 numbers. Thus, 10x16x16 possbilities are omitted. Next, consider 30. Only 35and 35 being picked as 2nd and 3rd number is accepted. Thus I will omit 16x16-1 possibilities. The next is 31. Only 34-35, 35-34 and 35-35 are possible. I omit 16x16-3 possibilities. This continues to apply up to 35, but it follows a pattern of 1 accepted for 30, 1+2 for 31, 1+2+3 for 32 and so on. (you can also work on the possibilities instead of this)Thus the actual working will be like this:Total possibilities=16^3=4096.Total number of ways the numbers do not fulfill criteria=10x16^2+255+253+250+246+241+235=4040Possi ble ways to fulfill criteria=4096-4040 =56Likelihood=56/4096For next part, do the same: narrowing down on the 1st and then 2nd number which fulfill the criteria. For 20, the sum of next 3 is

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