Question

The probability that a non-leap year has 53 sundays, is
(a)$\frac{2}{7}$
(b)$\frac{5}{7}$
(c)$\frac{6}{7}$
(d)$\frac{1}{7}$

Answer 1

SOLUTION :  

The correct option is (d) : 1/7

Given : A non leap year (An ordinary year)

Total number of days in ordinary year = 365 days .It  contain 52 weeks and 1 day

This one day can be any day of the week :  

Sunday, Monday, Tuesday, Wednesday, Thursday, Friday and Saturday.

Here, we have to make 53 Sundays so one  additional day should be Sunday.

Total number of days = 7  

Total number of outcomes = 7  

Let E = Event of getting a non leap year which has 53 Sundays

Number of favourable outcomes : 1 (Sunday)

Probability (E) = Number of favourable outcomes / Total number of outcomes

P(E) = 1/7  

Hence, Probability of getting a non leap year which has 53 Sundays, P(E) = 1/7 .

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

Hi there !
Here's the answer :

•°•°•°•°•°<><><<><>><><>°•°•°•°•°

In Non leap year,
there will be 365 days.

It has 52 weeks and 1 day

{365 days = (52×7) + 1}

In a Non leap year, 1 day will be left

There are definitely 52 Sundays.

To have 53 Sundays, the extra day must fall on Sunday.

=> There will be 53 Sundays,
only when this extra day is also a Sunday.

Now,

Let S be Sample space
n(S) - No. of total outcomes for the extra day

n(S) = 7

°•° S = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}

Let E be the Event that the extra day is a Sunday

E ={Sunday}

n(E) - No. of favourable outcomes for occurrence of Event E

n(E) = 1

$Probability = \dfrac{No.\: of\: favorable\: outcomes}{Total\: No.\: of\ Outcomes}$

$P(E) = \dfrac{n(E)}{n(S)}$

$P(E) = \frac{1}{7}$

•°• Required probability = $\sf \frac{1}{7}$

This answer exists in Option (d)

•°• Option (d) is Correct


•°•°•°•°•°<><><<><>><><>°•°•°•°•°

...