Range of lottery numbers











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In a choosing 6 from 49 lottery game - I don't want to pick the numbers 48 and 49, so I will only choose numbers between 1 and 47.



What is the probability that all 6 lottery numbers will fall in the range 1-47 ?






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    How many ways can you choose $6$ numbers from $1-47$? How many ways can you choose $6$ numbers from $1-49$?
    – lulu
    3 hours ago















up vote
0
down vote

favorite












In a choosing 6 from 49 lottery game - I don't want to pick the numbers 48 and 49, so I will only choose numbers between 1 and 47.



What is the probability that all 6 lottery numbers will fall in the range 1-47 ?






probability







share|cite|improve this question


















  • 5




    How many ways can you choose $6$ numbers from $1-47$? How many ways can you choose $6$ numbers from $1-49$?
    – lulu
    3 hours ago













up vote
0
down vote

favorite









up vote
0
down vote

favorite











In a choosing 6 from 49 lottery game - I don't want to pick the numbers 48 and 49, so I will only choose numbers between 1 and 47.



What is the probability that all 6 lottery numbers will fall in the range 1-47 ?






probability







share|cite|improve this question













In a choosing 6 from 49 lottery game - I don't want to pick the numbers 48 and 49, so I will only choose numbers between 1 and 47.



What is the probability that all 6 lottery numbers will fall in the range 1-47 ?






probability




probability






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asked 3 hours ago









Frankie139

214




214








  • 5




    How many ways can you choose $6$ numbers from $1-47$? How many ways can you choose $6$ numbers from $1-49$?
    – lulu
    3 hours ago














  • 5




    How many ways can you choose $6$ numbers from $1-47$? How many ways can you choose $6$ numbers from $1-49$?
    – lulu
    3 hours ago








5




5




How many ways can you choose $6$ numbers from $1-47$? How many ways can you choose $6$ numbers from $1-49$?
– lulu
3 hours ago




How many ways can you choose $6$ numbers from $1-47$? How many ways can you choose $6$ numbers from $1-49$?
– lulu
3 hours ago










1 Answer
1






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There are $2$ ways:



First:

Just select $6$ numbers from $47$, so that you can win, in $^{47}C_6$ ways.

Total possible selections will be from $49$, i.e. $^{49}C_6$

Divide both to get probability: $$frac{^{47}C_6}{^{49}C_6}$$
Second:
$4$ of the winning numbers will surely lie in $1$ to $47$. You just have to ensure that rest $2$ numbers also lie in $1$ to $47$.

So, now that $4$ numbers out of $1$ to $47$ are already occupied, you are left with $43$.

There are $^{43}C_2$ ways in which you win.

Divide by total ways, i.e. $^{49}C_2$: $$frac{^{43}C_2}{^{49}C_2}$$



Verify if you get the same result!






share|cite|improve this answer





















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    There are $2$ ways:



    First:

    Just select $6$ numbers from $47$, so that you can win, in $^{47}C_6$ ways.

    Total possible selections will be from $49$, i.e. $^{49}C_6$

    Divide both to get probability: $$frac{^{47}C_6}{^{49}C_6}$$
    Second:
    $4$ of the winning numbers will surely lie in $1$ to $47$. You just have to ensure that rest $2$ numbers also lie in $1$ to $47$.

    So, now that $4$ numbers out of $1$ to $47$ are already occupied, you are left with $43$.

    There are $^{43}C_2$ ways in which you win.

    Divide by total ways, i.e. $^{49}C_2$: $$frac{^{43}C_2}{^{49}C_2}$$



    Verify if you get the same result!






    share|cite|improve this answer

























      up vote
      0
      down vote













      There are $2$ ways:



      First:

      Just select $6$ numbers from $47$, so that you can win, in $^{47}C_6$ ways.

      Total possible selections will be from $49$, i.e. $^{49}C_6$

      Divide both to get probability: $$frac{^{47}C_6}{^{49}C_6}$$
      Second:
      $4$ of the winning numbers will surely lie in $1$ to $47$. You just have to ensure that rest $2$ numbers also lie in $1$ to $47$.

      So, now that $4$ numbers out of $1$ to $47$ are already occupied, you are left with $43$.

      There are $^{43}C_2$ ways in which you win.

      Divide by total ways, i.e. $^{49}C_2$: $$frac{^{43}C_2}{^{49}C_2}$$



      Verify if you get the same result!






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        There are $2$ ways:



        First:

        Just select $6$ numbers from $47$, so that you can win, in $^{47}C_6$ ways.

        Total possible selections will be from $49$, i.e. $^{49}C_6$

        Divide both to get probability: $$frac{^{47}C_6}{^{49}C_6}$$
        Second:
        $4$ of the winning numbers will surely lie in $1$ to $47$. You just have to ensure that rest $2$ numbers also lie in $1$ to $47$.

        So, now that $4$ numbers out of $1$ to $47$ are already occupied, you are left with $43$.

        There are $^{43}C_2$ ways in which you win.

        Divide by total ways, i.e. $^{49}C_2$: $$frac{^{43}C_2}{^{49}C_2}$$



        Verify if you get the same result!






        share|cite|improve this answer












        There are $2$ ways:



        First:

        Just select $6$ numbers from $47$, so that you can win, in $^{47}C_6$ ways.

        Total possible selections will be from $49$, i.e. $^{49}C_6$

        Divide both to get probability: $$frac{^{47}C_6}{^{49}C_6}$$
        Second:
        $4$ of the winning numbers will surely lie in $1$ to $47$. You just have to ensure that rest $2$ numbers also lie in $1$ to $47$.

        So, now that $4$ numbers out of $1$ to $47$ are already occupied, you are left with $43$.

        There are $^{43}C_2$ ways in which you win.

        Divide by total ways, i.e. $^{49}C_2$: $$frac{^{43}C_2}{^{49}C_2}$$



        Verify if you get the same result!







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        omega

        1,7122919




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