Probability that value of coins is not a multiple of 5 I have a probability to calculate from a course in discrete mathematics. I offer a solution.  Is my solution correct?
Problem
A jar has 70 cents in change: ten pennies and twelve nickels. Three coins are randomly drawn. What is the probability that the value of the coins is not a multiple of 5 if at least one of the coins drawn is a nickel?
Solution
If at least one of the selected coins is a nickel, the probability that at least one is a penny is the complement of the probability that both of the remaining coins selected are nickels. The probability that both of the remaining coins chosen are nickels is
\begin{equation*}
\dfrac{\binom{11}{2}}{\binom{21}{2}} = \frac{11}{42} .
\end{equation*}
So, the probability that at least one of the remaining two coins chosen is a penny is
\begin{equation*}
1 - \frac{11}{42} = \frac{31}{42} .
\end{equation*}
 A: $P(A\mid B)=\frac{Pr(A\cap B)}{Pr(B)}$
Let $A$ represent the event that the total value of the coins is a multiple of five.
Let $B$ represent the event that at least one of the coins is a nickel.
We calculate: $Pr(B)=1-Pr(B^c) = 1-Pr(\text{all three coins are pennies}) = 1-\dfrac{\binom{10}{3}}{\binom{22}{3}}$
$Pr(A\cap B)$, we notice by inspection that in order to have the total value to be a multiple of five we must have all coins as nickels (since 3 pennies, 2pennies 1 nickel, and 1 penny 2 nickels all have values not a multiple of five), so $A\subseteq B$, we can simplify:
$Pr(A\cap B)=Pr(A)=Pr(\text{all three coins are nickels}) = \dfrac{\binom{12}{3}}{\binom{22}{3}}$
Combining this information:
$$Pr(A\mid B) = \dfrac{\frac{\binom{12}{3}}{\binom{22}{3}}}{1-\frac{\binom{10}{3}}{\binom{22}{3}}}=\frac{\binom{12}{3}}{\binom{22}{3}-\binom{10}{3}} = \frac{11}{71}$$
Thus $Pr(\text{is NOT a multiple of five}\mid \text{at least one nickel})=Pr(A^c\mid B)=1-Pr(A\mid B)=\frac{60}{71}$

Note that the statement "given at least one of the coins is a nickel" is a different question than "given the first coin is a nickel"
Although the question doesn't mention whether we pull coins one at a time or all simultaneously, we may, to make calculations easier, assume we pull them one at a time in order to have a better idea of what is going on since making the action of pulling coins longer and more tedious doesn't in any way affect the probabilities.
Your calculations above were as though the first coin was a nickel, however this doesn't take into account the possibilities where the first coin was a penny and the second was a nickel or the third was a nickel.
