Latest Elitmus Aptitude Question SOLUTION: there are 23 chits numbered from 1 to 23. sonu pick up a chit and puts it back. monu picks another chit and puts it back. what is the probability that monu's chit number is greater

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(#M40152031) ELITMUS Numerical Ability question Probability Keep an EYE

there are 23 chits numbered from 1 to 23. sonu pick up a chit and puts it back. monu picks another chit and puts it back. what is the probability that monu's chit number is greater than that of sonu's.

Asked In Elitmus Ritika Jain (8 years ago)

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ajay kumar roshan  (8 years ago)

the no of cheat is 1 2 3 4 5................23,so the ans is 1/23*22/23+1/23*21/23+.......................1/23*1/23
1/23*1/23(22+21+20+19+....................1)
(1/23*1/23)(22*23)/2=11/23 ans

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Shivam Miglani  (8 years ago)

Easier Way : There is one possibility that both cards picked will be same.
Removing that we are left with 22 ways of choosing a card out of which only exactly half ways are of choosing a greater(smaller) card ..i.e. 11
That's why probability = 11/23.
Correct me if i am wrong.

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Abhinav Prakash  (8 years ago)

The no of ways sonu can pick up the chits=23
The no of ways monu can pick up the chits=23
therefore total no of ways chits can b picked up=23*23(maximum no of cases)

now if sonu picks 1 monu can pick 22 larger chits
if he picks 2 monu can pick 21 larger chits
and so on.....
therefore no of favourable cases = 22+21+20......+1
this is an ap with d=-1 a=23 n=23
sum = 23*11
therfore required probab=(23*11)/(23*23)=11/23

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Pravin Kumar  (8 years ago)

1/23 *( 22/22 + 21/22 + 20/22 + .... + 1/22) = 0.5

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Avishek Ghosh  (8 years ago)

1/2-1/23=22/46=11/23

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Harjaspreet singh  (8 years ago)

The answer would be 1/2
bcoz sonu picks a card and put it back now we left 22 cards ,total no of occurences is (23*22)
now no of occurrence for which monu picks greater than sonu is simple....think
(1,2),(1,3)..........(1,23)->23
+
(2,3),(2,4).........(2,23)->22
.....this way no of occur=1+2+3......23=11*23
p(n)= (11*23) /(22*23)=11/22=0.5

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