Lottery Probability Question: probability of exactly $x$ matching numbers
The question follows:
"In a lottery game $6$ balls are drawn randomly from $49$ balls. If you pick $6$ different numbers:
i) What is the probability that your numbers match those drawn?
ii) What is the probability that exactly $x$ of the numbers you choose match?"
So I've done part (i) and got $$frac{1}{binom{49}{6}}$$ which is roughly $7.2E-10$, but I'm not sure on how to approach (ii).
probability combinatorics
edited Nov 26 at 1:26
N. F. Taussig
43.5k93355
asked Nov 26 at 0:51
Jack Vilms
9
add a comment |
The question follows:
"In a lottery game $6$ balls are drawn randomly from $49$ balls. If you pick $6$ different numbers:
i) What is the probability that your numbers match those drawn?
ii) What is the probability that exactly $x$ of the numbers you choose match?"
So I've done part (i) and got $$frac{1}{binom{49}{6}}$$ which is roughly $7.2E-10$, but I'm not sure on how to approach (ii).
probability combinatorics
edited Nov 26 at 1:26
N. F. Taussig
43.5k93355
asked Nov 26 at 0:51
Jack Vilms
9
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Nov 26 at 1:28
add a comment |
The question follows:
"In a lottery game $6$ balls are drawn randomly from $49$ balls. If you pick $6$ different numbers:
i) What is the probability that your numbers match those drawn?
ii) What is the probability that exactly $x$ of the numbers you choose match?"
So I've done part (i) and got $$frac{1}{binom{49}{6}}$$ which is roughly $7.2E-10$, but I'm not sure on how to approach (ii).
probability combinatorics
edited Nov 26 at 1:26
N. F. Taussig
43.5k93355
asked Nov 26 at 0:51
Jack Vilms
9
The question follows:
"In a lottery game $6$ balls are drawn randomly from $49$ balls. If you pick $6$ different numbers:
i) What is the probability that your numbers match those drawn?
ii) What is the probability that exactly $x$ of the numbers you choose match?"
So I've done part (i) and got $$frac{1}{binom{49}{6}}$$ which is roughly $7.2E-10$, but I'm not sure on how to approach (ii).
probability combinatorics
probability combinatorics
edited Nov 26 at 1:26
N. F. Taussig
43.5k93355
asked Nov 26 at 0:51
Jack Vilms
9
edited Nov 26 at 1:26
N. F. Taussig
43.5k93355
asked Nov 26 at 0:51
Jack Vilms
9
edited Nov 26 at 1:26
N. F. Taussig
43.5k93355
edited Nov 26 at 1:26
N. F. Taussig
43.5k93355
edited Nov 26 at 1:26
N. F. Taussig
43.5k93355
43.5k93355
asked Nov 26 at 0:51
Jack Vilms
9
asked Nov 26 at 0:51
Jack Vilms
9
asked Nov 26 at 0:51
Jack Vilms
9
9
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Nov 26 at 1:28
add a comment |
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Nov 26 at 1:28
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Nov 26 at 1:28
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Nov 26 at 1:28
add a comment |
1 Answer
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votes
Your answer to the first question is correct.
Hint: If you match exactly $x$ of the $6$ selected numbers, you must also choose $6 - x$ of the remaining $49 - 6 = 43$ numbers.
answered Nov 26 at 1:31
N. F. Taussig
43.5k93355
1
Alright got the answer as (6 C x)*(43 C 6-x)/(49 C 6). Thanks for the help!
– Jack Vilms
Nov 26 at 2:36
That is correct.
– N. F. Taussig
Nov 26 at 3:11
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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active
oldest
votes
Your answer to the first question is correct.
Hint: If you match exactly $x$ of the $6$ selected numbers, you must also choose $6 - x$ of the remaining $49 - 6 = 43$ numbers.
answered Nov 26 at 1:31
N. F. Taussig
43.5k93355
1
Alright got the answer as (6 C x)*(43 C 6-x)/(49 C 6). Thanks for the help!
– Jack Vilms
Nov 26 at 2:36
That is correct.
– N. F. Taussig
Nov 26 at 3:11
add a comment |
Your answer to the first question is correct.
Hint: If you match exactly $x$ of the $6$ selected numbers, you must also choose $6 - x$ of the remaining $49 - 6 = 43$ numbers.
answered Nov 26 at 1:31
N. F. Taussig
43.5k93355
1
Alright got the answer as (6 C x)*(43 C 6-x)/(49 C 6). Thanks for the help!
– Jack Vilms
Nov 26 at 2:36
That is correct.
– N. F. Taussig
Nov 26 at 3:11
add a comment |
Your answer to the first question is correct.
Hint: If you match exactly $x$ of the $6$ selected numbers, you must also choose $6 - x$ of the remaining $49 - 6 = 43$ numbers.
answered Nov 26 at 1:31
N. F. Taussig
43.5k93355
Your answer to the first question is correct.
Hint: If you match exactly $x$ of the $6$ selected numbers, you must also choose $6 - x$ of the remaining $49 - 6 = 43$ numbers.
answered Nov 26 at 1:31
N. F. Taussig
43.5k93355
answered Nov 26 at 1:31
N. F. Taussig
43.5k93355
answered Nov 26 at 1:31
N. F. Taussig
43.5k93355
answered Nov 26 at 1:31
N. F. Taussig
43.5k93355
43.5k93355
1
Alright got the answer as (6 C x)*(43 C 6-x)/(49 C 6). Thanks for the help!
– Jack Vilms
Nov 26 at 2:36
That is correct.
– N. F. Taussig
Nov 26 at 3:11
add a comment |
1
Alright got the answer as (6 C x)*(43 C 6-x)/(49 C 6). Thanks for the help!
– Jack Vilms
Nov 26 at 2:36
That is correct.
– N. F. Taussig
Nov 26 at 3:11
1
1
Alright got the answer as (6 C x)*(43 C 6-x)/(49 C 6). Thanks for the help!
– Jack Vilms
Nov 26 at 2:36
Alright got the answer as (6 C x)*(43 C 6-x)/(49 C 6). Thanks for the help!
– Jack Vilms
Nov 26 at 2:36
That is correct.
– N. F. Taussig
Nov 26 at 3:11
That is correct.
– N. F. Taussig
Nov 26 at 3:11
add a comment |
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Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Nov 26 at 1:28