Out of 11 tickets marked with nos. 1 to 11, 3 tickets are drawn at random. Find the probability that the...
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I kinda tried it and got the answer 5/33.
But I feel like I am doing something wrong.
What I did was:.
Firstly I found favorable outcomes. My favourable outcomes were 25 such triplets. I think I am doing something wrong here only.
Then I found total outcomes by C(11,3).
Then I found the probability.
Please anyone correct me if I am wrong.
probability arithmetic-progressions elementary-probability
asked Jan 1 at 16:34
GarimaGarima
11
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add a comment |
$begingroup$
I kinda tried it and got the answer 5/33.
But I feel like I am doing something wrong.
What I did was:.
Firstly I found favorable outcomes. My favourable outcomes were 25 such triplets. I think I am doing something wrong here only.
Then I found total outcomes by C(11,3).
Then I found the probability.
Please anyone correct me if I am wrong.
probability arithmetic-progressions elementary-probability
asked Jan 1 at 16:34
GarimaGarima
11
$endgroup$
$begingroup$
I agree with you calculations. I missed one in my earlier comment.
$endgroup$
– saulspatz
Jan 1 at 16:40
$begingroup$
@lulu You just beat me to it.
$endgroup$
– saulspatz
Jan 1 at 16:42
add a comment |
$begingroup$
I kinda tried it and got the answer 5/33.
But I feel like I am doing something wrong.
What I did was:.
Firstly I found favorable outcomes. My favourable outcomes were 25 such triplets. I think I am doing something wrong here only.
Then I found total outcomes by C(11,3).
Then I found the probability.
Please anyone correct me if I am wrong.
probability arithmetic-progressions elementary-probability
asked Jan 1 at 16:34
GarimaGarima
11
$endgroup$
I kinda tried it and got the answer 5/33.
But I feel like I am doing something wrong.
What I did was:.
Firstly I found favorable outcomes. My favourable outcomes were 25 such triplets. I think I am doing something wrong here only.
Then I found total outcomes by C(11,3).
Then I found the probability.
Please anyone correct me if I am wrong.
probability arithmetic-progressions elementary-probability
probability arithmetic-progressions elementary-probability
asked Jan 1 at 16:34
GarimaGarima
11
asked Jan 1 at 16:34
GarimaGarima
11
asked Jan 1 at 16:34
GarimaGarima
11
asked Jan 1 at 16:34
GarimaGarima
11
asked Jan 1 at 16:34
GarimaGarima
11
11
$begingroup$
I agree with you calculations. I missed one in my earlier comment.
$endgroup$
– saulspatz
Jan 1 at 16:40
$begingroup$
@lulu You just beat me to it.
$endgroup$
– saulspatz
Jan 1 at 16:42
add a comment |
$begingroup$
I agree with you calculations. I missed one in my earlier comment.
$endgroup$
– saulspatz
Jan 1 at 16:40
$begingroup$
@lulu You just beat me to it.
$endgroup$
– saulspatz
Jan 1 at 16:42
$begingroup$
I agree with you calculations. I missed one in my earlier comment.
$endgroup$
– saulspatz
Jan 1 at 16:40
$begingroup$
I agree with you calculations. I missed one in my earlier comment.
$endgroup$
– saulspatz
Jan 1 at 16:40
$begingroup$
@lulu You just beat me to it.
$endgroup$
– saulspatz
Jan 1 at 16:42
$begingroup$
@lulu You just beat me to it.
$endgroup$
– saulspatz
Jan 1 at 16:42
add a comment |
2 Answers
2
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oldest
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$begingroup$
I expect the hard part is making sure you didn't miss some of the progressions. Let's count them by period.
Period $1$: $(1,2,3), (2,3,4), cdots, (9,10,11),$ so $boxed 9$
Period $2$: $(1,3,5), (2,4,6), cdots , (7,9,11),$ so $boxed 7$
Period $3$: $(1,4,7), (2,5,8), cdots, (5,8,11),$ so $boxed 5$
Period $4$: $(1,5,9), (2,6,10), (3, 7,11),$ so $boxed 3$
Period $5$: $(1,6,11),$ so $boxed 1$
Thus there are $1+3+5+7+9=25$ three term progressions. As there are $binom {11}3=165$ ways to choose three with no restriction, the answer is $$frac {25}{165}=boxed {frac 5{33}}$$
Confirming your result.
answered Jan 1 at 16:41
lulululu
42.5k25080
$endgroup$
$begingroup$
Thank you lulu. You know what it was so much relieving to get the correct answer and to get it confirmed by someone. So thanks again.
$endgroup$
– Garima
Jan 1 at 16:58
add a comment |
$begingroup$
There are ${11choose 3}=165$ ways to choose $3$ tickets. If the middle one has number $min[2..6]$ there are $m-1$ favorable triples, and for $m'in[7..10]$ we obtain the same as for $min[2..5]$. It follows that there are
$$2(1+2+3+4)+5=25$$
favorable triples. The requested probability then is $p={25over165}={5over33}$.
answered Jan 1 at 18:20
Christian BlatterChristian Blatter
174k8115327
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2 Answers
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2 Answers
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active
oldest
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$begingroup$
I expect the hard part is making sure you didn't miss some of the progressions. Let's count them by period.
Period $1$: $(1,2,3), (2,3,4), cdots, (9,10,11),$ so $boxed 9$
Period $2$: $(1,3,5), (2,4,6), cdots , (7,9,11),$ so $boxed 7$
Period $3$: $(1,4,7), (2,5,8), cdots, (5,8,11),$ so $boxed 5$
Period $4$: $(1,5,9), (2,6,10), (3, 7,11),$ so $boxed 3$
Period $5$: $(1,6,11),$ so $boxed 1$
Thus there are $1+3+5+7+9=25$ three term progressions. As there are $binom {11}3=165$ ways to choose three with no restriction, the answer is $$frac {25}{165}=boxed {frac 5{33}}$$
Confirming your result.
answered Jan 1 at 16:41
lulululu
42.5k25080
$endgroup$
$begingroup$
Thank you lulu. You know what it was so much relieving to get the correct answer and to get it confirmed by someone. So thanks again.
$endgroup$
– Garima
Jan 1 at 16:58
add a comment |
$begingroup$
I expect the hard part is making sure you didn't miss some of the progressions. Let's count them by period.
Period $1$: $(1,2,3), (2,3,4), cdots, (9,10,11),$ so $boxed 9$
Period $2$: $(1,3,5), (2,4,6), cdots , (7,9,11),$ so $boxed 7$
Period $3$: $(1,4,7), (2,5,8), cdots, (5,8,11),$ so $boxed 5$
Period $4$: $(1,5,9), (2,6,10), (3, 7,11),$ so $boxed 3$
Period $5$: $(1,6,11),$ so $boxed 1$
Thus there are $1+3+5+7+9=25$ three term progressions. As there are $binom {11}3=165$ ways to choose three with no restriction, the answer is $$frac {25}{165}=boxed {frac 5{33}}$$
Confirming your result.
answered Jan 1 at 16:41
lulululu
42.5k25080
$endgroup$
$begingroup$
Thank you lulu. You know what it was so much relieving to get the correct answer and to get it confirmed by someone. So thanks again.
$endgroup$
– Garima
Jan 1 at 16:58
add a comment |
$begingroup$
I expect the hard part is making sure you didn't miss some of the progressions. Let's count them by period.
Period $1$: $(1,2,3), (2,3,4), cdots, (9,10,11),$ so $boxed 9$
Period $2$: $(1,3,5), (2,4,6), cdots , (7,9,11),$ so $boxed 7$
Period $3$: $(1,4,7), (2,5,8), cdots, (5,8,11),$ so $boxed 5$
Period $4$: $(1,5,9), (2,6,10), (3, 7,11),$ so $boxed 3$
Period $5$: $(1,6,11),$ so $boxed 1$
Thus there are $1+3+5+7+9=25$ three term progressions. As there are $binom {11}3=165$ ways to choose three with no restriction, the answer is $$frac {25}{165}=boxed {frac 5{33}}$$
Confirming your result.
answered Jan 1 at 16:41
lulululu
42.5k25080
$endgroup$
I expect the hard part is making sure you didn't miss some of the progressions. Let's count them by period.
Period $1$: $(1,2,3), (2,3,4), cdots, (9,10,11),$ so $boxed 9$
Period $2$: $(1,3,5), (2,4,6), cdots , (7,9,11),$ so $boxed 7$
Period $3$: $(1,4,7), (2,5,8), cdots, (5,8,11),$ so $boxed 5$
Period $4$: $(1,5,9), (2,6,10), (3, 7,11),$ so $boxed 3$
Period $5$: $(1,6,11),$ so $boxed 1$
Thus there are $1+3+5+7+9=25$ three term progressions. As there are $binom {11}3=165$ ways to choose three with no restriction, the answer is $$frac {25}{165}=boxed {frac 5{33}}$$
Confirming your result.
answered Jan 1 at 16:41
lulululu
42.5k25080
edited Jan 1 at 16:47
answered Jan 1 at 16:41
lulululu
42.5k25080
answered Jan 1 at 16:41
lulululu
42.5k25080
answered Jan 1 at 16:41
lulululu
42.5k25080
42.5k25080
$begingroup$
Thank you lulu. You know what it was so much relieving to get the correct answer and to get it confirmed by someone. So thanks again.
$endgroup$
– Garima
Jan 1 at 16:58
add a comment |
$begingroup$
Thank you lulu. You know what it was so much relieving to get the correct answer and to get it confirmed by someone. So thanks again.
$endgroup$
– Garima
Jan 1 at 16:58
$begingroup$
Thank you lulu. You know what it was so much relieving to get the correct answer and to get it confirmed by someone. So thanks again.
$endgroup$
– Garima
Jan 1 at 16:58
$begingroup$
Thank you lulu. You know what it was so much relieving to get the correct answer and to get it confirmed by someone. So thanks again.
$endgroup$
– Garima
Jan 1 at 16:58
add a comment |
$begingroup$
There are ${11choose 3}=165$ ways to choose $3$ tickets. If the middle one has number $min[2..6]$ there are $m-1$ favorable triples, and for $m'in[7..10]$ we obtain the same as for $min[2..5]$. It follows that there are
$$2(1+2+3+4)+5=25$$
favorable triples. The requested probability then is $p={25over165}={5over33}$.
answered Jan 1 at 18:20
Christian BlatterChristian Blatter
174k8115327
$endgroup$
add a comment |
$begingroup$
There are ${11choose 3}=165$ ways to choose $3$ tickets. If the middle one has number $min[2..6]$ there are $m-1$ favorable triples, and for $m'in[7..10]$ we obtain the same as for $min[2..5]$. It follows that there are
$$2(1+2+3+4)+5=25$$
favorable triples. The requested probability then is $p={25over165}={5over33}$.
answered Jan 1 at 18:20
Christian BlatterChristian Blatter
174k8115327
$endgroup$
add a comment |
$begingroup$
There are ${11choose 3}=165$ ways to choose $3$ tickets. If the middle one has number $min[2..6]$ there are $m-1$ favorable triples, and for $m'in[7..10]$ we obtain the same as for $min[2..5]$. It follows that there are
$$2(1+2+3+4)+5=25$$
favorable triples. The requested probability then is $p={25over165}={5over33}$.
answered Jan 1 at 18:20
Christian BlatterChristian Blatter
174k8115327
$endgroup$
There are ${11choose 3}=165$ ways to choose $3$ tickets. If the middle one has number $min[2..6]$ there are $m-1$ favorable triples, and for $m'in[7..10]$ we obtain the same as for $min[2..5]$. It follows that there are
$$2(1+2+3+4)+5=25$$
favorable triples. The requested probability then is $p={25over165}={5over33}$.
answered Jan 1 at 18:20
Christian BlatterChristian Blatter
174k8115327
answered Jan 1 at 18:20
Christian BlatterChristian Blatter
174k8115327
answered Jan 1 at 18:20
Christian BlatterChristian Blatter
174k8115327
answered Jan 1 at 18:20
Christian BlatterChristian Blatter
174k8115327
174k8115327
add a comment |
add a comment |
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$begingroup$
I agree with you calculations. I missed one in my earlier comment.
$endgroup$
– saulspatz
Jan 1 at 16:40
$begingroup$
@lulu You just beat me to it.
$endgroup$
– saulspatz
Jan 1 at 16:42