Latest Elitmus Aptitude Question SOLUTION: ram draws a card randomly among cards numbered 1-23 and keep it back.Then sam draws a card among those then what is the probality that Sam has drawn a card greater than Ram.

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(#M40037876) ELITMUS Numerical Ability question Keep an EYE

ram draws a card randomly among cards numbered 1-23 and keep it back.Then sam draws a card among those then what is the probality that Sam has drawn a card greater than Ram.

Asked In Elitmus Sonali Ghadai (9 years ago)

Unsolved Read Solution (9)

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Read Solution (9) :

Gaurav Malhotra  (9 years ago)

@VIVEK JAIN hw u solved.... its being
=> 1/23*22/23 + 1/23*21/23 + 1/23*20/23 + .... + 1/23*1/23

then after solving it the ans is 11/23

Is this solution Helpfull? Yes (35) | No (2)  

veer kumar  (9 years ago)

ram can draw 1 or 2 or 3 or 4......or 23 => p(ram draws 1 card)=1/23
now for shayam can have choices to draw card greater than ram
if ram draws card 1 then shayam has 22 choices so p(E)=1/23*22/23
" " " " 2 " " " 21 " so p(E)=1/23*21/23
so on........
finally when ram draws card 22 then shayam has 1 choice so p(E)=1/23*1/23
so total probability will be sum of all individual events
i.e p(E)=1/23(22/23+21/23+......+1/23)
=>1/23^2(22+21+20+.....1)
=>1/23^2*((22*(22+1)/2)
=>11/23 ans

Is this solution Helpfull? Yes (29) | No (1)  

Vivek Jain  (9 years ago)

if ram draws 1 then sam has 22 choices
if ram draws 2 then sam has 21 choices
So..
the probality that Sam has drawn a card greater than Ram
=> 1/23*22/23 + 1/23*21/23 + 1/23*20/23 + .... + 1/23*1/23
=> 1/23*[(22+21+...+10)/23]
=> 1/23*{22/2*[1+22]}/23
=> 252/529 or 84/143

Is this solution Helpfull? Yes (11) | No (16)  

Chandrashekhar   (9 years ago)

(1/23)*({22+21+20+19+..............+1}/23)

Is this solution Helpfull? Yes (4) | No   

ramanjulu  (9 years ago)

If ram draws 1 then Sam has 22 choices. If ram draws 2 then Sam has 21 choices and so on..

So, the probality that Sam has drawn a card greater than Ram is

=> 1/23*22/23 + 1/23*21/23 + 1/23*20/23 + .... + 1/23*1/23

=> 1/23*[(22+21+...+10)/23]

=> 1/23*{22/2*[1+22]}/23



=> 252/529 or 84/143

Is this solution Helpfull? Yes (2) | No (3)  

ayush srivastava  (9 years ago)

(1/23)*({22+21+20+19+..............+1}/23)

Is this solution Helpfull? Yes (1) | No (1)  

LOKESH RANJAN  (9 years ago)

1/23(22+21+20+...+1)/23

Is this solution Helpfull? Yes (1) | No   

krishna kanth  (9 years ago)

as sam should be more than ram
probability of sam is 10/23 and
probability of ram is 12/23
total probability is 10*12/(23*23)=120/529

Is this solution Helpfull? Yes | No (4)  

gaurav  (5 years ago)

If ram draws 1 then Sam has 22 choices. If ram draws 2 then Sam has 21 choices and so on..

So, the probality that Sam has drawn a card greater than Ram is

=> 1/23*22/23 + 1/23*21/23 + 1/23*20/23 + .... + 1/23*1/23

=> 1/23*[(22+21+...+10)/23]

=> 1/23*{22/2*[1+22]}/23



=> 252/529 or 84/143

Is this solution Helpfull? Yes | No (1)  

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