probability of drawing marbles based on coin flip 
An urn contains 4 red and 3 white marbles. A random marble $M_1$ is drawn and a fair coin is flipped. If the flip is heads then $M_1$ is put back into the urn. On the other hand, if the flip is tails, the marble $M_1$ is not put back into the urn. Now another random marble $M_2$ is drawn from the urn.
(i) What is $\mathrm{Pr}(M_2 = \text{red})$?
  (ii) What is $\mathrm{Pr}(M_1 = \text{red}\mid  M_2 = \text{red})$?
  (iii) What is $\mathrm{Pr}(\text{flip is heads}\mid M_2 = \text{white})$?
Solutions: 
  (i) $\dfrac{4}{7}$;   (ii) $\dfrac{15}{28}$;   (iii) $\dfrac{1}{2}$

(iii)'s answer is a half because the the marble color is independent of the coin flip, correct?
Why is (i)'s answer $\dfrac{4}{7}$, why not $\dfrac{3}{6}$ or $\dfrac{4}{6}$ if the coin flip outcome is tails?
As for (ii), I'm completely lost.
Can someone explain please?
 A: (i) Every marble is just as likely to be picked for $M_2$ as any other marble.  That is why the probability is $\frac{4}{7}$.  If you want to see this in more detail, lets look at all the possible ways of getting $M_2$ = red.  Flip heads: white then red, red then red, flip tails: White then red, red then red.  Total probability:
$$\frac{1}{2} \frac{3}{7} \frac{4}{7} + \frac{1}{2} \frac{4}{7} \frac{4}{7} + \frac{1}{2} \frac{3}{7} \frac{4}{6} + \frac{1}{2} \frac{4}{7} \frac{3}{6} = \frac{4}{7}$$
(ii) Using $P(A|B) = P(A \cap B) / P(B)$
$$ P(M_1 = \text{red} \cap M_2 = \text{red}) = \frac{1}{2} \frac{4}{7} \frac{3}{6} + \frac{1}{2} \frac{4}{7} \frac{4}{7} = \frac{15}{49}$$
Giving $\frac{15}{49}/\frac{4}{7} = \frac{15}{28}$ as the answer.
(iii) 
$$ P(\text{flip = heads} \cap M_2 = \text{white}) = \frac{1}{2} \left(\frac{4}{7} \frac{3}{7} + \frac{3}{7} \frac{3}{7}\right) = \frac{3}{14}$$
$P(M_2 = \text{white}) = \frac{3}{7}$ by similar argument to (i) and so using the conditional probability rule, the final answer is $\frac{3}{14} / \frac{3}{7} = \frac{1}{2}$
A: (i) 
Given there are 4 red and 3 white marbles, there is a $\frac 4 7$ probability that $M_1 =$ red is drawn and a $\frac37$ probability that $M_1 =$ white. Now here is the tricky bit. A fair coin is flipped. 


*

*If the coin flip is heads, then the urn will contain 4 red and 3 white for the 2nd draw.

*If the coin flip is tails, then there is a $\frac47$ probability that the urn will contain 3 red and 3 white; and a $\frac37$ probability that the urn will contain 4 red and 2 white.


Both these events have a $50/50$ probability of occurring. So taking into account all this
$$P(M_2 \text{is red}) = 0.5\begin{pmatrix}\frac47\end{pmatrix} + 0.5\begin{bmatrix}\begin{pmatrix}\frac47\end{pmatrix}\begin{pmatrix}\frac36\end{pmatrix} +\begin{pmatrix}\frac37\end{pmatrix}\begin{pmatrix}\frac46\end{pmatrix}\end{bmatrix} = \frac47$$
(ii)
$$P(M_1 \text{is red} \cap M_2 \text{is red})\\= 0.5\begin{pmatrix}\frac47\end{pmatrix}\begin{pmatrix}\frac47\end{pmatrix} + 
\begin{bmatrix}
0.5\begin{pmatrix}\frac47\end{pmatrix}\begin{pmatrix}\frac36\end{pmatrix}\end{bmatrix}=\frac{15}{49}
$$
Taking the conditional probability formula,
$$
\begin{align*}
P(M_1 \text{is red} | M_2 \text{is red}) &= \frac{P(M_1 \text{is red} \cap M_2 \text{is red})}{P( M_2 \text{is red})}\\
&= \frac{\frac{15}{49}}{\frac47}\\
&=\frac{15}{28}
\end{align*}
$$
