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.
There are many ways of making a prediction.
Let us denote by $p$ the mean of the Bernoulli trials. We have two complementary hypothesis:
- $H_0$ is the hypothesis $p = 1/2$
- $H_1$ is the hypothesis $p = 3/4$
We could define the following test:
- We accept $H_0$ if and only if the empirical mean $\overline{X}:= \frac{1}{n}\sum_{i=1}^n X_i$ is less or equal then $5/8$ (i.e.: the empirical mean is nearer to $1/2$ then to $3/4$).
What is the probability of the prediction being wrong? $$\mathbb{P}(\text{ wrong estimate }| p = 1/2) = \mathbb{P}(\overline{X} > 5/8 | p = 1/2)$$
We find:
Chebychev: $$\mathbb{P}(\overline{X} > 5/8 | p = 1/2) \le \mathbb{P}(|\overline{X} - 1/2|^2 > 1/64 | p = 1/2) \le \frac{1/4}{n (1/64)} = 0.05 \text{ if n = $320$ } $$
Chernoff: $$\mathbb{P}(\overline{X} - 1/2 > 1/8 | p = 1/2) \le e^{-2n(1/64)} \approx 0.024 \text{ if $n = 119$ }$$
Similarly one can deal with:
$$ \mathbb{P}(\text{ wrong estimate }| p = 3/4) = \mathbb{P}(\overline{X} \le 5/8 | p = 3/4) $$
So the $\epsilon$ you were missing in the variance is actually imposed by the test you choose.
P.s.: you might see that with this test you get that Chernoff errs with probability less then $5\%$ already after $96$ trials.
Furthermore you should notice that since Chernoff does not depend on the distribution (but only on the boundedness of the distribution) you do not have to check the Chernoff estimate again in the second case ($p= 3/4$).
Best Answer
The first four look OK. For the fifth, you need to subtract $\binom{26}{5}$, the number of all red hands, from the total number of hands.
For the sixth, you must subtract the number with no repetitions from the total. The number with no repetitions is $(26)(25)(24)(10)(9)(8)(7)$. The total is $26^310^4$. So the answer is $26^310^4-(26)(25)(24)(10)(9)(8)(7)$.
The seventh and eighth look right. The added question is done incorrectly. First of all there is the implicit assumption that there are $4$ men and $4$ women. But even assuming that, $\frac{3}{7}$ is clearly much too large.
My way of doing it is to observe that the seats for the men can be chosen in $\binom{8}{4}$ equally likely ways. Of these, only $5$ have the men next to each other.
You could also use "permutations" to solve the problem. That multiplies both "top" and "bottom" by $(4!)^2$.