Lucky2783 wrote:

At the beginning of a show, a lottery representative put one hundred balls in a bag. Any ball is equally likely to be selected.

Each ball is either blue or yellow. If someone picks one ball at random, it probably will be a yellow one.

How many blue balls has the representative put in the bag?

(1) If four yellow balls had been repainted blue, it would have been more likely to select a blue ball rather than a yellow one.

(2) After the representative removes fifty balls, the probability of selecting a blue ball is the same as it was initially.

From task we know that probability of blue more than probability of yellow P(Y) > P(B)

Total 100 balls

1) From this statement we can make infer that there is 3 possible variants:

Y=53, B=47

probability of yellow equal to \(\frac{53}{100}\) and probability of blue equal to \(\frac{47}{100}\)

ANd if we repaint 4 yellow balls to blue it'll be \(P(Y) = \frac{49}{100}\) and \(P(B) = \frac{51}{100}\)

Next variants: Y=52, B=48 and Y=51, B=49

Insufficient

2) This statement says that we remove 50 balls and received the same probability (so we removed equal quantity of borh colors).

So it should be symmetric remove. And we can have symmetric remove only if case in which we have even number of balls.

But it can be any variant that include even numbers Y=60 B=40; Y=72 B=28 etc

Insufficient

1+2) From first statement we have 3 variants: two with odd numbers and one with even numbers. And from second statement we know that we should have even numbers

So answer is Y=52, B=48: \(P(Y)=\frac{52}{100}\) and \(P(B) = \frac{48}{100}\)

If we repaint 4 balls we received such probabilities:

\(P(Y)=\frac{48}{100}\) and \(P(B) = \frac{52}{100}\)

And if we remove 25 balls of each color we received \(P(Y) = \frac{26}{50}\) and \(P(B)=\frac{24}{50}\)

Sufficient.

Answer is C

Awesome.The key here was to pick set with even no. balls, as set with odd number of balls won't result in realistic numbers as far as balls are concerned.

+1 for this solution.