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You have a bag with $347$ black balls and $278$ white balls. Without looking, you pick up two balls from the bag and apply the following rule. If both balls are of the same colour, you throw them both away. Otherwise, you throw away the black ball and return the white ball to the bag. You keep repeating this process. If at some stage there is exactly one ball left in the bag, which of the following is true?

  1. The ball in the bag is definitely white.
  2. The ball in the bag is definitely black.
  3. Both colours are possible, but the probability of it being white is greater.
  4. Both colours are possible, but the probability of it being black is greater.
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i think b is r8 answer
as 347b  278w
bb  atlast 1 will remain black as (346/2=173)

bw 69 b 278 w then 278 w then 0 ball

wb same 0 ball

ww  0 ball
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3 Answers

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Answer: B
Three cases are possible:
1) both black balls--thow both
2) both white balls-throw both
3)one white and one black--black throw and white remain
we can observe that overallveffect of this pass is to either
*DECREASE WHITE BALLS BY 2 or 0 in each pickup
*DECREASE BLACK BALLS BY 1 OR 2 in each pickup.
 If we start with even no of white balls then definitely we can conclude that the last balls remains is black ball or in other words it is impossible to start with even no and after decreasing 2 in passes we get odd number.

Therefore, the last ball in the bag is definitely BLACK ball..
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Answer C)

Pick

B B - both ball thrown up

B W - Black ball throw away

W B - Black ball throw away

W W - both ball thrown up

So, in 4 times of throw waste of black ball 3 times and waste of white balls 1 time

So, probability ratio of Black ball with white ball =3:1

As we taking two ball in each pick ratio will be (347/2) : (278/2) ≈173:139

So, in ratio no of white ball is more as it should be

So, white ball should remain in last

2 Comments

didn't get it. Could you please elaborate it a little bit.
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where is ur problem?
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