GATE2013 EE: GA-6

Question

What is the chance that a leap year, selected at random, will contain $53$ Saturdays?

• A. $2/7$
• B. $3/7$
• C. $1/7$
• D. $5/7$

Answer 1

Sample space is $S :\{\text{Monday-Tuesday, Tuesday-Wednesday, Wednesday-Thursday}\ldots \text{Sunday-Monday}\}$
Number of elements in $S = n(S) = 7$
What we want is a set $A$ (say) that comprises of the elements $\text{Saturday-Sunday and Friday-Saturday}$
Number of elements in set $A = n(A) = 2$
By definition, probability of occurrence of $A =\dfrac{n(A)}{n(S)} = {2}/{7}$

Therefore, probability that a leap year has $53$ Saturdays is $\dfrac{2}{7}.$

Correct Answer: $A$

Comments

I think this question can be solved with intuition also. Like let's say that the first day of the year is thrusday then 53 sats will come. Also if the first day is saturday itself then also 53 ssats will come. Hence we will have total 2 days of starting for which saturdays will come. That are thursday and saturday hence the probablity is 2/7.

 

Answer 2

I think there is  another way .

Since it is leap year .. number of days  = 366

we have 52 weeks in a year .. 52 x 7 = 364

therefore we have already had 52 Saturdays  in these 364 days .. now probability For 53 Saturdays ,

=number of remaining days / number of days in a week = 2/7

the same can be applied to any day of the week .

Comments

=number of remaining days / number of days in a week = 2/7

Is this part like  ??

  [ __ __ ]

Two days remaining ..chance for the day to be saturday is 1/7

We want 53 saturday so either first slot is SAT or second slot is SAT

hence 1/7 + 1/7 = 2/7

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$2\ odd\ days\ can\ be:$

$\left \{ Sun-Mon \right \}$ $\left \{ Mon-Tue \right \}$ $\left \{ Tue-Wed \right \}$ $\left \{ Wed-Thu \right \}$ $\left \{ Thu-Fri \right \}$ $\left \{ Fri-Sat \right \}$ $\left \{ Sat-Sun \right \}$

$Probability=\dfrac{fav}{total}=\dfrac{2}{7}$