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f you randomly pick two of these coins and flip both, what is the probability that at most one of them will come up heads?

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First of all, you have an incorrect answer for (1). If you condition on the probability of drawing each respective coin and multiply by the probability of turning up heads, you have $$ \frac{1}{3}\bigg(\frac{5}{11}+\frac{4}{11}+\frac{3}{11}\bigg) = \frac{4}{11}$$

Then for (2), consider the probability of both coins turning up heads, and each case in which you can draw two coins. Taking the sum of each $$ 1 - P(\text{both chosen turning heads}) \cdot P(\text{picking these two coins}) $$
for each possible combination of two coins will give you your desired result.

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Surely you mean $4/11$ for the first one: the average of the three individual probabilities.

For the second one, consider the first two coins. What is the probability of at most coming up heads? It is one minus the probability that both come up heads—i.e., $1-(5/11)(3/11) = 1-15/121 = 106/121$. Now figure it out for the other two possible pairings. Finally, take the average of these three results to obtain the desired probability.

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