Another approach is to use Bayes theorem: For two independent events A and B,
$$
P(A|B)= \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|A^c)P(A^c)}
$$
Question 1:
Let event A be 'Chosen coin is double headed' and Event B be 'coin lands on head all 7 flips".
- P($A$)=P(Chosen coin is double headed)= $ \frac{1}{100}$
- P($A^c$)=P(Chosen coin is not double headed)= $ \frac{99}{100}$
- P($B|A$)=P(All 7 heads given coin is double headed)=1
- P($B$|$A^c$)=P(All 7 heads given coin is not double headed)= $ \frac{1}{2^7} $
Therefore,
$$
P(A|B)= \frac{\frac{1}{100} * 1}
{ \frac{1}{100} * 1 + \frac{1}{2^7} *\frac{99}{100} }
\approx 0.5639
$$
Question 2(a):
Let event A be 'One of the 100 coins is Double head' and Event B be 'coin lands on head all 7 flips".
- P($A$)=P(One of the 100 coins in Double head)= $ \frac{1}{2}$
- P($A^c$)=P(None of the 100 coins is Double head)= $ \frac{1}{2}$
- P($B|A$)=P(All 7 heads given one of the coin is double headed)= $\frac{1}{100} + \frac{1}{2^7} * \frac{99}{100} $
- P($B$|$A^c$)=P(All 7 heads given no coin is double headed)= $ \frac{1}{2^7} $
Therefore,
$$
P(A|B)= \frac{(\frac{1}{100} + \frac{1}{2^7} * \frac{99}{100} ) * \frac{1}{2}}
{ (\frac{1}{100} + \frac{1}{2^7} * \frac{99}{100} ) * \frac{1}{2} + \frac{1}{2^7} *\frac{1}{2} }
\approx 0.6942
$$
Question 2(b):
Let event A be 'Chosen coin is Double head' and Event B be 'coin lands on head all 7 flips".
- P($A$)=P(One of the 100 coins in Double head)= $ \frac{1}{2} * \frac{1}{100} + 0$
- P($A^c$)=P(None of the 100 coins is Double head)= $ \frac{1}{2} + \frac{1}{2} * \frac{99}{100}$
- P($B|A$)=P(All 7 heads given coin is double headed)=1
- P($B$|$A^c$)=P(ll 7 heads given coin is not double headed)= $ \frac{1}{2^7} $
Therefore,
$$
P(A|B)= \frac{1*(\frac{1}{2} * \frac{1}{100} + 0)}
{ 1*(\frac{1}{2} * \frac{1}{100} + 0) + \frac{1}{2^7} * ( \frac{1}{2} + \frac{1}{2} * \frac{99}{100}) }
\approx 0.3194
$$
As pointed out at Number of tosses to ensure $95\%$ that the coin selected is double-headed, the existing answer to this question is incorrect, as it doesn't take into account the given information on how the coin was chosen.
The existing answer answers the question how many times you'd have to throw heads in a row so that the probability for a fair coin to produce that result would be less than $5\%$. I understand the question to ask instead how many times you'd have to throw heads in a row so that the conditional probability for the coin to be fair, given how it was chosen, would be less than $5\%$.
As calculated by @Idonknow in the question linked to above, the answer is $8$. The conditional probability for the coin to be double-headed after $n$ heads in a row is
\begin{eqnarray*}
P(\text{double}\mid\text{$n$ heads})
&=&
\frac{P(\text{double}\land\text{$n$ heads})}{P(\text{$n$ heads})}
\\
&=&
\frac{P(\text{$n$ heads}\mid\text{double})P(\text{double})}{P(\text{$n$ heads})}\;.
\\
&=&
\frac{1\cdot\frac1{10}}{1\cdot\frac1{10}+\left(\frac12\right)^n\cdot\frac9{10}}
\\
&=&
\frac1{1+9\cdot2^{-n}}\;.
\end{eqnarray*}
For this to be $\ge95\%$, we need $1+9\cdot2^{-n}\le\frac1{0.95}$ and thus
$$
n\ge\log_2\frac9{\frac1{0.95}-1}=\log_2171\approx7.4\;,
$$
so we need $8$ consecutive heads to be $95\%$ sure that the selected coin is the double-headed coin.
Note that the number of fair coins appears in the numerator of the argument of the logarithm. That means that every time we double the number of fair coins, we need one more throw of heads to reach the same level of certainty. For instance, if we had $4\cdot9=36$ fair coins and one double-headed one to choose from, we'd need $10$ consecutive throws of heads to reach $95\%$ certainty that we chose the double-headed one.
Best Answer
If $DH$ stands for double-headed coin, and $H$ for Heads, you need $$p(DH|H)=\frac{p(DH\cap H)}{p(H)}$$ $$=\frac{\frac 12\times 1}{\frac 12\times 1+\frac 12\times\frac 12}$$ $$=\frac 23$$