Ball drawing problem with a condition (with replacing and without replacing) You have $9$ balls in a bag: $4$ blue, $2$ purple, $3$ green. If you draw $3$ balls and I tell you that you've drawn at least $1$ blue, what is the probability that you've drawn $3$ blue balls?

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*solve for both, replacing and not replacing the balls.

My attempts so far: using Bayes' rule we get $P(3\ blue | al1\ blue) = \frac{P(al1\ blue | 3\ blue) \cdot P(3\ blue)}{P(al1\ blue)}$, here "$al1$" is "at least one."
Hence, without replacing I get: $$P(3\ blue | al1 \, blue) = \frac{1 \cdot \left(\frac{4}{9} \cdot \frac{3}{8} \cdot\frac{2}{7}\right)}{\left(1-all\ non\ blue\right)}=\frac{1 \cdot \left(\frac{4}{9} \cdot \frac{3}{8} \cdot \frac{2}{7}\right)}{\left(1-\frac{5}{9} \cdot \frac{4}{8}\cdot\frac{3}{7}\right)}=\frac{2}{37}.$$
With replacing, the formula is the same but numerators and denominators don't decrease and we get $\dfrac{16}{151}$.
This doesn't feel right, what are more intuitive ways to
think about this problem and hence solve it?
As this doesn't feel right, I have tried taking another approach which feels much better yet I am not 100% sure: there are $3$ possibilities: the first, the second and the third ball was blue. In case of without replacing summing them up will be:
$$\frac{3\cdot2}{8\cdot7} + \frac{4\cdot2}{8\cdot7} + \frac{4\cdot3}{9\cdot8} = \frac{105}{252},$$
and in case of replacements, again, the approach is the same but numbers dont decrease

As an alternative approach without replacement,

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*there in a sense ${9 \choose 3} = 84$ equally likely ways of choosing three balls from nine

*${4 \choose 3}=4$ are all blue

*${5 \choose 3}=10$ are none blue

*$84-10=74$ have at least one blue

so the conditional probability you want has probability $\frac{4}{74}=\frac{2}{37}$ as you found.
With replacement,

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*there in a sense $9^3 = 729$ equally likely ways of choosing the three balls from nine

*$4^3=64$ are all blue

*$5^3=125$ are none blue

*$729-125=604$ have at least one blue

so the conditional probability you want has probability $\frac{64}{604}=\frac{16}{151}$ as you also found.
