Probability question at least one without replacement.
1
Title says it all, it’s a bit confusing to understand.
A bag contains 13 marbles 6 red, 3 yellow, and 4 green. If 3 marbles are chosen at random (without replacement) then what is the probability that at least one of them is yellow.
This is what I got. 1-(3/13)(2/12)(1/11).
I know that at least one is 1-p. Since it’s asking if 3 are chosen at random without replacement then what’s the probability of 3 yellow, I figured if you picked up a marble out of a bag for the first time your chance of getting a yellow marble would be 3/13 since there’s no replacement it would decrease the second time and the third. So the first chance would be 3/13 the second would be 2/12 and the third would be 1/11
probability combinatorics
asked Dec 4 '18 at 16:14
J.sant
83
add a comment |
1
Title says it all, it’s a bit confusing to understand.
A bag contains 13 marbles 6 red, 3 yellow, and 4 green. If 3 marbles are chosen at random (without replacement) then what is the probability that at least one of them is yellow.
This is what I got. 1-(3/13)(2/12)(1/11).
I know that at least one is 1-p. Since it’s asking if 3 are chosen at random without replacement then what’s the probability of 3 yellow, I figured if you picked up a marble out of a bag for the first time your chance of getting a yellow marble would be 3/13 since there’s no replacement it would decrease the second time and the third. So the first chance would be 3/13 the second would be 2/12 and the third would be 1/11
probability combinatorics
asked Dec 4 '18 at 16:14
J.sant
83
math.stackexchange.com/questions/697433/…
– lab bhattacharjee
Dec 4 '18 at 16:19
What have you tried so far?
– user3482749
Dec 4 '18 at 16:39
This is what I got. 1-(3/13)*(2/12)*(1/11). Is that correct
– J.sant
Dec 4 '18 at 16:41
I was asking for a thought process, not a number.
– user3482749
Dec 4 '18 at 16:51
Well, I know that at least one is 1-p. Since it’s asking if 3 are chosen at random without replacement then what’s the probability of 3 yellow, I figured if you picked up a marble out of a bag for the first time your chance of getting a yellow marble would be 3/13 since there’s no replacement it would decrease the second time and the third. So the first chance would be 3/13 the second would be 2/12 and the third would be 1/11
– J.sant
Dec 4 '18 at 16:58
add a comment |
1
1
1
Title says it all, it’s a bit confusing to understand.
A bag contains 13 marbles 6 red, 3 yellow, and 4 green. If 3 marbles are chosen at random (without replacement) then what is the probability that at least one of them is yellow.
This is what I got. 1-(3/13)(2/12)(1/11).
I know that at least one is 1-p. Since it’s asking if 3 are chosen at random without replacement then what’s the probability of 3 yellow, I figured if you picked up a marble out of a bag for the first time your chance of getting a yellow marble would be 3/13 since there’s no replacement it would decrease the second time and the third. So the first chance would be 3/13 the second would be 2/12 and the third would be 1/11
probability combinatorics
asked Dec 4 '18 at 16:14
J.sant
83
Title says it all, it’s a bit confusing to understand.
A bag contains 13 marbles 6 red, 3 yellow, and 4 green. If 3 marbles are chosen at random (without replacement) then what is the probability that at least one of them is yellow.
This is what I got. 1-(3/13)(2/12)(1/11).
I know that at least one is 1-p. Since it’s asking if 3 are chosen at random without replacement then what’s the probability of 3 yellow, I figured if you picked up a marble out of a bag for the first time your chance of getting a yellow marble would be 3/13 since there’s no replacement it would decrease the second time and the third. So the first chance would be 3/13 the second would be 2/12 and the third would be 1/11
probability combinatorics
probability combinatorics
asked Dec 4 '18 at 16:14
J.sant
83
asked Dec 4 '18 at 16:14
J.sant
83
edited Dec 4 '18 at 17:29
asked Dec 4 '18 at 16:14
J.sant
83
asked Dec 4 '18 at 16:14
J.sant
83
asked Dec 4 '18 at 16:14
J.sant
83
83
math.stackexchange.com/questions/697433/…
– lab bhattacharjee
Dec 4 '18 at 16:19
What have you tried so far?
– user3482749
Dec 4 '18 at 16:39
This is what I got. 1-(3/13)*(2/12)*(1/11). Is that correct
– J.sant
Dec 4 '18 at 16:41
I was asking for a thought process, not a number.
– user3482749
Dec 4 '18 at 16:51
Well, I know that at least one is 1-p. Since it’s asking if 3 are chosen at random without replacement then what’s the probability of 3 yellow, I figured if you picked up a marble out of a bag for the first time your chance of getting a yellow marble would be 3/13 since there’s no replacement it would decrease the second time and the third. So the first chance would be 3/13 the second would be 2/12 and the third would be 1/11
– J.sant
Dec 4 '18 at 16:58
add a comment |
math.stackexchange.com/questions/697433/…
– lab bhattacharjee
Dec 4 '18 at 16:19
What have you tried so far?
– user3482749
Dec 4 '18 at 16:39
This is what I got. 1-(3/13)*(2/12)*(1/11). Is that correct
– J.sant
Dec 4 '18 at 16:41
I was asking for a thought process, not a number.
– user3482749
Dec 4 '18 at 16:51
Well, I know that at least one is 1-p. Since it’s asking if 3 are chosen at random without replacement then what’s the probability of 3 yellow, I figured if you picked up a marble out of a bag for the first time your chance of getting a yellow marble would be 3/13 since there’s no replacement it would decrease the second time and the third. So the first chance would be 3/13 the second would be 2/12 and the third would be 1/11
– J.sant
Dec 4 '18 at 16:58
math.stackexchange.com/questions/697433/…
– lab bhattacharjee
Dec 4 '18 at 16:19
math.stackexchange.com/questions/697433/…
– lab bhattacharjee
Dec 4 '18 at 16:19
What have you tried so far?
– user3482749
Dec 4 '18 at 16:39
What have you tried so far?
– user3482749
Dec 4 '18 at 16:39
This is what I got. 1-(3/13)*(2/12)*(1/11). Is that correct
– J.sant
Dec 4 '18 at 16:41
This is what I got. 1-(3/13)*(2/12)*(1/11). Is that correct
– J.sant
Dec 4 '18 at 16:41
I was asking for a thought process, not a number.
– user3482749
Dec 4 '18 at 16:51
I was asking for a thought process, not a number.
– user3482749
Dec 4 '18 at 16:51
Well, I know that at least one is 1-p. Since it’s asking if 3 are chosen at random without replacement then what’s the probability of 3 yellow, I figured if you picked up a marble out of a bag for the first time your chance of getting a yellow marble would be 3/13 since there’s no replacement it would decrease the second time and the third. So the first chance would be 3/13 the second would be 2/12 and the third would be 1/11
– J.sant
Dec 4 '18 at 16:58
Well, I know that at least one is 1-p. Since it’s asking if 3 are chosen at random without replacement then what’s the probability of 3 yellow, I figured if you picked up a marble out of a bag for the first time your chance of getting a yellow marble would be 3/13 since there’s no replacement it would decrease the second time and the third. So the first chance would be 3/13 the second would be 2/12 and the third would be 1/11
– J.sant
Dec 4 '18 at 16:58
add a comment |
1 Answer
1
active
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HINT
- Find the probability that none are yellow.
- Remember that if $A$ is any event, then $mathbb{P}[A] = 1-mathbb{P}[bar{A}]$
UPDATE
- The first time you grab a yellow ball with a chance of 3/13. You want the negation of this, so the chance of that is $1-3/13 = 10/30$.
- The second time, you grab a yellow (assuming first time you did not) with a chance 3/12. Third time it will be 3/11.
- You want to draw yellow in neither of the tries, the probability of that is $$
left[1 - frac{3}{13}right]
times left[1 - frac{3}{12}right]
times left[1 - frac{3}{11}right]
= frac{10 cdot 9 cdot 8}{13 cdot 12 cdot 11}
= frac{60}{143}
$$
- Finally you want the probability that any draw results in a yellow, which is $$1 - frac{60}{143} = frac{83}{143} approx 58%.$$
answered Dec 4 '18 at 16:15
gt6989b
33.1k22452
This is what I got. 1-(3/13)*(2/12)*(1/11). Is that correct?
– J.sant
Dec 4 '18 at 16:35
1
@J.sant No, you have to calculate $1-P("color{blue}{textrm{No yellow}} textrm{marble is chosen}")$. The probability is approximately $58%$
– callculus
Dec 4 '18 at 17:35
1
So 1-10/13. That’s the probability of none yellow. Is that correct?
– J.sant
Dec 4 '18 at 18:22
1
@J.sant Not really. You chose your first non-yellow marble: $frac{10}{13}$. Then you choose your second non-yellow marble: $frac{9}{12}$. And lastly you choose your third non-yellow marble: $frac{8}{11}$. Now subtract the product from $1$.
– callculus
Dec 4 '18 at 18:53
I’m confused on how to subtract 1- a fraction 1-10/13*9/12*8/11 do you multiply the numerator and denomination then subtract the 1??
– J.sant
Dec 4 '18 at 20:01
|
show 1 more comment
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1 Answer
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1 Answer
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1
HINT
- Find the probability that none are yellow.
- Remember that if $A$ is any event, then $mathbb{P}[A] = 1-mathbb{P}[bar{A}]$
UPDATE
- The first time you grab a yellow ball with a chance of 3/13. You want the negation of this, so the chance of that is $1-3/13 = 10/30$.
- The second time, you grab a yellow (assuming first time you did not) with a chance 3/12. Third time it will be 3/11.
- You want to draw yellow in neither of the tries, the probability of that is $$
left[1 - frac{3}{13}right]
times left[1 - frac{3}{12}right]
times left[1 - frac{3}{11}right]
= frac{10 cdot 9 cdot 8}{13 cdot 12 cdot 11}
= frac{60}{143}
$$
- Finally you want the probability that any draw results in a yellow, which is $$1 - frac{60}{143} = frac{83}{143} approx 58%.$$
answered Dec 4 '18 at 16:15
gt6989b
33.1k22452
This is what I got. 1-(3/13)*(2/12)*(1/11). Is that correct?
– J.sant
Dec 4 '18 at 16:35
1
@J.sant No, you have to calculate $1-P("color{blue}{textrm{No yellow}} textrm{marble is chosen}")$. The probability is approximately $58%$
– callculus
Dec 4 '18 at 17:35
1
So 1-10/13. That’s the probability of none yellow. Is that correct?
– J.sant
Dec 4 '18 at 18:22
1
@J.sant Not really. You chose your first non-yellow marble: $frac{10}{13}$. Then you choose your second non-yellow marble: $frac{9}{12}$. And lastly you choose your third non-yellow marble: $frac{8}{11}$. Now subtract the product from $1$.
– callculus
Dec 4 '18 at 18:53
I’m confused on how to subtract 1- a fraction 1-10/13*9/12*8/11 do you multiply the numerator and denomination then subtract the 1??
– J.sant
Dec 4 '18 at 20:01
|
show 1 more comment
1
HINT
- Find the probability that none are yellow.
- Remember that if $A$ is any event, then $mathbb{P}[A] = 1-mathbb{P}[bar{A}]$
UPDATE
- The first time you grab a yellow ball with a chance of 3/13. You want the negation of this, so the chance of that is $1-3/13 = 10/30$.
- The second time, you grab a yellow (assuming first time you did not) with a chance 3/12. Third time it will be 3/11.
- You want to draw yellow in neither of the tries, the probability of that is $$
left[1 - frac{3}{13}right]
times left[1 - frac{3}{12}right]
times left[1 - frac{3}{11}right]
= frac{10 cdot 9 cdot 8}{13 cdot 12 cdot 11}
= frac{60}{143}
$$
- Finally you want the probability that any draw results in a yellow, which is $$1 - frac{60}{143} = frac{83}{143} approx 58%.$$
answered Dec 4 '18 at 16:15
gt6989b
33.1k22452
This is what I got. 1-(3/13)*(2/12)*(1/11). Is that correct?
– J.sant
Dec 4 '18 at 16:35
1
@J.sant No, you have to calculate $1-P("color{blue}{textrm{No yellow}} textrm{marble is chosen}")$. The probability is approximately $58%$
– callculus
Dec 4 '18 at 17:35
1
So 1-10/13. That’s the probability of none yellow. Is that correct?
– J.sant
Dec 4 '18 at 18:22
1
@J.sant Not really. You chose your first non-yellow marble: $frac{10}{13}$. Then you choose your second non-yellow marble: $frac{9}{12}$. And lastly you choose your third non-yellow marble: $frac{8}{11}$. Now subtract the product from $1$.
– callculus
Dec 4 '18 at 18:53
I’m confused on how to subtract 1- a fraction 1-10/13*9/12*8/11 do you multiply the numerator and denomination then subtract the 1??
– J.sant
Dec 4 '18 at 20:01
|
show 1 more comment
1
1
1
HINT
- Find the probability that none are yellow.
- Remember that if $A$ is any event, then $mathbb{P}[A] = 1-mathbb{P}[bar{A}]$
UPDATE
- The first time you grab a yellow ball with a chance of 3/13. You want the negation of this, so the chance of that is $1-3/13 = 10/30$.
- The second time, you grab a yellow (assuming first time you did not) with a chance 3/12. Third time it will be 3/11.
- You want to draw yellow in neither of the tries, the probability of that is $$
left[1 - frac{3}{13}right]
times left[1 - frac{3}{12}right]
times left[1 - frac{3}{11}right]
= frac{10 cdot 9 cdot 8}{13 cdot 12 cdot 11}
= frac{60}{143}
$$
- Finally you want the probability that any draw results in a yellow, which is $$1 - frac{60}{143} = frac{83}{143} approx 58%.$$
answered Dec 4 '18 at 16:15
gt6989b
33.1k22452
HINT
- Find the probability that none are yellow.
- Remember that if $A$ is any event, then $mathbb{P}[A] = 1-mathbb{P}[bar{A}]$
UPDATE
- The first time you grab a yellow ball with a chance of 3/13. You want the negation of this, so the chance of that is $1-3/13 = 10/30$.
- The second time, you grab a yellow (assuming first time you did not) with a chance 3/12. Third time it will be 3/11.
- You want to draw yellow in neither of the tries, the probability of that is $$
left[1 - frac{3}{13}right]
times left[1 - frac{3}{12}right]
times left[1 - frac{3}{11}right]
= frac{10 cdot 9 cdot 8}{13 cdot 12 cdot 11}
= frac{60}{143}
$$
- Finally you want the probability that any draw results in a yellow, which is $$1 - frac{60}{143} = frac{83}{143} approx 58%.$$
answered Dec 4 '18 at 16:15
gt6989b
33.1k22452
edited Dec 4 '18 at 20:25
answered Dec 4 '18 at 16:15
gt6989b
33.1k22452
answered Dec 4 '18 at 16:15
gt6989b
33.1k22452
answered Dec 4 '18 at 16:15
gt6989b
33.1k22452
33.1k22452
This is what I got. 1-(3/13)*(2/12)*(1/11). Is that correct?
– J.sant
Dec 4 '18 at 16:35
1
@J.sant No, you have to calculate $1-P("color{blue}{textrm{No yellow}} textrm{marble is chosen}")$. The probability is approximately $58%$
– callculus
Dec 4 '18 at 17:35
1
So 1-10/13. That’s the probability of none yellow. Is that correct?
– J.sant
Dec 4 '18 at 18:22
1
@J.sant Not really. You chose your first non-yellow marble: $frac{10}{13}$. Then you choose your second non-yellow marble: $frac{9}{12}$. And lastly you choose your third non-yellow marble: $frac{8}{11}$. Now subtract the product from $1$.
– callculus
Dec 4 '18 at 18:53
I’m confused on how to subtract 1- a fraction 1-10/13*9/12*8/11 do you multiply the numerator and denomination then subtract the 1??
– J.sant
Dec 4 '18 at 20:01
|
show 1 more comment
This is what I got. 1-(3/13)*(2/12)*(1/11). Is that correct?
– J.sant
Dec 4 '18 at 16:35
1
@J.sant No, you have to calculate $1-P("color{blue}{textrm{No yellow}} textrm{marble is chosen}")$. The probability is approximately $58%$
– callculus
Dec 4 '18 at 17:35
1
So 1-10/13. That’s the probability of none yellow. Is that correct?
– J.sant
Dec 4 '18 at 18:22
1
@J.sant Not really. You chose your first non-yellow marble: $frac{10}{13}$. Then you choose your second non-yellow marble: $frac{9}{12}$. And lastly you choose your third non-yellow marble: $frac{8}{11}$. Now subtract the product from $1$.
– callculus
Dec 4 '18 at 18:53
I’m confused on how to subtract 1- a fraction 1-10/13*9/12*8/11 do you multiply the numerator and denomination then subtract the 1??
– J.sant
Dec 4 '18 at 20:01
This is what I got. 1-(3/13)*(2/12)*(1/11). Is that correct?
– J.sant
Dec 4 '18 at 16:35
This is what I got. 1-(3/13)*(2/12)*(1/11). Is that correct?
– J.sant
Dec 4 '18 at 16:35
1
1
@J.sant No, you have to calculate $1-P("color{blue}{textrm{No yellow}} textrm{marble is chosen}")$. The probability is approximately $58%$
– callculus
Dec 4 '18 at 17:35
@J.sant No, you have to calculate $1-P("color{blue}{textrm{No yellow}} textrm{marble is chosen}")$. The probability is approximately $58%$
– callculus
Dec 4 '18 at 17:35
1
1
So 1-10/13. That’s the probability of none yellow. Is that correct?
– J.sant
Dec 4 '18 at 18:22
So 1-10/13. That’s the probability of none yellow. Is that correct?
– J.sant
Dec 4 '18 at 18:22
1
1
@J.sant Not really. You chose your first non-yellow marble: $frac{10}{13}$. Then you choose your second non-yellow marble: $frac{9}{12}$. And lastly you choose your third non-yellow marble: $frac{8}{11}$. Now subtract the product from $1$.
– callculus
Dec 4 '18 at 18:53
@J.sant Not really. You chose your first non-yellow marble: $frac{10}{13}$. Then you choose your second non-yellow marble: $frac{9}{12}$. And lastly you choose your third non-yellow marble: $frac{8}{11}$. Now subtract the product from $1$.
– callculus
Dec 4 '18 at 18:53
I’m confused on how to subtract 1- a fraction 1-10/13*9/12*8/11 do you multiply the numerator and denomination then subtract the 1??
– J.sant
Dec 4 '18 at 20:01
I’m confused on how to subtract 1- a fraction 1-10/13*9/12*8/11 do you multiply the numerator and denomination then subtract the 1??
– J.sant
Dec 4 '18 at 20:01
|
show 1 more comment
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math.stackexchange.com/questions/697433/…
– lab bhattacharjee
Dec 4 '18 at 16:19
What have you tried so far?
– user3482749
Dec 4 '18 at 16:39
This is what I got. 1-(3/13)*(2/12)*(1/11). Is that correct
– J.sant
Dec 4 '18 at 16:41
I was asking for a thought process, not a number.
– user3482749
Dec 4 '18 at 16:51
Well, I know that at least one is 1-p. Since it’s asking if 3 are chosen at random without replacement then what’s the probability of 3 yellow, I figured if you picked up a marble out of a bag for the first time your chance of getting a yellow marble would be 3/13 since there’s no replacement it would decrease the second time and the third. So the first chance would be 3/13 the second would be 2/12 and the third would be 1/11
– J.sant
Dec 4 '18 at 16:58