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A randomly selected year is containing 53 Mondays then probability that it is a leap year
- a)2 / 5
- b)3 / 4
- c)1 / 4
- d)2 / 7
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A randomly selected year is containing 53 Mondays then probability tha...
The correct option is A
- Selected year will be a non leap year with a probability 3/4
- Selected year will be a leap year with a probability 1/4
- A selected leap year will have 53 Mondays with probability 2/7
- A selected non leap year will have 53 Mondays with probability 1/7
- E→ Event that randomly selected year contains 53 Mondays
P(E) = (3/4 × 1/7) + (1/4 × 2/7)
P(Leap Year/ E) = (2/28) / (5/28) = 2/5
P(Leap Year/ E) = (2/28) / (5/28) = 2/5
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Community Answer
A randomly selected year is containing 53 Mondays then probability tha...
Probability of a Leap Year with 53 Mondays
Explanation:
We know that a leap year occurs once every four years. A leap year has 366 days instead of the usual 365 days. This extra day is added to the month of February, making it 29 days instead of 28 days.
Step 1: Find the total number of days in a year:
365 days in a non-leap year
366 days in a leap year
Step 2: Find the total number of Mondays in a year:
There are 52 weeks in a year, which means there are 52 Mondays in a non-leap year. However, in some years, there is an extra day that falls on a Monday, making it 53 Mondays in that year.
Step 3: Find the probability of a randomly selected year being a leap year given that it has 53 Mondays:
Let P(A) be the probability of selecting a leap year. Let P(B) be the probability of selecting a year with 53 Mondays. We want to find P(A|B), the probability of selecting a leap year given that it has 53 Mondays:
P(A|B) = P(A and B) / P(B)
P(A and B) is the probability of selecting a leap year with 53 Mondays. This is the same as the probability of selecting a leap year and having 53 Mondays in that year.
We know that there are 7 possible days of the week that the year can start on (Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday). For each of these days, there is exactly one leap year that has 53 Mondays. Therefore, there are a total of 7 leap years that have 53 Mondays.
P(A and B) = 7 / total number of years with 53 Mondays
To find the total number of years with 53 Mondays, we need to consider both leap years and non-leap years.
In a non-leap year, there are 52 Mondays. Therefore, we need to find the number of non-leap years that have 53 Mondays. This can only happen if the year starts on a Monday and ends on a Monday (since there are 52 weeks in a year). This occurs once every 7 years, so there are a total of 2 non-leap years with 53 Mondays in a 14-year cycle.
In a leap year, there are 53 Mondays. Therefore, we need to find the number of leap years that have 53 Mondays. This can only happen if the year starts on a Monday, and the next year starts on a Saturday (since there are 365 days in a leap year). This occurs once every 28 years, so there is a total of 1 leap year with 53 Mondays in a 28-year cycle.
Therefore, the total number of years with 53 Mondays in a 28-year cycle is 2 + 1 = 3.
Total number of years with 53 Mondays = 2 non-leap years + 1 leap year = 3 years
P(B) is the probability of selecting a year with 53 Mondays. This is the same as the probability of selecting a non-leap year with 53 Mondays plus the probability of selecting a leap year with 53 Mondays.
Probability of selecting
Explanation:
We know that a leap year occurs once every four years. A leap year has 366 days instead of the usual 365 days. This extra day is added to the month of February, making it 29 days instead of 28 days.
Step 1: Find the total number of days in a year:
365 days in a non-leap year
366 days in a leap year
Step 2: Find the total number of Mondays in a year:
There are 52 weeks in a year, which means there are 52 Mondays in a non-leap year. However, in some years, there is an extra day that falls on a Monday, making it 53 Mondays in that year.
Step 3: Find the probability of a randomly selected year being a leap year given that it has 53 Mondays:
Let P(A) be the probability of selecting a leap year. Let P(B) be the probability of selecting a year with 53 Mondays. We want to find P(A|B), the probability of selecting a leap year given that it has 53 Mondays:
P(A|B) = P(A and B) / P(B)
P(A and B) is the probability of selecting a leap year with 53 Mondays. This is the same as the probability of selecting a leap year and having 53 Mondays in that year.
We know that there are 7 possible days of the week that the year can start on (Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday). For each of these days, there is exactly one leap year that has 53 Mondays. Therefore, there are a total of 7 leap years that have 53 Mondays.
P(A and B) = 7 / total number of years with 53 Mondays
To find the total number of years with 53 Mondays, we need to consider both leap years and non-leap years.
In a non-leap year, there are 52 Mondays. Therefore, we need to find the number of non-leap years that have 53 Mondays. This can only happen if the year starts on a Monday and ends on a Monday (since there are 52 weeks in a year). This occurs once every 7 years, so there are a total of 2 non-leap years with 53 Mondays in a 14-year cycle.
In a leap year, there are 53 Mondays. Therefore, we need to find the number of leap years that have 53 Mondays. This can only happen if the year starts on a Monday, and the next year starts on a Saturday (since there are 365 days in a leap year). This occurs once every 28 years, so there is a total of 1 leap year with 53 Mondays in a 28-year cycle.
Therefore, the total number of years with 53 Mondays in a 28-year cycle is 2 + 1 = 3.
Total number of years with 53 Mondays = 2 non-leap years + 1 leap year = 3 years
P(B) is the probability of selecting a year with 53 Mondays. This is the same as the probability of selecting a non-leap year with 53 Mondays plus the probability of selecting a leap year with 53 Mondays.
Probability of selecting
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A randomly selected year is containing 53 Mondays then probability that it is a leap yeara)2 / 5b)3 / 4c)1 / 4d)2 / 7Correct answer is option 'A'. Can you explain this answer?
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