Latest TCS Aptitude Question SOLUTION: an urn contains m white balls ad n black balls,a ball is drawn at random and is put back into the urn along with k additional balls of the same colour as that of the ball drawn.a

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(#M40038787) TCS Numerical Ability question probability Keep an EYE

an urn contains m white balls ad n black balls,a ball is drawn at random and is put back into the urn along with k additional balls of the same colour as that of the ball drawn.a ball is again drawn at random,the probability of drawing a white ball is?
a)increases with k
b)does not depend on k
c)decreases with k
d)cannot be determined without additional information

Asked In TCS satish (9 years ago)

Unsolved Read Solution (12)

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Read Solution (12) :

Sankalan Gain  (9 years ago)

In the 1st turn prob. of drawing a white ball= m/(m+n)& that of a black ball= n/(m+n).

If white ball is drawn in 1st turn,then prob.of drawing a white ball =(m/(m+n))*((m+k)/(m+n+k)).

If black ball is drawn in 1st turn,then prob.of drawing a white ball =(n/(m+n))*(m/(m+n+k)).

So prob. of drawing a white ball=((m/(m+n))*((m+k)/(m+n+k)))+((n/(m+n))*(m/(m+n+k)))= m/(m+n)i.e does not depend on k (Ans - b)

Is this solution Helpfull? Yes (39) | No (3)  

Niveditha Ravichandran  (9 years ago)

ans: cannot be determined without additional information

previously a ball is drawn and replaced (colour unknown- either black or white)

k additional balls represents same colour as previously drawn ball - can be black or white

so additional info is required

Is this solution Helpfull? Yes (13) | No (3)  

surbhi dubey  (9 years ago)

what is the right answer

Is this solution Helpfull? Yes (3) | No   

Souradip Nath  (9 years ago)

ans: "decreases with k"


as the no of ball increases, the denominator part of our soln. also increases. Hence the overall probability value decreases.

Is this solution Helpfull? Yes (2) | No (10)  

Aayushi Gupta  (9 years ago)

previous one is incorrect....

Case 1: firstly white ball is taken out and then agn white :
prob. = m/(m+n)*(m+k)/(m+n+k)
Case 2: firstly black ball is taken out and then white :
prob. = n/(m+n)*m/(m+n+k)
adding both cases :
prob. obtaind = m/(m+n)
independant of k...hence option b

Is this solution Helpfull? Yes (2) | No   

amit agrawal  (9 years ago)

cannot be determined
if first ball drawn is black den probability decreases with increase of k otherwise probability increases.
hence probability depends on first draw.

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

nagasai  (9 years ago)

correct ans :b)

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Aayushi Gupta  (9 years ago)

Case 1: firstly white ball is taken out and then agn white :
prob. = m/(m+n)*m/(m+n+k)
Case 2: firstly black ball is taken out and then white :
prob. = n/(m+n)*m/(m+n+k)
adding both cases :
prob. obtaind = m/(m+n+k)
thus as k increases , overall probability decreases...hence option c

Is this solution Helpfull? Yes | No (2)  

sanyogita holkar  (9 years ago)

d...ans is d

Is this solution Helpfull? Yes | No (1)  

hazeera  (9 years ago)

sankalan gain answer is correct

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RAJ KUMAR  (8 years ago)

admin sir u plese give the right answer....

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Hashavardhini  (8 years ago)

Option b) does not depend on k

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