Latest Elitmus Aptitude Question SOLUTION: in a particular state, the god of rain announces that there will be raining on exactly 20 days in each calender year. if the prbobality of raining on a particular day of year 2011

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(#M40157010) ELITMUS Numerical Ability question probability asked on 10 july 2016 Keep an EYE

in a particular state, the god of rain announces that there will be raining on exactly 20 days in each calender year. if the prbobality of raining on a particular day of year 2011, 2012 and 2013 are P, Q and R respectively then which of the following relation holds true:
a) 2Q>=P+R b) Q^2 >=P*R c) 2/Q >= 1/P + 1/Q d) all of these

Asked In Elitmus abhishek chaubey (7 years ago)

Solved Naveen babu Read Solution (11)

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Naveen babu  (7 years ago) Best Solution

Probability of raining on a particular day is 20/365(No. of favorable days/total no of days), for a leap year its probability is 20/266.
Now we have P=20/365, Q=20/366, R=20/365.... now verify options, only option (3) is correct

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n1sh2  (7 years ago)

Ans: d)All of these
Explanation:
Since during each year it rains for 20 days,
then probability of rain for each year is as follows :
P->YEAR 2011 -> 20/365=0.0547 ;
Q->YEAR 2012(Leap year) -> 20/366=0.0546 ;
R->YEAR 2013 ->20/365=0.0547
Applying these values for each relation in the option we get,
For a)2(.054)=.054+.054 --> 0.108=0.108;(similarly for the other relations,we get equal values )
b)(.054)^2=.054*.054
c)2/.054 =1/.054+1/.054
So my friends all the relations hold true and you get the magic answer. :P

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Abhay choudhary  (7 years ago)

option C is right.....

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Hemanth sai  (7 years ago)

probability of raining on a particular day is=1/20
so, p=1/20,q=1/20,r=1/20.
all the above conditions satisfies this values

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sandeep  (7 years ago)

in 2011 , 2012 and 2013 only one leap year
=> 2012/4=0
so
probability 20 days in each calendar year
so probability 20/2011, 20/2012, 20/2013
now checking the relation
a) 2Q = 22*40/2011=40/366, P+R= 40/365 (condition false or inequalities)
b) Q^2= (20/366)^2 , P*R= 40/(365*365)(condition false or inequalities)
c) 2/Q= (2*366)/20 , 1/P + 1/Q =(365/20) + (365/20) (condition true or inequalities)
so option c is true

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kirti shobhakant mandal  (7 years ago)

a) is the solution

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Prabhat Dubey  (7 years ago)

Here 2012 is leap year means probablity Q=20/366 and for 2011 and 2013 P=R=20/365
so by checking all options we get that all are satisfying so
Ans = (d) all of these

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suvasree chatterjee  (7 years ago)

Is any1 attempt this Q in xm ,i wants to kno correct ans plz help me

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sachin kumar  (7 years ago)

Option c answer... don't go in decimal.

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Amit Kumar Singh  (7 years ago)

C would be answer

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sheetal singh  (7 years ago)

how you will know that which one is a leap year ??

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