# Hypergeometric Probability Question Involving Drawing 100s of Balls

I was reviewing past finals to practice for an upcoming one, and saw this question with no answer:

A bowl contains 100 red balls, 200 white balls, and 300 blue balls. The balls are taken/drawn from the bowl, one at a time, until the bowl is empty.

(a) What is the probability that the ball taken out on draw number 101 is red?

(b) What is the probability that none of the first 100 balls drawn is red, and that the ball taken out on draw number 101 is red?

For (a), my guess is 1/6 -- but I'm not sure of actual answer and its logic.

For (b), would the answer be ((500 C 100)*(100 C 1)) / (600 C 101)?

For (a), my guess is 1/6 -- but I'm not sure of actual answer and its logic.

Every individual ball has equal probability for being the 101th ball drawn, and $$100$$ among the $$600$$ are red.

For (b), would the answer be ((500 C 100)*(100 C 1)) / (600 C 101)?

Not quite. You want the probability that all among the first 100 draws are from the 500 non-red and one from the 100 red are drawn immediately after, when selecting among 600 balls for those positions. The denominator must be evaluated in the same manner as the numerator (ie: treat the 101th place as a special position).

$$\dfrac{\dbinom{500}{100}}{\dbinom{600}{100}}\cdot\dfrac{\dbinom{100}{1}}{\dbinom{500}{1}}$$

Alternatively, when selecting 100 positions among the 600 possible, you want the probability for obtaining position 101, and 99 among the subsequent 499 positions.$$\dfrac{\dbinom{1}{1}\dbinom{499}{99}}{\dbinom{600}{100}}$$

This is so because colors have no preference for positions,
so a red ball will have a Pr of $$\frac16$$ at any position

For part (b), I misread it as given that the first $$100$$ balls are not red, .....
$$\dfrac{\binom{500}{400}}{\binom{600}{500}}\cdot\frac15\;\;$$ The $$\frac15$$ part is coming on the same logic as in part (a)
For part b) $$P=\frac{500\times 499\times 498\times......\times 401}{600\times 599\times 598\times......\times 501}\times \frac{100}{500}=\frac{1}{5}\times \frac{\binom{500}{400}}{ \binom{600}{500}}$$.
• Sirling's approx gives $P=\frac{1}{\sqrt{24}} \times (\frac{5}{4})^{400}\times (\frac{5}{6})^{600}$ Commented May 24, 2021 at 21:46