what is the probability of winning a lottery which requires selection of 5 numbers from1 to 40

Calculate the numerator: There is only one winning combination. So the numerator of the desired probability is 1. Calculate the denominator: There are "40 numbers, choose 5", or 40C5 possible combinations. So the denominator of the desired probability is 40C5. So the probability is 1/40C5=1/658008

Answer: 1/658008

For this kind of probability problem, assuming the number is removed from play once chosen, the probability increases by one each time.

$$\left({\frac{{\mathtt{1}}}{{\mathtt{40}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{39}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{38}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{37}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{36}}}}\right)$$

The bottom five numbers multiply to 78960960, which means that the odds are 1/78960960

Mine was assuming the numbers can be reused

The number of 5 digit combinations (order doesn't matter) from 40 is

40C5

$${\left({\frac{{\mathtt{40}}{!}}{{\mathtt{5}}{!}{\mathtt{\,\times\,}}({\mathtt{40}}{\mathtt{\,-\,}}{\mathtt{5}}){!}}}\right)} = {\mathtt{658\,008}}$$

only one of those wins so the chance is     $$\frac{1}{658008}$$

Just like FoxTears said!

Thanks Foxtears.

Foxtears, I don't think that yours was assuming that the numbers could be reused.

It was just assuming that the order they were chosen in was of no consequence.

GoldenLeaf I believe yours would have been correct if the numbers had to be chosen in a specific order.

GoldenLeaf's approach was almost right, but he should have had 5, 4, 3, 2 and 1 in the numerators.

There are 5 possible choices for the first number, so probability of this is 5/40.  For each of these

there are 4 possible choices for the 2nd number, so probability of both 1st and 2nd is (5/40)*(4/39).  For each of these ... etc. to get the overall probability as

$${\mathtt{p}} = \left({\frac{{\mathtt{5}}}{{\mathtt{40}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{4}}}{{\mathtt{39}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{38}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{2}}}{{\mathtt{37}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{36}}}}\right) \Rightarrow {\mathtt{p}} = {\mathtt{0.000\: \!001\: \!519\: \!738\: \!361\: \!8}}$$

or p≈1.52*10^-6  

1/p = 658008

Yes of course - that did not occur to me.  Thanks Alan.

Calling Bertie, CPhill and Rom (Troll alert! General quarters!)

Along with Melody and Alan, we need input from Bertie, CPhill, and Rom, too.

With the top brains represented here that Troll might comeback and bite one or more.

I know he’ll bite FoxTears for saying, “Mine was assuming the numbers can be reused” (your formula (40C5) contradicts this).

The Troll will follow-up, by asking what is the probability of (something) of (five, six or seven) mathematicians getting a probability question correct.

Then he’ll say about the same as walking, blindfolded, through a large buffalo heard and stepping in four piles with both feet.

What’s annoying: he’s usually right.

CPhill might not be able to make it. He might be trying to get the “Roman Zero” from inside Sisyphus’s boulder.

Maybe he could take a break and visit. He’s not been around for a day or two. I hope he wasn’t squashed by the boulder.

No matter: cartoon physics seem to apply in that realm. If he does visit, maybe one of those angels will help him get the “Roman Zero.”

By: Troll Detector