# How many balls here can't be in the bag?

A bag contains colored balls of which at least $90\%$ are red. Balls are drawn from the bag one by one and their color noted. It is found that $49$ of the first $50$ balls drawn are red. Thereafter $7$ out of every $8$ balls drawn are red. The number of balls in the bag CAN NOT BE

1. $170$
2. $210$
3. $250$
4. $194$

and the answer turns out to be $250$.

MY TRY: $\dfrac{\binom{0.9n}{1}\binom{0.9n-1}{1}\cdots\binom{0.9n-48}{1}\times\binom{0.1n}{1}}{\binom{n}{1}\binom{n-1}{1}\cdots\binom{n-48}{1}\times\binom{n-49}{1}}=\dfrac{49}{50}$ and similarly for 7 & 8.

How can we get the answer?

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The total number of red balls drawn after $k$ subsequent draws is $49 + 7k$. The total number of balls drawn is $50 + 8k$.

The ratio of the is equal to the proportion of the red balls in the bag, which is at least $0.9$. In other words,

$$\frac{49 + 7k}{50 + 8k} \ge 0.9$$ $$49 + 7k \ge 45 + 7.2k$$ $$0.2k \le 4$$ $$k \le 20$$

Hence, the total number of balls drawn,

$$50 + 8k \le 50 + 20\cdot8 = 210$$

This says that at most $210$ balls can be drawn. So there must be at most $210$ balls in the bag, and so there cannot be $250$ balls in the bag (because $250 > 210$).

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nice answer@yiyuan – Sameh Shenawy Apr 5 '14 at 7:58
Thank you so much, Sir. So good answer! – Silent Apr 5 '14 at 7:59

Set up the ratio: $\frac{49 + 7x}{50 + 8x} = \frac{9}{10}$. Solving it explicitly gives you $x = 20$, so (2) is a valid answer.

If $x = 15$, we get $170$ as the denominator, and $90.6\%$.

When $x = 18$, we get $194$ as the denominator and $90.2\%$

When $x = 25$, we get 250 in the denominator and $89.6\%$.

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Thank you so much :) – Silent Apr 5 '14 at 7:58