<s>[INST] Probability regarding random marbles chosen from a bag I am unsure if my thought process for these questions is correct.

Susie has a bag of marbles containing 3 Red, 7 Green, and 10 Blue marbles.



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*What is the probability of picking 5 marbles and getting at least one red marble? Calculate the probability (a) with replacement, and (b) without replacement.


What I have tried so far is that with replacement, it would be $$1- \left(\frac{17}{20}\right)^5 = 0.556$$
For without replacement, I tried $$1- \left(\frac{17}{20} \times \frac{16}{19} \times \frac{15}{18} \times \frac{14}{17} \cdot \frac{13}{16}\right) = 0.601$$



*Calculate the probability of picking $8$ marbles and getting exactly $4$ green and $4$ blue (a) with replacement and (b) without replacement.


What I have tried for this so far is that with replacement, it would be $$\left(\frac{7}{20}\right)^4 \cdot \left(\frac{10}{20}\right)^4 \cdot 8^2 = 0.066$$
For without replacement, I am unsure so I just assumed it would follow the same as 1(b) based off of 1(a).
 [/INST] You solved both parts of the first problem correctly.
For the second part of the first problem, where you are selecting without replacement, you could also express the answer in the form
$$\Pr(\text{at least one red}) = 1 - \Pr(\text{no reds}) = 1 - \frac{\dbinom{17}{5}}{\dbinom{20}{5}}$$
where the denominator counts the number of ways of selecting five of the $20$ marbles in the bag and the numerator of the subtracted term counts the number of ways of selecting $5$ of the $20 - 3 = 17$ non-red marbles in the bag.

Susie has a bag of marbles containing $3$ red, $7$ green, and $10$ blue marbles.  Calculate the probability of picking $8$ marbles and getting exactly $4$ green and $4$ blue with replacement.

Since the selections are made with replacement, the probability of selecting a particular color is the same for each outcome.  We can use the multinomial distribution.  The probability of selecting exactly $r$ red, $g$ green, and $b$ blue marbles from $3$ red, $7$ green, and $10$ blue marbles when $n = r + g + b$ marbles are selected from the bag with replacement is
$$\Pr(r, g, b) = \binom{n}{r, g, b}\left(\frac{3}{20}\right)^r\left(\frac{7}{20}\right)^g\left(\frac{10}{20}\right)^b$$
Substituting $8$ for $n$, $0$ for $r$, $4$ for $g$, and $4$ for $b$ yields
$$\Pr(r, g, b) = \binom{8}{0, 4, 4}\left(\frac{3}{20}\right)^0\left(\frac{7}{20}\right)^4\left(\frac{10}{20}\right)^4$$

Susie has a bag of marbles containing $3$ red, $7$ green, and $10$ blue marbles.  Calculate the probability of picking $8$ marbles and getting exactly $4$ green and $4$ blue without replacement.

Since the marbles are selected without replacement, we can use the multivariate hypergeometric distribution.  The probability of selecting exactly $r$ red, $g$ green, and $b$ blue marbles when $n = r + g + b$ marbles are selected from the $3$ red, $7$ green, and $10$ blue marbles in the bag without replacement is
$$\Pr(r, g, b) = \frac{\dbinom{3}{r}\dbinom{7}{g}\dbinom{10}{b}}{\dbinom{20}{n}}$$
Can you finish?
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