The probability of an ordinary year having 53 Tuesdays is: -
(a) \[\dfrac{2}{7}\]
(b) \[\dfrac{1}{7}\]
(c) \[\dfrac{3}{7}\]
(d) \[\dfrac{4}{7}\]
Answer
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Hint: Find out the total number of weeks in one year and calculate the total number of days covered by these weeks by using the information that “1 week has 7 days”. Subtract the total number of calculated days from a total of 365 days. Now, from the obtained difference find the probability of getting Tuesday by assuming the 7 days as the total sample space.
Complete step-by-step solution
Here, we have to find the probability of an ordinary year having 53 Tuesdays.
Now, we know that one year has 52 weeks and 365 days in total. These 52 weeks will contain 52 Tuesdays. Since 1 week has 7 days, therefore, we have,
Number of days in 52 weeks = 52 \[\times \] 7 = 364
Now, there will be a day left after 52 weeks and this day may be any of the seven days of the week. So, the total number of sample space can be given as: -
n (S) = Monday, Tuesday, Wednesday, Thursday, Friday, Saturday = 7
Hence, there are a total of 7 possible outcomes for this \[{{365}^{th}}\] day. We have to determine the probability of getting a Tuesday. Clearly, we can see that among these days there is only one Tuesday. So, the favorable outcome can be given as: -
n (E) = Tuesday = 1
Now, we know that the probability of an event to occur is the ratio of the number of favorable outcomes to the total number of outcomes.
\[\Rightarrow \] Required probability = \[\dfrac{n\left( E \right)}{n\left( S \right)}=\dfrac{1}{7}\]
Hence, option (b) is the correct answer.
Note: One may note that we do not have to care about those 52 Tuesdays in 52 weeks because 52 Tuesdays will occur for sure. We have to consider that 1 day which will be left after 52 weeks. It is not necessary that the \[{{52}^{th}}\] week will end on Sunday. It can be any of the seven days and therefore we have considered the total number of sample spaces equal to 7. Remember that in case of a leap year there will be 2 days left after 52 weeks.
Complete step-by-step solution
Here, we have to find the probability of an ordinary year having 53 Tuesdays.
Now, we know that one year has 52 weeks and 365 days in total. These 52 weeks will contain 52 Tuesdays. Since 1 week has 7 days, therefore, we have,
Number of days in 52 weeks = 52 \[\times \] 7 = 364
Now, there will be a day left after 52 weeks and this day may be any of the seven days of the week. So, the total number of sample space can be given as: -
n (S) = Monday, Tuesday, Wednesday, Thursday, Friday, Saturday = 7
Hence, there are a total of 7 possible outcomes for this \[{{365}^{th}}\] day. We have to determine the probability of getting a Tuesday. Clearly, we can see that among these days there is only one Tuesday. So, the favorable outcome can be given as: -
n (E) = Tuesday = 1
Now, we know that the probability of an event to occur is the ratio of the number of favorable outcomes to the total number of outcomes.
\[\Rightarrow \] Required probability = \[\dfrac{n\left( E \right)}{n\left( S \right)}=\dfrac{1}{7}\]
Hence, option (b) is the correct answer.
Note: One may note that we do not have to care about those 52 Tuesdays in 52 weeks because 52 Tuesdays will occur for sure. We have to consider that 1 day which will be left after 52 weeks. It is not necessary that the \[{{52}^{th}}\] week will end on Sunday. It can be any of the seven days and therefore we have considered the total number of sample spaces equal to 7. Remember that in case of a leap year there will be 2 days left after 52 weeks.
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