Latest self Aptitude Question SOLUTION: what is the chance that a leap year selected at random will have 53 sundays

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~Sandeep~

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(#M40010440) SELF aptitude question Keep an EYE

what is the chance that a leap year selected at random will have 53 sundays

dola (1 year ago)

Puzzle Status: Unsolved Read Solution (1)

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Madhu  (1 year ago)

the leap year has 52 weeks and 2 days totally 366 days.
52 weeks contain 52 sundays.
2 days which contain sundays are sunday,monday (or)saturday,sunday out of 7 days.
the probability leap yr contain 53 sundays is=2/7

so the answer is 2/7

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(#M40021890) TCS MATHS PUZZLE Keep an EYE

what is the chance that a leap year selected at random contains 53 fridays?

Asked In TCS Submited By-- reddy siri (10 Months ago)

Puzzle Status: Solved     By »» Rehman

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Rehman  (10 Months ago)

A leap year has 366 days, therefore 52 weeks(i.e. 52 fridays) and 2 days.

The remaining 2 days may be any of the following :

(i) Sunday and Monday

(ii) Monday and Tuesday

(iii) Tuesday and Wednesday

(iv) Wednesday and Thursday

(v) Thursday and Friday

(vi) Friday and Saturday

(vii) Saturday and Sunday

For having 53 fridays in a year, one of the remaining 2 days must be a friday.

n(S) = 7

n(E) = 2

P(E) = n(E) / n(S) = 2 / 7

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jyotikumari  (10 Months ago)

there are 366days in a leap year.so left 2 days may be friday out of 7 days hence 2/7 is the probablity

Is this solution Helpfull? Yes (4) | No (2)  

sravani  (10 Months ago)

no.of days in leap year=366=52weeks+2days.
2days are consecutive and have 7 possibilities,2 combinations contain friday.
so ans is 2/7.

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rekha  (10 Months ago)

ans is 2/7

Is this solution Helpfull? Yes (2) | No (2)  

anandi kamble  (10 Months ago)

Ans-n(S)=7,n(E)=2
P(E)=n(E)/n(S)
=2/7

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gopal kumar jha  (10 Months ago)

in one leap year ,there is 2 odd days
and in one weeks 7 days
chance of friday with two odd days(2)/no. of days in a weeks(7)
ans 2/7

Is this solution Helpfull? Yes (1) | No