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Assume that there are 100 balls, 70 red and 30 blue.

If you pick 5 with replacement, what’s the probability that all 5 are red?

If you pick 5 without replacement, what’s the probability that all 5 are red?

If you pick 10 without replacement, what’s the probability that

**5 are red and 5 are blue**?

This question tests combinations with and without permutations.

- Since we have repetition, we can essentially estimate the probability through a binomial distribution \(Bin(5,7/10)\):

\[ \Pr(5 \text{ red with repetition}) = \binom{5}{5} (7/10)^5 \approx 0.17 \]

- To get all 5 red, initially, we have 70 options, then 69, then 68, etc. To get any 5 balls, the combinations are 100, then 99, then 98. etc. So we can estimate this probability as:

\[ \Pr(5 \text{ red without repetition}) = \frac{70 * 69 * 68 * 67 * 66}{100 * 99 * 98 * 97 * 96} \]

Alternatively, we can use the binomial coefficient as follows:

\[ \Pr(5 \text{ red without repetition}) = \frac{\binom{70}{5}}{\binom{100}{5}} = \frac{\frac{70!}{5!65!}}{\frac{100!}{5! 95!}} = \frac{70 * 69 * 68 * 67 * 66}{100 * 99 * 98 * 97 * 96} \approx 0.16 \]

You can read the second equation as *the number of ways I can choose 5 out of 70 reds over the number of ways that I can choose 5 out of 100 balls*.

- Similar to b, we can estimate this as follows:

\[ \Pr(5 \text{ red without repitition}) = \frac{\binom{70}{5} \binom{30}{5}}{\binom{100}{10}} \approx 0.004 \]

Counting, Combinations, Repetition, Binomial

- Trailing by two: should we go for two or three? Easy (Independence, Decision making)
- Median probability Medium (Binomial, Uniform, CDF)
- Consecutive tails Easy (Permutations, Repetition)
- Paths to destination Easy (Counting, Combinations)
- Largest number rolled Medium (Counting, Permutations, Repetition)