Basic, but confusing, counting problem

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SUMMARY

This discussion addresses the probability of selecting exactly 4 correct numbers in a lottery scenario where 6 numbers are drawn from a total of 49. The correct formula for this calculation is (43 Choose 2)(6 Choose 4)/(49 Choose 6). The reasoning behind this involves understanding that (43 Choose 2) represents the selection of 2 incorrect numbers from the 43 non-winning options, while (6 Choose 4) accounts for the selection of 4 winning numbers from the 6 available. This combination is essential for determining the total number of valid lottery tickets that meet the criteria.

PREREQUISITES
  • Understanding of combinatorial mathematics, specifically combinations
  • Familiarity with the concept of probability in lottery scenarios
  • Basic knowledge of the binomial coefficient notation (n Choose k)
  • Ability to interpret mathematical expressions and formulas
NEXT STEPS
  • Study the concept of combinations in depth, focusing on the binomial coefficient
  • Learn about probability theory, particularly in relation to lottery and gambling scenarios
  • Explore intuitive explanations of permutations and combinations using visual aids
  • Review examples of similar counting problems to reinforce understanding
USEFUL FOR

Students studying probability and combinatorics, educators teaching mathematical concepts, and anyone interested in understanding lottery odds and counting principles.

Bre Ntt
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This is a pretty basic counting problem, but it is confusing me to no end. I know the answer (from the back of the book), but I just don't understand the answer.

Homework Statement


Find the probability of getting exactly 4 numbers correct in a lottery where 6 numbers are chosen from 49 numbers (no repetitions.)


The answer is (43 Choose 2)(6 Choose 4)/(49 Choose 6)

(43 Choose 2)(6 Choose 4) is apparently the number of possible tickets with exactly 4 correct answers. But I'm not sure exactly why. I think (43 Choose 2) is the number of possible tickets which have 2 numbers which do not match the winning numbers. Why multiply this by (6 Choose 4) though? I'm thinking it has something to do with considering how many ways you can choose 4 numbers from the 6 winning numbers, but I don't understand why one multiplies it by the number of ways possible tickets which have exactly 2-non-matching numbers.


I can't figure out an intuitive way to grasp this counting process. It almost makes sense to me, but not quite. Can anyone explain this with an intuitive counting argument?
 
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Imagine there are 49 balls in a bag, 6 are red and 43 are blue.
You pick six balls out of the bag.
What are the odds you'll get exactly 4 red balls.
Now do you get it?

There are 43 numbers that are incorrect and six correct numbers, total 49.
2 of your numbers come from the 43 incorrect ones.
4 of your numbers come from the 6 correct ones.

Have a look at:
http://betterexplained.com/articles/easy-permutations-and-combinations/
 
That way of thinking about it helps. Thank you :)
 

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