
Find the probability of getting 53 Sundays in a non-leap year.
A.$\dfrac{2}{7}$
B.$\dfrac{1}{7}$
C.$\dfrac{1}{5}$
D.$\dfrac{1}{9}$
Answer
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Hint: To find the probability of having 53 Sundays first of all we need to know the minimum number of Sundays in a year for that we need to calculate the number of weeks first. After that we can find the probability of having extra Sundays. Probability is calculated by- $P(E)=\dfrac{n(E)}{n(S)}$ where E denotes Event and S denotes Sample space and n(E) is number of elements in set E and n(S) is number of elements in set S.
Complete step-by-step answer:
We know the number of days in a non-leap year is 365. If we wish to calculate the number of weeks in a year we can calculate by dividing 365 by 7. The quotient will give a number of weeks and the remainder will give an extra number of days that could not complete a week.
After dividing 365 by 7 we get 52 as quotient and 1 as remainder. Therefore, in a year we have 52 weeks and 1 day. Every week has Sunday, so we have a minimum 52 Sundays in a year but that extra day can be a Sunday, Monday or any other day of the week.
Sample space is the set of all possible outcomes. The remaining one day can be anyone of the days of a week. Therefore, we have the sample space,
$S=\left\{ sun,mon,tue,wed,thur,fri,sat \right\}$
An Event is the subset of sample space containing favorable outcomes. In this question our event is the probability of having Sunday on the remaining one day of the 365 days. Therefore, we have E as $E=\left\{ sun \right\}$ .
Now probability is given by- $P(E)=\dfrac{n(E)}{n(S)}$ …(i) where n(E) is number of elements in set E and n(S) is number of elements in set S. Therefore, $n(E)=1$ and $n(S)=7$
From equation (i) we have, $P(E)=\dfrac{1}{7}$ .
Hence, option (b) is the correct answer.
Note: If we do not wish to see probability in terms of sets then we have a simpler definition of probability i.e. P(E)=no. of favourable outcomes/All possible outcomes . Sample space is the set of possible outcomes and Event is the set of favorable outcomes. Our favorable outcome was having a Sunday and all possible outcomes was having any of the seven days of a week. We can also see that Event is a subset of Sample space as Sunday was both in sample space and event.
Complete step-by-step answer:
We know the number of days in a non-leap year is 365. If we wish to calculate the number of weeks in a year we can calculate by dividing 365 by 7. The quotient will give a number of weeks and the remainder will give an extra number of days that could not complete a week.
After dividing 365 by 7 we get 52 as quotient and 1 as remainder. Therefore, in a year we have 52 weeks and 1 day. Every week has Sunday, so we have a minimum 52 Sundays in a year but that extra day can be a Sunday, Monday or any other day of the week.
Sample space is the set of all possible outcomes. The remaining one day can be anyone of the days of a week. Therefore, we have the sample space,
$S=\left\{ sun,mon,tue,wed,thur,fri,sat \right\}$
An Event is the subset of sample space containing favorable outcomes. In this question our event is the probability of having Sunday on the remaining one day of the 365 days. Therefore, we have E as $E=\left\{ sun \right\}$ .
Now probability is given by- $P(E)=\dfrac{n(E)}{n(S)}$ …(i) where n(E) is number of elements in set E and n(S) is number of elements in set S. Therefore, $n(E)=1$ and $n(S)=7$
From equation (i) we have, $P(E)=\dfrac{1}{7}$ .
Hence, option (b) is the correct answer.
Note: If we do not wish to see probability in terms of sets then we have a simpler definition of probability i.e. P(E)=no. of favourable outcomes/All possible outcomes . Sample space is the set of possible outcomes and Event is the set of favorable outcomes. Our favorable outcome was having a Sunday and all possible outcomes was having any of the seven days of a week. We can also see that Event is a subset of Sample space as Sunday was both in sample space and event.
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