What is the probability that there would be 53 sundays and 53 fridays in a leap year

Text Solution

Solution : There are 366 days in a leap year. Also, there are 52 weeks and 2 days. The prossible combinations of two days are as follows: Sunday-Monday, Monday-Tuesday, Tuesday-Wednesday, Wednesday-Thursday, Thurday-Friday, Friday-Saturday, Saturday-Sunday Now, let E be the event in which leap year will have 53 Fridays or 53 Saturdays. We observe that there are three possibilities for event E, viz. Thursday-Friday, Friday-Saturday, and Saturday-Sunday. Since each pair of 2 equally likely, `P(E) = (1)/(7) + (1)/(7) + (1)/(7) = (3)/(7)`

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Given:

Probability of leap year having 53 Sundays is to be determined.

Concepts used:

Number of days in leap year = 366

Calculation:

A week has 7 days and total days are 366.

⇒ Number of Sundays in a leap year = 366/7 = 52 Sundays + 2 days

⇒ Total outcomes with 2 days = (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday) = 7

⇒ Number of outcomes without Sundays = 5

⇒ Probability of leap year with 53 Sundays = 2/7

∴ Probability of leap year with 53 Sundays is 2/7.

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Answer

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Hint: Here we have to find the probability that a leap year will have 53 Fridays or 53 Saturdays. We know that the leap year has 366 days and also leap year has 52 weeks and 2 extra days. We will find the sample space for these two days. Then we will count the number of possibilities that these two days are either Friday or Saturday or both. From there, we will find the required probability.

Complete step-by-step answer:

We know that the leap year has 366 days.We will find the number of weeks in a leap year. For that, we will divide the total days by 7.Therefore,Number of weeks \[ = \dfrac{{366}}{7} = 52\] weeks \[ + 2\] days Thus, leap year has 52 weeks and 2 extra days.We will find the sample space for these two extra days.The sample space for these two extra days is given below:\[S = \left\{ \begin{array}{l}\left( {Sunday,Monday} \right),\left( {Monday,Tuesday} \right),\left( {Tuesday,Wednesday} \right),\left( {Wednesday,Thursday} \right),\\\left( {Friday,Saturday} \right),\left( {Friday,Saturday} \right),\left( {Saturday,Sunday} \right)\end{array} \right\}\] Therefore, there are 7 total cases or possibility for these 2 extra days.Thus, \[n\left( S \right) = 7\].But according to question, we need these 2 days have either Friday or Saturday.Let \[E\] be the event that the leap year has 53 Fridays or 53 Saturdays.Therefore, from the sample space, we get\[E = \left\{ {\left( {Friday,Saturday} \right),\left( {Friday,Saturday} \right),\left( {Saturday,Sunday} \right)} \right\}\]Therefore, \[n\left( E \right) = 3\] Now, we will find the required probability using the formula \[P\left( E \right) = \dfrac{{n\left( E \right)}}{{n\left( S \right)}}\] .Now, we will substitute the value of \[n\left( E \right)\] and \[n\left( S \right)\] in the above formula. Therefore,\[P\left( E \right) = \dfrac{3}{7}\]Hence, the required probability is \[\dfrac{3}{7}\].

Thus, the correct option is B.

Note: Here we have obtained the probability of a leap year having 53 Fridays or 53 Saturdays. Probability is defined as the ratio of desired or favorable outcome to the total number of outcomes.

If we add the probability of all the events from the sample space, we get the probability as one. Range of probability is from 0 to 1, where 0 and 1 is included i.e. range of probability is \[\left[ {0,1} \right]\]. If we get the probability as zero, then that means that event is impossible.

3/7We know that a leap year has 366 days (i.e. 7 \[\times\] 52 + 2) = 52 weeks and 2 extra daysThe sample space for these 2 extra days is given below:S = {(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)}There are 7 cases.

∴ n(S) = 7

Let E be the event that the leap year has 53 Fridays or 53 Saturdays.E = { (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)}

i.e. n(E) = 3

\[\therefore P\left( E \right) = \frac{n\left( E \right)}{n\left( S \right)} = \frac{3}{7}\]

Hence, the probability that a leap year has 53 Fridays or 53 Saturdays is \[\frac{3}{7}\] .

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