To win a prize, a person must select 3 numbers from 13 numbers. Find the probability of winning if a person buys one ticket. (Note: The numbers can be selected in any order.) Round to four decimal places, if necessary.

Question

texaschic101 · Answer

I am not sure....I keep wanting to say 3/13...but I do not know for sure. If that is correct, 3/13 = .2308 ???

texaschic101 · Answer

@genius12 .....is this correct ?

texaschic101 · Answer

@KingGeorge ...did I do this right ?

kinggeorge · Answer

I don't think that's the correct to it. It should be significantly less than 3/13. Although it's a fine place to start.

anonymous · Answer

@schwanna13 This is a simple "Combinations" problem. As soon as you something like "numbers can be selected in any order", you know it involves combinatorics of some kind. Here we win by choosing a combination of 3 specific numbers from a total of 13. There is only 1 combination that will get us the win. The total number of combinations is given by:$$\bf \left(\begin{matrix}13 \ 3\end{matrix}\right)=286$$This means that a player can "choose" 286 unique combinations of 3 numbers. Since there is just 1 combination that gives us the win out of a total of 286 possible combinations, the probability is given by:$$\bf P(winning)=\frac{ n(A) }{ n(S) }=\frac{ 1 }{ 286 }$$

@schwanna13 
@KingGeorge 
@texaschic101

texaschic101 · Answer

I had a feeling somebody else would get this....thanks for teaching me this :)