Probability of flood.
$begingroup$
I was solving a question of probability and the question was " If the probability of occurrence of flood in any year is 1/20 then what is probability that flood will occur in 3rd year exactly?
This is easy to solve and the answer that I got is 361/8000. But then one question strikes my mind what if exactly is not written? What if we need to find the probability that flood will occur in 3rd year?
On the previous question my approach was 19/20 × 19/20 × 1/20. But I am sure that this can not be applied to latter question. I think this will be solved by probability distribution however I am not sure about this. If anyone know how to solve this please provide guidance.
probability probability-distributions
asked Jan 11 at 7:36
Kishan ThackerKishan Thacker
214
$endgroup$
add a comment |
$begingroup$
I was solving a question of probability and the question was " If the probability of occurrence of flood in any year is 1/20 then what is probability that flood will occur in 3rd year exactly?
This is easy to solve and the answer that I got is 361/8000. But then one question strikes my mind what if exactly is not written? What if we need to find the probability that flood will occur in 3rd year?
On the previous question my approach was 19/20 × 19/20 × 1/20. But I am sure that this can not be applied to latter question. I think this will be solved by probability distribution however I am not sure about this. If anyone know how to solve this please provide guidance.
probability probability-distributions
asked Jan 11 at 7:36
Kishan ThackerKishan Thacker
214
$endgroup$
add a comment |
$begingroup$
I was solving a question of probability and the question was " If the probability of occurrence of flood in any year is 1/20 then what is probability that flood will occur in 3rd year exactly?
This is easy to solve and the answer that I got is 361/8000. But then one question strikes my mind what if exactly is not written? What if we need to find the probability that flood will occur in 3rd year?
On the previous question my approach was 19/20 × 19/20 × 1/20. But I am sure that this can not be applied to latter question. I think this will be solved by probability distribution however I am not sure about this. If anyone know how to solve this please provide guidance.
probability probability-distributions
asked Jan 11 at 7:36
Kishan ThackerKishan Thacker
214
$endgroup$
I was solving a question of probability and the question was " If the probability of occurrence of flood in any year is 1/20 then what is probability that flood will occur in 3rd year exactly?
This is easy to solve and the answer that I got is 361/8000. But then one question strikes my mind what if exactly is not written? What if we need to find the probability that flood will occur in 3rd year?
On the previous question my approach was 19/20 × 19/20 × 1/20. But I am sure that this can not be applied to latter question. I think this will be solved by probability distribution however I am not sure about this. If anyone know how to solve this please provide guidance.
probability probability-distributions
probability probability-distributions
asked Jan 11 at 7:36
Kishan ThackerKishan Thacker
214
asked Jan 11 at 7:36
Kishan ThackerKishan Thacker
214
edited Jan 11 at 9:02
Kishan Thacker
asked Jan 11 at 7:36
Kishan ThackerKishan Thacker
214
asked Jan 11 at 7:36
Kishan ThackerKishan Thacker
214
asked Jan 11 at 7:36
Kishan ThackerKishan Thacker
214
214
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In any year the probability is $frac1{20}$, so also for the $3$rd year.
answered Jan 11 at 8:21
drhabdrhab
103k545136
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add a comment |
$begingroup$
First of all, in your previous computation your approach is correct but the result is wrong:
$$19/20 cdot 19/20 cdot 1/20=frac{361}{8000}.$$
As regards your variation, if you would like to evaluate the probability that the flood occurs within the third year, i.e. the first year or the second year or the third year, then it should be:
$$1/20+(19/20cdot 1/20)+(19/20 cdot 19/20 cdot 1/20)=frac{1141}{8000}.$$
Note that the probability that the flood occurs the third year (we remove the word "exactly"), or any other year, is just $1/20$.
answered Jan 11 at 8:01
Robert ZRobert Z
99.8k1068140
$endgroup$
$begingroup$
Sorry 381 was typing error. That has been corrected.
$endgroup$
– Kishan Thacker
Jan 11 at 8:07
$begingroup$
I am bit embarrassed about the question as just now I realised that I was puzzeled after altering the question. And the answer to my altered question was none other then 1/20 itself. 😥
$endgroup$
– Kishan Thacker
Jan 11 at 8:11
$begingroup$
I think you have misunderstood my question I think 1141/8000 is probability of occurrence of flood exaxactly on 1st or exactly on 2nd or exactly on 3rd year.
$endgroup$
– Kishan Thacker
Jan 11 at 8:13
$begingroup$
Yes if you simply remove the word "exactly" the probability is $1/20$. If you mean within the third year then it is $1141/8000$
$endgroup$
– Robert Z
Jan 11 at 8:20
$begingroup$
BTW your correction is not complete: it is $361/8000$ not $361/800$.
$endgroup$
– Robert Z
Jan 11 at 8:23
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In any year the probability is $frac1{20}$, so also for the $3$rd year.
answered Jan 11 at 8:21
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
In any year the probability is $frac1{20}$, so also for the $3$rd year.
answered Jan 11 at 8:21
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
In any year the probability is $frac1{20}$, so also for the $3$rd year.
answered Jan 11 at 8:21
drhabdrhab
103k545136
$endgroup$
In any year the probability is $frac1{20}$, so also for the $3$rd year.
answered Jan 11 at 8:21
drhabdrhab
103k545136
answered Jan 11 at 8:21
drhabdrhab
103k545136
answered Jan 11 at 8:21
drhabdrhab
103k545136
answered Jan 11 at 8:21
drhabdrhab
103k545136
103k545136
add a comment |
add a comment |
$begingroup$
First of all, in your previous computation your approach is correct but the result is wrong:
$$19/20 cdot 19/20 cdot 1/20=frac{361}{8000}.$$
As regards your variation, if you would like to evaluate the probability that the flood occurs within the third year, i.e. the first year or the second year or the third year, then it should be:
$$1/20+(19/20cdot 1/20)+(19/20 cdot 19/20 cdot 1/20)=frac{1141}{8000}.$$
Note that the probability that the flood occurs the third year (we remove the word "exactly"), or any other year, is just $1/20$.
answered Jan 11 at 8:01
Robert ZRobert Z
99.8k1068140
$endgroup$
$begingroup$
Sorry 381 was typing error. That has been corrected.
$endgroup$
– Kishan Thacker
Jan 11 at 8:07
$begingroup$
I am bit embarrassed about the question as just now I realised that I was puzzeled after altering the question. And the answer to my altered question was none other then 1/20 itself. 😥
$endgroup$
– Kishan Thacker
Jan 11 at 8:11
$begingroup$
I think you have misunderstood my question I think 1141/8000 is probability of occurrence of flood exaxactly on 1st or exactly on 2nd or exactly on 3rd year.
$endgroup$
– Kishan Thacker
Jan 11 at 8:13
$begingroup$
Yes if you simply remove the word "exactly" the probability is $1/20$. If you mean within the third year then it is $1141/8000$
$endgroup$
– Robert Z
Jan 11 at 8:20
$begingroup$
BTW your correction is not complete: it is $361/8000$ not $361/800$.
$endgroup$
– Robert Z
Jan 11 at 8:23
add a comment |
$begingroup$
First of all, in your previous computation your approach is correct but the result is wrong:
$$19/20 cdot 19/20 cdot 1/20=frac{361}{8000}.$$
As regards your variation, if you would like to evaluate the probability that the flood occurs within the third year, i.e. the first year or the second year or the third year, then it should be:
$$1/20+(19/20cdot 1/20)+(19/20 cdot 19/20 cdot 1/20)=frac{1141}{8000}.$$
Note that the probability that the flood occurs the third year (we remove the word "exactly"), or any other year, is just $1/20$.
answered Jan 11 at 8:01
Robert ZRobert Z
99.8k1068140
$endgroup$
$begingroup$
Sorry 381 was typing error. That has been corrected.
$endgroup$
– Kishan Thacker
Jan 11 at 8:07
$begingroup$
I am bit embarrassed about the question as just now I realised that I was puzzeled after altering the question. And the answer to my altered question was none other then 1/20 itself. 😥
$endgroup$
– Kishan Thacker
Jan 11 at 8:11
$begingroup$
I think you have misunderstood my question I think 1141/8000 is probability of occurrence of flood exaxactly on 1st or exactly on 2nd or exactly on 3rd year.
$endgroup$
– Kishan Thacker
Jan 11 at 8:13
$begingroup$
Yes if you simply remove the word "exactly" the probability is $1/20$. If you mean within the third year then it is $1141/8000$
$endgroup$
– Robert Z
Jan 11 at 8:20
$begingroup$
BTW your correction is not complete: it is $361/8000$ not $361/800$.
$endgroup$
– Robert Z
Jan 11 at 8:23
add a comment |
$begingroup$
First of all, in your previous computation your approach is correct but the result is wrong:
$$19/20 cdot 19/20 cdot 1/20=frac{361}{8000}.$$
As regards your variation, if you would like to evaluate the probability that the flood occurs within the third year, i.e. the first year or the second year or the third year, then it should be:
$$1/20+(19/20cdot 1/20)+(19/20 cdot 19/20 cdot 1/20)=frac{1141}{8000}.$$
Note that the probability that the flood occurs the third year (we remove the word "exactly"), or any other year, is just $1/20$.
answered Jan 11 at 8:01
Robert ZRobert Z
99.8k1068140
$endgroup$
First of all, in your previous computation your approach is correct but the result is wrong:
$$19/20 cdot 19/20 cdot 1/20=frac{361}{8000}.$$
As regards your variation, if you would like to evaluate the probability that the flood occurs within the third year, i.e. the first year or the second year or the third year, then it should be:
$$1/20+(19/20cdot 1/20)+(19/20 cdot 19/20 cdot 1/20)=frac{1141}{8000}.$$
Note that the probability that the flood occurs the third year (we remove the word "exactly"), or any other year, is just $1/20$.
answered Jan 11 at 8:01
Robert ZRobert Z
99.8k1068140
edited Jan 11 at 8:24
answered Jan 11 at 8:01
Robert ZRobert Z
99.8k1068140
answered Jan 11 at 8:01
Robert ZRobert Z
99.8k1068140
answered Jan 11 at 8:01
Robert ZRobert Z
99.8k1068140
99.8k1068140
$begingroup$
Sorry 381 was typing error. That has been corrected.
$endgroup$
– Kishan Thacker
Jan 11 at 8:07
$begingroup$
I am bit embarrassed about the question as just now I realised that I was puzzeled after altering the question. And the answer to my altered question was none other then 1/20 itself. 😥
$endgroup$
– Kishan Thacker
Jan 11 at 8:11
$begingroup$
I think you have misunderstood my question I think 1141/8000 is probability of occurrence of flood exaxactly on 1st or exactly on 2nd or exactly on 3rd year.
$endgroup$
– Kishan Thacker
Jan 11 at 8:13
$begingroup$
Yes if you simply remove the word "exactly" the probability is $1/20$. If you mean within the third year then it is $1141/8000$
$endgroup$
– Robert Z
Jan 11 at 8:20
$begingroup$
BTW your correction is not complete: it is $361/8000$ not $361/800$.
$endgroup$
– Robert Z
Jan 11 at 8:23
add a comment |
$begingroup$
Sorry 381 was typing error. That has been corrected.
$endgroup$
– Kishan Thacker
Jan 11 at 8:07
$begingroup$
I am bit embarrassed about the question as just now I realised that I was puzzeled after altering the question. And the answer to my altered question was none other then 1/20 itself. 😥
$endgroup$
– Kishan Thacker
Jan 11 at 8:11
$begingroup$
I think you have misunderstood my question I think 1141/8000 is probability of occurrence of flood exaxactly on 1st or exactly on 2nd or exactly on 3rd year.
$endgroup$
– Kishan Thacker
Jan 11 at 8:13
$begingroup$
Yes if you simply remove the word "exactly" the probability is $1/20$. If you mean within the third year then it is $1141/8000$
$endgroup$
– Robert Z
Jan 11 at 8:20
$begingroup$
BTW your correction is not complete: it is $361/8000$ not $361/800$.
$endgroup$
– Robert Z
Jan 11 at 8:23
$begingroup$
Sorry 381 was typing error. That has been corrected.
$endgroup$
– Kishan Thacker
Jan 11 at 8:07
$begingroup$
Sorry 381 was typing error. That has been corrected.
$endgroup$
– Kishan Thacker
Jan 11 at 8:07
$begingroup$
I am bit embarrassed about the question as just now I realised that I was puzzeled after altering the question. And the answer to my altered question was none other then 1/20 itself. 😥
$endgroup$
– Kishan Thacker
Jan 11 at 8:11
$begingroup$
I am bit embarrassed about the question as just now I realised that I was puzzeled after altering the question. And the answer to my altered question was none other then 1/20 itself. 😥
$endgroup$
– Kishan Thacker
Jan 11 at 8:11
$begingroup$
I think you have misunderstood my question I think 1141/8000 is probability of occurrence of flood exaxactly on 1st or exactly on 2nd or exactly on 3rd year.
$endgroup$
– Kishan Thacker
Jan 11 at 8:13
$begingroup$
I think you have misunderstood my question I think 1141/8000 is probability of occurrence of flood exaxactly on 1st or exactly on 2nd or exactly on 3rd year.
$endgroup$
– Kishan Thacker
Jan 11 at 8:13
$begingroup$
Yes if you simply remove the word "exactly" the probability is $1/20$. If you mean within the third year then it is $1141/8000$
$endgroup$
– Robert Z
Jan 11 at 8:20
$begingroup$
Yes if you simply remove the word "exactly" the probability is $1/20$. If you mean within the third year then it is $1141/8000$
$endgroup$
– Robert Z
Jan 11 at 8:20
$begingroup$
BTW your correction is not complete: it is $361/8000$ not $361/800$.
$endgroup$
– Robert Z
Jan 11 at 8:23
$begingroup$
BTW your correction is not complete: it is $361/8000$ not $361/800$.
$endgroup$
– Robert Z
Jan 11 at 8:23
add a comment |
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