![]() ![]() ![]() This gives you the total number of non-unqiue ways to choose these numbers. Think of this top part of the probability (numerator) as 4p4 since you have 4 numbers to pick from and you want to pick 4 numbers, the number of ways you can pick 4 numbers from 4 numbers is 4*3*2*1. For the lottery question, another way to think of it is as below. I'm having a hard time explaining it all though so would love feedback. Which, as we already have seen, is the same as Only one combination of them is what we have chosen. There are (60 C 4) ways of choosing four numbers (all different) from 60. So your and Sal's ways are the same thing just expressed in different ways. Let's look at the different ways of arriving to the answer. This sounds like a tautology but your intuition is right because it is right. Can anyone do a better job of explaining why my intuition was right? This makes some sense to me, but I tend to come at these things backwards, using intuition and numbers first and then going back over the solution to see whether or not I could claim any larger guiding principles and/or formulas. It seems that I could say this represents the number of possible arrangements of any particular set of 4 numbers distributed over, or divided by, the total number of possible ways to draw all sets of 4 numbers out of 60. I am trying to draw some sort of larger logical conclusion out of this, but am having some trouble not confusing myself. I can just calculate this and get the same answer that Sal gets, but what do I make of the fact that this expression is equal to So the probability of drawing any particular set of 4 numbers out of 60, if we cannot draw any number twice, is: The last two draws follow logically: 2 acceptable numbers out of 58, and then 1 acceptable number out of 57. On my second draw, there are now 3 numbers left that I want, out of a possible 59 total remaining, so I begin to multiply: If I am hoping to draw 4 particular numbers randomly out of 60, then I can say that on my first draw there are 4 numbers that I could hope to get, out of a total of 60, so I begin withĤ/60 as my chances of getting one of those numbers on that first draw. Is there any reason why I could not solve the problem this way? Because I did and it turned out ok, but I don't always trust my own leaps of logic: Likewise you cannot take the number of possible favorable combination outcomes and divide that by the number of total possible permutation outcomes and get the correct probability. You cannot take the number of possible favorable permutation outcomes and divide that by the number of total possible combination outcomes and get the correct probability. You would not say that after moving 3 inches you are halfway to traveling 6 meters, even though 3/6 = ½. This is the same answer we got using permutations.Ĭonsider combinations and permutations to be different “units”. Our answer using combinations would be the number of favorable outcomes/the number of possible outcomes which would be 1/487,635. The number of possible combinations of 4 numbers taken out of 60 different numbers is 60!/((60-4)!*4!). Using combinations, there is only one (1) combination of numbers that gives us that favorable outcome (that one way, achu). ![]() Our answer using permutations would be the number of favorable outcomes/the number of possible outcomes which would be (4*3*2*1)/(60*59*58*57). The total number of different permutations of 4 numbers taken out of 60 different numbers is 60!/((60-4)!), which can be written as 60*59*58*57. Another way to say this is that there are 4! different ways to order the four numbers –or- there are 4! different permutations of the four numbers that give us the favorable outcome. Let us do it both ways, using the permutations first.Īs you mentioned, there a 4! ways of writing the four numbers. He could have taken the number of possible permutations with a favorable outcome and divided that by the total possible number of permutations –or-he could have taken the number of possible combinations with a favorable outcome and divided that by the total number of possible combinations (which is what he did). Sal could have solved this problem in two ways. ![]()
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