Latest TCS Aptitude Question SOLUTION: lottery balls are numbered and colored balls. in the famous lotto, six numbered balls are drawn at random from 49 balls. In each ticket one needs to guess the six numbers that woul

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(#M40039045) TCS Numerical Ability question Keep an EYE

lottery balls are numbered and colored balls. in the famous lotto, six numbered balls are drawn at random from 49 balls. In each ticket one needs to guess the six numbers that would be drawn.
an urn contains m white and n black balls. a ball is drawn at random and is put back in to the urn along with k additional balls of the same color as that of the ball drawn. A ball is again drawn at random. probability of drawing a white ball now:
1) does not depend on k
2) cant be determined
3) decreases with k
4) increases with k

Asked In TCS shikha panwar (9 years ago)

Unsolved Read Solution (18)

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Read Solution (18) :

sajid  (9 years ago)

does not depend on k

Is this solution Helpfull? Yes (15) | No (2)  

abhisek banerjee  (9 years ago)

i took tcs open seame and right answer is deos not depend on k.Can anyone please tell why this is the right solution?

Is this solution Helpfull? Yes (7) | No (5)  

Dhanalakshmi M  (9 years ago)

cannot be determined.

Is this solution Helpfull? Yes (4) | No (3)  

anchal gupta  (9 years ago)

what is the right ans

Is this solution Helpfull? Yes (4) | No (1)  

Neha Dubey  (9 years ago)

Cannot be determined because we are not sure about which ball was picked in the first attempt

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

Nitin Singhal  (8 years ago)

Let W1 (B1) be the event that a white (a black) ball is drawn in the first draw and let W be the event that a white ball is drawn in the second draw. Then


P (W) = P (B1). P (W| B1) + P (W1). P (W| W1)


= n/m + n. m/m + n + k + m/m + n. m + k/m + n + k


= m(n + m + k)/(m + n) (m + n + k) = m/m + n

Is this solution Helpfull? Yes (2) | No (1)  

Prasanth  (9 years ago)

ans is 4.
Since additional balls are but back into the urn so the probability of getting a same color (whiite) increases with k.

Is this solution Helpfull? Yes (1) | No (6)  

Ankita Ghosh  (9 years ago)

cannot be determined is given as wrong solution in openseesame test.which is the right solution?

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

Gayatri badlani  (9 years ago)

gyatribadlani@gmail.com

Is this solution Helpfull? Yes (1) | No   

Premkumar  (9 years ago)

decreases with k

Is this solution Helpfull? Yes | No (8)  

Sridhar VG  (9 years ago)

Ans:4
If the k additional balls of a same colour are drawn it depends upon the ball drawn at random.For eg if the number 6 (white ball) is drawn that total number is added to the urn....
increases by k

Is this solution Helpfull? Yes | No (2)  

Anjali Mondal  (9 years ago)

increases with k
(as the ball increases automatically based on selection)

Is this solution Helpfull? Yes | No (2)  

C.P.Shrikrishnan  (9 years ago)

Take an example of 3 whilte balls and 2 black balls
If K picks a white ball then the number of white balls will be 2
And it credits an addition ball along with the ball that is taken into the urn.
So the number of white balls in the urn will be 5.
The probability of picking an white ball again purely relies on the value of K.

Is this solution Helpfull? Yes | No   

sejal patel  (9 years ago)

cant be determine.

Is this solution Helpfull? Yes | No   

ankit gupta  (9 years ago)

i think answer provided on tcs sesame is wrong...it should be cant be determined

Is this solution Helpfull? Yes | No (1)  

Avinash Kumar Barnwal  (8 years ago)

Dear admin! Can U plz upload the soln of this qn?

Is this solution Helpfull? Yes | No   

sarfaraz nawaz qureshi  (8 years ago)

THE CORRECT ANSWER IS "CANT BE DETERMINED"

Is this solution Helpfull? Yes | No   

harshit gupta  (6 years ago)

m white balls and n black balls so in first time probability of white ball m/n+m.
probability of black ball n/n+m.
second time probability of white ball if first time ball was white (m+k)/(m+n+k)
if first time ball was black then probability of white ball m/(m+n+k)
so final probability for white ball ((m+k)/(m+n+k) + m/(m+n+k))=(2m+k)/(m+n+k)
now subtract k from both and the resultant will be free from k

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