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An urn contains four (4) red chips and six (6) white chips. Two (2) chips are drawn out and discarded and a third chip is drawn. What is the probability that the third chip is red?

Would Hyper geometric Distribution be the best possible method for solving this?

Aash
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    No. Since we aren't told anything about the two chips that are discarded, the probability that the third chip is red is the same as the probability that the first chip is red: $\frac25$ – saulspatz Jun 21 '20 at 09:11

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The best solution is obviously the one written in the comment by @salulpatz: the probability that the first, the second, the third a.s.o. to estract a red chip is always $\frac{4}{10}$

But if you do not realize that, in a simple example like this exercise you can calculate it...

After drawing 2 chips, the urn will be the following

  • $\{2R;3W\}$ with a probability of $\frac{\binom{4}{2}}{\binom{10}{2}}=\frac{2}{15}$

  • $\{3R;5W\}$ with a probability of $\frac{\binom{4}{1}\binom{6}{1}}{\binom{10}{2}}=\frac{8}{15}$

  • $\{4R;4W\}$ with a probability of $\frac{\binom{6}{2}}{\binom{10}{2}}=\frac{5}{15}$

Then now the probability to draw a Red Chip is obviously

$$\mathbb{P}[R]=\frac{2}{15}\frac{2}{8}+\frac{8}{15}\frac{3}{8}+\frac{5}{15}\frac{4}{8}=\frac{2}{5}$$

as already known before this tedious calculations

tommik
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