What is probability that: (a)a non leap year has 53 Sundays. (b)a leap year has 53 fridays.
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This one really bowled me over. I don't understand from where I should begin.
I actually thought that a non leap year will have 52 weeks and 1 day. But I couldn't think how to find the probability .
I think we are more concerned with having 53 Sundays in (a) part. In a non leap year there are 52 weeks and 1 day extra, this means that there are 52 Sunday already and the given condition wants that the 1 day which was extra should also be Sunday ( then there would be 53 Sundays in total). There is only one possible way for this. So favourable outcome becomes one. But I am not clear about total outcomes.
Please someone help
probability experimental-mathematics
asked Jan 1 at 17:07
GarimaGarima
11
$endgroup$
add a comment |
$begingroup$
This one really bowled me over. I don't understand from where I should begin.
I actually thought that a non leap year will have 52 weeks and 1 day. But I couldn't think how to find the probability .
I think we are more concerned with having 53 Sundays in (a) part. In a non leap year there are 52 weeks and 1 day extra, this means that there are 52 Sunday already and the given condition wants that the 1 day which was extra should also be Sunday ( then there would be 53 Sundays in total). There is only one possible way for this. So favourable outcome becomes one. But I am not clear about total outcomes.
Please someone help
probability experimental-mathematics
asked Jan 1 at 17:07
GarimaGarima
11
$endgroup$
$begingroup$
For $(a)$, you have to find the probability that the one extra day is a Sunday and for $(b)$ that one of those two days is a Friday.
$endgroup$
– Mohammad Zuhair Khan
Jan 1 at 17:19
1
$begingroup$
I would personally write a script to check this rather than do this with pen/paper. The Gregorian calendar repeats once every $400$ years. You can look at the $400$ years and break them apart as leap years or non-leap years. Precisely $94$ of the years in the repeating cycle are leap years and the rest are non-leap years. Check how many of the nonleap years have $53$ sundays and divide the result by $306$, and check how many of the leap years have $53$ fridays and divide the result by $94$.
$endgroup$
– JMoravitz
Jan 1 at 17:20
$begingroup$
It is worth pointing out that a non-leap year has 53 of a specific day of the week if and only if Jan1 of that year is that day of the week. Meanwhile a leap year has $53$ of a specific day of the week if and only if Jan1 or Jan2 is that day of the week.
$endgroup$
– JMoravitz
Jan 1 at 17:22
1
$begingroup$
You need to be clearer about what you are asking. If, say, you assume that Jan $1$ is equally likely to occur on any day of the week, then the problem is straight forward. Alas, that assumption is not accurate. See, e.g., this
$endgroup$
– lulu
Jan 1 at 17:24
$begingroup$
The question is this much only. I think we are more concerned with having 53 Sundays in (a) part
$endgroup$
– Garima
Jan 1 at 17:33
add a comment |
$begingroup$
This one really bowled me over. I don't understand from where I should begin.
I actually thought that a non leap year will have 52 weeks and 1 day. But I couldn't think how to find the probability .
I think we are more concerned with having 53 Sundays in (a) part. In a non leap year there are 52 weeks and 1 day extra, this means that there are 52 Sunday already and the given condition wants that the 1 day which was extra should also be Sunday ( then there would be 53 Sundays in total). There is only one possible way for this. So favourable outcome becomes one. But I am not clear about total outcomes.
Please someone help
probability experimental-mathematics
asked Jan 1 at 17:07
GarimaGarima
11
$endgroup$
This one really bowled me over. I don't understand from where I should begin.
I actually thought that a non leap year will have 52 weeks and 1 day. But I couldn't think how to find the probability .
I think we are more concerned with having 53 Sundays in (a) part. In a non leap year there are 52 weeks and 1 day extra, this means that there are 52 Sunday already and the given condition wants that the 1 day which was extra should also be Sunday ( then there would be 53 Sundays in total). There is only one possible way for this. So favourable outcome becomes one. But I am not clear about total outcomes.
Please someone help
probability experimental-mathematics
probability experimental-mathematics
asked Jan 1 at 17:07
GarimaGarima
11
asked Jan 1 at 17:07
GarimaGarima
11
edited Jan 1 at 17:38
Garima
asked Jan 1 at 17:07
GarimaGarima
11
asked Jan 1 at 17:07
GarimaGarima
11
asked Jan 1 at 17:07
GarimaGarima
11
11
$begingroup$
For $(a)$, you have to find the probability that the one extra day is a Sunday and for $(b)$ that one of those two days is a Friday.
$endgroup$
– Mohammad Zuhair Khan
Jan 1 at 17:19
1
$begingroup$
I would personally write a script to check this rather than do this with pen/paper. The Gregorian calendar repeats once every $400$ years. You can look at the $400$ years and break them apart as leap years or non-leap years. Precisely $94$ of the years in the repeating cycle are leap years and the rest are non-leap years. Check how many of the nonleap years have $53$ sundays and divide the result by $306$, and check how many of the leap years have $53$ fridays and divide the result by $94$.
$endgroup$
– JMoravitz
Jan 1 at 17:20
$begingroup$
It is worth pointing out that a non-leap year has 53 of a specific day of the week if and only if Jan1 of that year is that day of the week. Meanwhile a leap year has $53$ of a specific day of the week if and only if Jan1 or Jan2 is that day of the week.
$endgroup$
– JMoravitz
Jan 1 at 17:22
1
$begingroup$
You need to be clearer about what you are asking. If, say, you assume that Jan $1$ is equally likely to occur on any day of the week, then the problem is straight forward. Alas, that assumption is not accurate. See, e.g., this
$endgroup$
– lulu
Jan 1 at 17:24
$begingroup$
The question is this much only. I think we are more concerned with having 53 Sundays in (a) part
$endgroup$
– Garima
Jan 1 at 17:33
add a comment |
$begingroup$
For $(a)$, you have to find the probability that the one extra day is a Sunday and for $(b)$ that one of those two days is a Friday.
$endgroup$
– Mohammad Zuhair Khan
Jan 1 at 17:19
1
$begingroup$
I would personally write a script to check this rather than do this with pen/paper. The Gregorian calendar repeats once every $400$ years. You can look at the $400$ years and break them apart as leap years or non-leap years. Precisely $94$ of the years in the repeating cycle are leap years and the rest are non-leap years. Check how many of the nonleap years have $53$ sundays and divide the result by $306$, and check how many of the leap years have $53$ fridays and divide the result by $94$.
$endgroup$
– JMoravitz
Jan 1 at 17:20
$begingroup$
It is worth pointing out that a non-leap year has 53 of a specific day of the week if and only if Jan1 of that year is that day of the week. Meanwhile a leap year has $53$ of a specific day of the week if and only if Jan1 or Jan2 is that day of the week.
$endgroup$
– JMoravitz
Jan 1 at 17:22
1
$begingroup$
You need to be clearer about what you are asking. If, say, you assume that Jan $1$ is equally likely to occur on any day of the week, then the problem is straight forward. Alas, that assumption is not accurate. See, e.g., this
$endgroup$
– lulu
Jan 1 at 17:24
$begingroup$
The question is this much only. I think we are more concerned with having 53 Sundays in (a) part
$endgroup$
– Garima
Jan 1 at 17:33
$begingroup$
For $(a)$, you have to find the probability that the one extra day is a Sunday and for $(b)$ that one of those two days is a Friday.
$endgroup$
– Mohammad Zuhair Khan
Jan 1 at 17:19
$begingroup$
For $(a)$, you have to find the probability that the one extra day is a Sunday and for $(b)$ that one of those two days is a Friday.
$endgroup$
– Mohammad Zuhair Khan
Jan 1 at 17:19
1
1
$begingroup$
I would personally write a script to check this rather than do this with pen/paper. The Gregorian calendar repeats once every $400$ years. You can look at the $400$ years and break them apart as leap years or non-leap years. Precisely $94$ of the years in the repeating cycle are leap years and the rest are non-leap years. Check how many of the nonleap years have $53$ sundays and divide the result by $306$, and check how many of the leap years have $53$ fridays and divide the result by $94$.
$endgroup$
– JMoravitz
Jan 1 at 17:20
$begingroup$
I would personally write a script to check this rather than do this with pen/paper. The Gregorian calendar repeats once every $400$ years. You can look at the $400$ years and break them apart as leap years or non-leap years. Precisely $94$ of the years in the repeating cycle are leap years and the rest are non-leap years. Check how many of the nonleap years have $53$ sundays and divide the result by $306$, and check how many of the leap years have $53$ fridays and divide the result by $94$.
$endgroup$
– JMoravitz
Jan 1 at 17:20
$begingroup$
It is worth pointing out that a non-leap year has 53 of a specific day of the week if and only if Jan1 of that year is that day of the week. Meanwhile a leap year has $53$ of a specific day of the week if and only if Jan1 or Jan2 is that day of the week.
$endgroup$
– JMoravitz
Jan 1 at 17:22
$begingroup$
It is worth pointing out that a non-leap year has 53 of a specific day of the week if and only if Jan1 of that year is that day of the week. Meanwhile a leap year has $53$ of a specific day of the week if and only if Jan1 or Jan2 is that day of the week.
$endgroup$
– JMoravitz
Jan 1 at 17:22
1
1
$begingroup$
You need to be clearer about what you are asking. If, say, you assume that Jan $1$ is equally likely to occur on any day of the week, then the problem is straight forward. Alas, that assumption is not accurate. See, e.g., this
$endgroup$
– lulu
Jan 1 at 17:24
$begingroup$
You need to be clearer about what you are asking. If, say, you assume that Jan $1$ is equally likely to occur on any day of the week, then the problem is straight forward. Alas, that assumption is not accurate. See, e.g., this
$endgroup$
– lulu
Jan 1 at 17:24
$begingroup$
The question is this much only. I think we are more concerned with having 53 Sundays in (a) part
$endgroup$
– Garima
Jan 1 at 17:33
$begingroup$
The question is this much only. I think we are more concerned with having 53 Sundays in (a) part
$endgroup$
– Garima
Jan 1 at 17:33
add a comment |
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$begingroup$
For $(a)$, you have to find the probability that the one extra day is a Sunday and for $(b)$ that one of those two days is a Friday.
$endgroup$
– Mohammad Zuhair Khan
Jan 1 at 17:19
1
$begingroup$
I would personally write a script to check this rather than do this with pen/paper. The Gregorian calendar repeats once every $400$ years. You can look at the $400$ years and break them apart as leap years or non-leap years. Precisely $94$ of the years in the repeating cycle are leap years and the rest are non-leap years. Check how many of the nonleap years have $53$ sundays and divide the result by $306$, and check how many of the leap years have $53$ fridays and divide the result by $94$.
$endgroup$
– JMoravitz
Jan 1 at 17:20
$begingroup$
It is worth pointing out that a non-leap year has 53 of a specific day of the week if and only if Jan1 of that year is that day of the week. Meanwhile a leap year has $53$ of a specific day of the week if and only if Jan1 or Jan2 is that day of the week.
$endgroup$
– JMoravitz
Jan 1 at 17:22
1
$begingroup$
You need to be clearer about what you are asking. If, say, you assume that Jan $1$ is equally likely to occur on any day of the week, then the problem is straight forward. Alas, that assumption is not accurate. See, e.g., this
$endgroup$
– lulu
Jan 1 at 17:24
$begingroup$
The question is this much only. I think we are more concerned with having 53 Sundays in (a) part
$endgroup$
– Garima
Jan 1 at 17:33