Probability with pins in a box
Clash Royale CLAN TAG#URR8PPP
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The question is as follows:
There are 7 red pins, 7 black pins, and 7 green pins in a box. Pick 5 at random without replacement. What is the probability that at least one color was not picked?
My thinking is that this could happen in two ways. Either the 5 pins chosen were all of the same color, or the 5 pins chosen were two different colors. In each case, at least one color is excluded.
Clearly, there are $binom215$ ways of choosing the 5 pins from the box.
In the first case, we choose the pins so that they all have the same color. This equates to $$binom75 + binom75 + binom75= 3binom75.$$
In the second case, the 5 pins are chosen from two colors, excluding one. So, we have $14$ pins to choose from and we choose $5.$ We can either pick from the red and black, the black and green, or the red and green. This then equates to
$$binom145 + binom145 + binom145 = 3binom145.$$
The final answer would then be
$$frac3binom75+3binom145binom215 approx 2982.$$
Would this be the correct answer and the correct approach? Thanks in advance!
probability combinatorics binomial-coefficients
edited Sep 11 at 2:15
N. F. Taussig
42.1k93254
asked Sep 11 at 2:03
StatGuy
509110
add a comment |
up vote
1
down vote
favorite
The question is as follows:
There are 7 red pins, 7 black pins, and 7 green pins in a box. Pick 5 at random without replacement. What is the probability that at least one color was not picked?
My thinking is that this could happen in two ways. Either the 5 pins chosen were all of the same color, or the 5 pins chosen were two different colors. In each case, at least one color is excluded.
Clearly, there are $binom215$ ways of choosing the 5 pins from the box.
In the first case, we choose the pins so that they all have the same color. This equates to $$binom75 + binom75 + binom75= 3binom75.$$
In the second case, the 5 pins are chosen from two colors, excluding one. So, we have $14$ pins to choose from and we choose $5.$ We can either pick from the red and black, the black and green, or the red and green. This then equates to
$$binom145 + binom145 + binom145 = 3binom145.$$
The final answer would then be
$$frac3binom75+3binom145binom215 approx 2982.$$
Would this be the correct answer and the correct approach? Thanks in advance!
probability combinatorics binomial-coefficients
edited Sep 11 at 2:15
N. F. Taussig
42.1k93254
asked Sep 11 at 2:03
StatGuy
509110
4
When you choose $5$ pens from the $14$ red and black pens, you may choose all five to be black, a case which was already counted in $binom75$. So you have some double counting.
– Mike Earnest
Sep 11 at 2:06
Ahhh. Thank you for that comment. I can't believe I missed that.
– StatGuy
Sep 11 at 2:07
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The question is as follows:
There are 7 red pins, 7 black pins, and 7 green pins in a box. Pick 5 at random without replacement. What is the probability that at least one color was not picked?
My thinking is that this could happen in two ways. Either the 5 pins chosen were all of the same color, or the 5 pins chosen were two different colors. In each case, at least one color is excluded.
Clearly, there are $binom215$ ways of choosing the 5 pins from the box.
In the first case, we choose the pins so that they all have the same color. This equates to $$binom75 + binom75 + binom75= 3binom75.$$
In the second case, the 5 pins are chosen from two colors, excluding one. So, we have $14$ pins to choose from and we choose $5.$ We can either pick from the red and black, the black and green, or the red and green. This then equates to
$$binom145 + binom145 + binom145 = 3binom145.$$
The final answer would then be
$$frac3binom75+3binom145binom215 approx 2982.$$
Would this be the correct answer and the correct approach? Thanks in advance!
probability combinatorics binomial-coefficients
edited Sep 11 at 2:15
N. F. Taussig
42.1k93254
asked Sep 11 at 2:03
StatGuy
509110
The question is as follows:
There are 7 red pins, 7 black pins, and 7 green pins in a box. Pick 5 at random without replacement. What is the probability that at least one color was not picked?
My thinking is that this could happen in two ways. Either the 5 pins chosen were all of the same color, or the 5 pins chosen were two different colors. In each case, at least one color is excluded.
Clearly, there are $binom215$ ways of choosing the 5 pins from the box.
In the first case, we choose the pins so that they all have the same color. This equates to $$binom75 + binom75 + binom75= 3binom75.$$
In the second case, the 5 pins are chosen from two colors, excluding one. So, we have $14$ pins to choose from and we choose $5.$ We can either pick from the red and black, the black and green, or the red and green. This then equates to
$$binom145 + binom145 + binom145 = 3binom145.$$
The final answer would then be
$$frac3binom75+3binom145binom215 approx 2982.$$
Would this be the correct answer and the correct approach? Thanks in advance!
probability combinatorics binomial-coefficients
probability combinatorics binomial-coefficients
edited Sep 11 at 2:15
N. F. Taussig
42.1k93254
asked Sep 11 at 2:03
StatGuy
509110
edited Sep 11 at 2:15
N. F. Taussig
42.1k93254
asked Sep 11 at 2:03
StatGuy
509110
edited Sep 11 at 2:15
N. F. Taussig
42.1k93254
edited Sep 11 at 2:15
N. F. Taussig
42.1k93254
edited Sep 11 at 2:15
N. F. Taussig
42.1k93254
42.1k93254
asked Sep 11 at 2:03
StatGuy
509110
asked Sep 11 at 2:03
StatGuy
509110
asked Sep 11 at 2:03
StatGuy
509110
509110
4
When you choose $5$ pens from the $14$ red and black pens, you may choose all five to be black, a case which was already counted in $binom75$. So you have some double counting.
– Mike Earnest
Sep 11 at 2:06
Ahhh. Thank you for that comment. I can't believe I missed that.
– StatGuy
Sep 11 at 2:07
add a comment |
4
When you choose $5$ pens from the $14$ red and black pens, you may choose all five to be black, a case which was already counted in $binom75$. So you have some double counting.
– Mike Earnest
Sep 11 at 2:06
Ahhh. Thank you for that comment. I can't believe I missed that.
– StatGuy
Sep 11 at 2:07
4
4
When you choose $5$ pens from the $14$ red and black pens, you may choose all five to be black, a case which was already counted in $binom75$. So you have some double counting.
– Mike Earnest
Sep 11 at 2:06
When you choose $5$ pens from the $14$ red and black pens, you may choose all five to be black, a case which was already counted in $binom75$. So you have some double counting.
– Mike Earnest
Sep 11 at 2:06
Ahhh. Thank you for that comment. I can't believe I missed that.
– StatGuy
Sep 11 at 2:07
Ahhh. Thank you for that comment. I can't believe I missed that.
– StatGuy
Sep 11 at 2:07
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
There is no need to split into two cases.
Ban a color. There are 3 ways to do this.
Now choose 5 pens from the rest 14 pens. There are $binom145$ ways to do that.
However we must be aware of double counting. If we chose 5 pens from the same color, then it's counted twice. Therefore we must subtract $binom75times3$.
Therefore the probability is $frac3times(binom145-binom75)binom215$
answered Sep 11 at 2:18
abc...
2,177529
Thank you. The question says AT LEAST one color. So, two colors can be banned as well. Does this take that in to account?
– StatGuy
Sep 11 at 2:26
1
Yes. all 5 pens are from the same is counted, and in fact, twice.
– abc...
Sep 11 at 2:50
1
choose 5 pens from 14 pens does not necessarily choose both colors.
– abc...
Sep 11 at 2:51
2
It is ye olde Principle of Inclusion and Exclusion at work once more.
– Graham Kemp
Sep 11 at 3:19
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
There is no need to split into two cases.
Ban a color. There are 3 ways to do this.
Now choose 5 pens from the rest 14 pens. There are $binom145$ ways to do that.
However we must be aware of double counting. If we chose 5 pens from the same color, then it's counted twice. Therefore we must subtract $binom75times3$.
Therefore the probability is $frac3times(binom145-binom75)binom215$
answered Sep 11 at 2:18
abc...
2,177529
Thank you. The question says AT LEAST one color. So, two colors can be banned as well. Does this take that in to account?
– StatGuy
Sep 11 at 2:26
1
Yes. all 5 pens are from the same is counted, and in fact, twice.
– abc...
Sep 11 at 2:50
1
choose 5 pens from 14 pens does not necessarily choose both colors.
– abc...
Sep 11 at 2:51
2
It is ye olde Principle of Inclusion and Exclusion at work once more.
– Graham Kemp
Sep 11 at 3:19
add a comment |
up vote
2
down vote
accepted
There is no need to split into two cases.
Ban a color. There are 3 ways to do this.
Now choose 5 pens from the rest 14 pens. There are $binom145$ ways to do that.
However we must be aware of double counting. If we chose 5 pens from the same color, then it's counted twice. Therefore we must subtract $binom75times3$.
Therefore the probability is $frac3times(binom145-binom75)binom215$
answered Sep 11 at 2:18
abc...
2,177529
Thank you. The question says AT LEAST one color. So, two colors can be banned as well. Does this take that in to account?
– StatGuy
Sep 11 at 2:26
1
Yes. all 5 pens are from the same is counted, and in fact, twice.
– abc...
Sep 11 at 2:50
1
choose 5 pens from 14 pens does not necessarily choose both colors.
– abc...
Sep 11 at 2:51
2
It is ye olde Principle of Inclusion and Exclusion at work once more.
– Graham Kemp
Sep 11 at 3:19
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
There is no need to split into two cases.
Ban a color. There are 3 ways to do this.
Now choose 5 pens from the rest 14 pens. There are $binom145$ ways to do that.
However we must be aware of double counting. If we chose 5 pens from the same color, then it's counted twice. Therefore we must subtract $binom75times3$.
Therefore the probability is $frac3times(binom145-binom75)binom215$
answered Sep 11 at 2:18
abc...
2,177529
There is no need to split into two cases.
Ban a color. There are 3 ways to do this.
Now choose 5 pens from the rest 14 pens. There are $binom145$ ways to do that.
However we must be aware of double counting. If we chose 5 pens from the same color, then it's counted twice. Therefore we must subtract $binom75times3$.
Therefore the probability is $frac3times(binom145-binom75)binom215$
answered Sep 11 at 2:18
abc...
2,177529
answered Sep 11 at 2:18
abc...
2,177529
answered Sep 11 at 2:18
abc...
2,177529
answered Sep 11 at 2:18
abc...
2,177529
2,177529
Thank you. The question says AT LEAST one color. So, two colors can be banned as well. Does this take that in to account?
– StatGuy
Sep 11 at 2:26
1
Yes. all 5 pens are from the same is counted, and in fact, twice.
– abc...
Sep 11 at 2:50
1
choose 5 pens from 14 pens does not necessarily choose both colors.
– abc...
Sep 11 at 2:51
2
It is ye olde Principle of Inclusion and Exclusion at work once more.
– Graham Kemp
Sep 11 at 3:19
add a comment |
Thank you. The question says AT LEAST one color. So, two colors can be banned as well. Does this take that in to account?
– StatGuy
Sep 11 at 2:26
1
Yes. all 5 pens are from the same is counted, and in fact, twice.
– abc...
Sep 11 at 2:50
1
choose 5 pens from 14 pens does not necessarily choose both colors.
– abc...
Sep 11 at 2:51
2
It is ye olde Principle of Inclusion and Exclusion at work once more.
– Graham Kemp
Sep 11 at 3:19
Thank you. The question says AT LEAST one color. So, two colors can be banned as well. Does this take that in to account?
– StatGuy
Sep 11 at 2:26
Thank you. The question says AT LEAST one color. So, two colors can be banned as well. Does this take that in to account?
– StatGuy
Sep 11 at 2:26
1
1
Yes. all 5 pens are from the same is counted, and in fact, twice.
– abc...
Sep 11 at 2:50
Yes. all 5 pens are from the same is counted, and in fact, twice.
– abc...
Sep 11 at 2:50
1
1
choose 5 pens from 14 pens does not necessarily choose both colors.
– abc...
Sep 11 at 2:51
choose 5 pens from 14 pens does not necessarily choose both colors.
– abc...
Sep 11 at 2:51
2
2
It is ye olde Principle of Inclusion and Exclusion at work once more.
– Graham Kemp
Sep 11 at 3:19
It is ye olde Principle of Inclusion and Exclusion at work once more.
– Graham Kemp
Sep 11 at 3:19
add a comment |
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4
When you choose $5$ pens from the $14$ red and black pens, you may choose all five to be black, a case which was already counted in $binom75$. So you have some double counting.
– Mike Earnest
Sep 11 at 2:06
Ahhh. Thank you for that comment. I can't believe I missed that.
– StatGuy
Sep 11 at 2:07