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Let's consider first one bag:

The probability to grab one green marble is Pg=1/5

The probability to grab one red marble is Pr=1-Pg=4/5

For the 4 bags, it's a binomial distribution: probability to get k green out of n bags,

P(X=k)=nCk pk (1-p)n-k = nCk Pgk Prn-k

Now, the probability to grab at least 2 green marbles from 4 bags is

1 - (Probability to get no green marble from 4 bags) - (Probability to get just one green marble from 4 bags)

Probability to get no green marble is = 4C0 (4/5)4 = (4/5)4

(n=4, k=0, no green marble from each bag, 4C0 Pr4)

Probability to get just one green marble is = 4C1 (1/5) (4/5)3

(n=4, k=1, one green marble from one bag and red marbles from the other ones, 4C1 Pg Pr3)

Probability to grab at least 2 green marbles from 4 bags is

1-(4/5)4-4*(1/5)(4/5)3 = 0.1801

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Q: There are 4 bags each containing 4 red marbles and 1 green marble If you grab one marble from each bag what is the probability of grabbing at least 2 green marbles?
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