Find the probability that A drew it on the first draw A and B draw coins in turn without replacement from a bag containing $3$ dimes and $4$ nickels. A draws first. It is known that A drew the first dime. Find the probability that A drew it on the first draw.
I know that the probability of drawing the first dime on the first draw must be $\frac{3}{7}$. Is this the correct answer?
 A: P$[$A draws dime on first draw draws first dime$|$ A draws first dime$]=\dfrac{P(\mbox{A draws dime on the first draw })}{P(\mbox{A draws first dime})}$
So, $$P(\mbox{A draws dime on first draw})=\dfrac37$$
Since, there are only $3$ dimes,  in order for $A$ to draw the first dime, this must happen on $A$'s first,second or third draw.  Thus, we need$$P(\mbox{A draws first dime})=P(\mbox{A draws dime on first draw})+P(\mbox{A draws first dime on second draw})+P(\mbox{A draws first dime on third draw.})$$
$$P(\mbox{A draws dime on second draw})=\dfrac47\cdot\dfrac36\cdot\dfrac35=\dfrac{6}{35}$$Because $A$'s first draw is one of the four non-dimes and $B$'s first draw is on of th three remaining non-dimes after $A$'s draw, and $A$'s second draw is one of the three dimes of the five remaining coins. Similarly, $$P(\mbox{A draws first dime on the third draw})=\dfrac47\cdot\dfrac36\cdot\dfrac25\cdot\dfrac14=\dfrac{1}{35}$$
Then, $$P(\mbox{A draws first dime})=\dfrac37+\dfrac{6}{35}+\dfrac{1}{35}=\dfrac{22}{35}$$
$$P(\mbox{A draws dime on first draw}|\mbox{A draws first dime})=\dfrac{\dfrac37}{\dfrac{22}{35}}=\dfrac{15}{22}$$
A: Hint: The probability of drawing a dime on the first draw is $\frac{3}7$, but what if the first A's draw is a nickel, and then B draws another nickel? A can obtain the first dime in its second try...
A: If we don't know A drew the first dime, $\frac{3}{7}$ is the correct answer.
But given that A drew the first dime, it turns to be a conditional probability.
Let $E_1=${A drew the first dime}, $E_2$={A drew a dime on the first draw}, then
$$P(E_2|E_1)=\frac{P(E_1\cap E_2)}{P(E_1)}=\frac{\frac{3}{7}}{\frac{3}{7}+\frac{4}{7}*\frac{3}{6}*\frac{3}{5}+\frac{4}{7}*\frac{3}{6}*\frac{2}{5}*\frac{1}{4}*\frac{3}{3}}=\frac{15}{22}$$
A: Let us model the situation. The space $\Omega$ is the space of all possibilities to extract the seven coins in order,
$$\Omega=\Big\{\omega=(\omega_1,\omega_2,\dots,\omega_7)\ :\ \omega_j\in\{D,N\}\text{ and there are 3D and 4N components }\Big\}\ .$$ 
Each element (better, one element subset of $\Omega$) has the same probability, this defines a probability $\Bbb P$ on $\Omega$ (with its power set as sigma-algebra). The elements of $\omega$ are in a bijection with the subsets with $3$ elements of $\{1,2,3,4,5,6,7\}$, by mapping $\omega$ to the positions in $\omega$ where there is a D component.
Now we let the players A and B extract alternatively for each "path of the chances" $\omega\in\Omega$, so A will extract for a fixed $\omega$ its components $\omega_1,\omega_3,\omega_5,\omega_7$. An B the other ones. The question asks for the conditional probabiltiy $$\Bbb P(S|T)\ ,$$
where $T$ is the set corresponding to "player extracts the first D", i.e. the first D occurs on an odd indexed place, and $S$ is the the event $\{\omega\ :\ \omega_1=D\}$. It is clear that $S\subset T$. The event $T$ corresponds to the cases:
D******
NND****
NNNND** = NNNNDDD

So we ( need to) compute 
$$\Bbb P(S|T)
= \frac{\Bbb P(S\cap T)}{\Bbb P(T)}
= \frac{\Bbb P(S)}{\Bbb P(T)}
= \frac{\frac 37}
{\frac 37 
 +\frac 47\cdot \frac 36\cdot \frac 35
 +\frac 47\cdot \frac 36\cdot \frac 25\cdot \frac 14\cdot \frac 33}
=
\frac {15}{22}
\ .
$$
