We will calculate a probability under the assumption that the sisters are guessing. The probability of at least one correct match must be equal to one minus the probability of no correct matches. Let $F_i$ be the event that the sisters fail to match for the ith card shown. The probability of no correct matches is $\mathrm{P}\left(F_1 \cap F_2 \cap \ldots \cap F_{10}\right)$, where $\mathrm{P}\left(F_i\right)=\frac{2}{3}$ for each $i$. If we assume that successive attempts at matching cards are independent, we can multiply together the probabilities for these independent events, and so obtain
$$
\mathrm{P}\left(F_1 \cap F_2 \cap \ldots \cap F_{10}\right)=\mathrm{P}\left(F_1\right) \mathrm{P}\left(F_2\right) \ldots \mathrm{P}\left(F_{10}\right)=\left(\frac{2}{3}\right)^{10}=0.0173 .
$$
Hence the probability of at least one match is $1-0.0173=0.9827$.

Let $A$ be the event that you give the correct answer. Let $B$ be the event that you knew the answer. We want to find $\mathrm{P}(A)$. But $\mathrm{P}(A)=\mathrm{P}(A \cap B)+\mathrm{P}\left(A \cap B^c\right)$ where $\mathrm{P}(A \cap B)=\mathrm{P}(A \mid B) \mathrm{P}(B)=1 \times 0.75=0.75$ and $\mathrm{P}\left(A \cap B^c\right)=\mathrm{P}\left(A \mid B^c\right) \mathrm{P}\left(B^c\right)=\frac{1}{5} \times 0.25=0.05$. Hence $\mathrm{P}(A)=0.75+0.05=0.8$.

Let $E_i$ be the event that baby $i$ is reunited with its mother. We need $\mathrm{P}\left(E_1 \cup E_2 \cup E_3\right)$, where we can use the result
$$
\operatorname{Pr}(A \cup B \cup C)=\operatorname{Pr}(A)+\operatorname{Pr}(B)+\operatorname{Pr}(C)-\operatorname{Pr}(A \cap B)-\operatorname{Pr}(A \cap C)-\operatorname{Pr}(B \cap C)+\operatorname{Pr}(A \cap B \cap C) .
$$
for any $\mathrm{A}, \mathrm{B}, \mathrm{C}$. The individual probabilities are $\mathrm{P}\left(E_1\right)=\mathrm{P}\left(E_2\right)=\mathrm{P}\left(E_3\right)=\frac{1}{3}$. The pairwise joint probabilities are equal to $\frac{1}{6}$, since $\mathrm{P}\left(E_1 E_2\right)=\mathrm{P}\left(E_2 \mid E_1\right) \mathrm{P}\left(E_1\right)=\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)$, and the triplet $\mathrm{P}\left(E_1 E_2 E_3\right)=\frac{1}{6}$ similarly. Hence our final answer is
$$
\frac{1}{3}+\frac{1}{3}+\frac{1}{3}-\frac{1}{6}-\frac{1}{6}-\frac{1}{6}+\frac{1}{6}=\frac{2}{3}
$$

$\begin{aligned} L\left(\theta_1, \ldots, \theta_n \mid x_1, \ldots, x_n\right) & =\prod_{i=1}^n f\left(x_i \mid \theta_i\right) \ & =\prod_{i=1}^n \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2}\left(x_i-\theta_i\right)^2} .\end{aligned}$

Taking the logarithm of the likelihood function, we have

$\begin{aligned} \ell\left(\theta_1, \ldots, \theta_n \mid x_1, \ldots, x_n\right) & =\log L\left(\theta_1, \ldots, \theta_n \mid x_1, \ldots, x_n\right) \ & =-\frac{n}{2} \log 2 \pi-\frac{1}{2} \sum_{i=1}^n\left(x_i-\theta_i\right)^2 .\end{aligned}$

To find the maximum likelihood estimate of $\left(\theta_1,\ldots,\theta_n\right)$, we differentiate $\ell$ with respect to each $\theta_i$, set the derivatives equal to zero, and solve the resulting system of equations. Specifically,

$\frac{\partial \ell}{\partial \theta_i}=\sum_{j=1}^n\left(x_j-\theta_j\right) \delta_{i j}=x_i-\theta_i=0$

where $\delta_{ij}$ is the Kronecker delta. Thus, the maximum likelihood estimate of $\theta_i$ is $\hat{\theta}_i = x_i$, for $i=1,\ldots,n$.

(b). For $P_\theta=N(\theta, 1)$, determine the maximum likelihood estimate of $\theta$ when $\theta_1 = \cdots = \theta_n$.

When $\theta_1 = \cdots = \theta_n = \theta$, the likelihood function is given by

$L\left(\theta \mid x_1, \ldots, x_n\right)=\prod_{i=1}^n f\left(x_i \mid \theta\right)=\prod_{i=1}^n \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2}\left(x_i-\theta\right)^2}$

Taking the logarithm of the likelihood function, we have

$\begin{aligned} \ell\left(\theta \mid x_1, \ldots, x_n\right) & =\log L\left(\theta \mid x_1, \ldots, x_n\right) \ & =-\frac{n}{2} \log 2 \pi-\frac{1}{2} \sum_{i=1}^n\left(x_i-\theta\right)^2 .\end{aligned}$

To find the maximum likelihood estimate of $\theta$, we differentiate $\ell$ with respect to $\theta$, set the derivative equal to zero, and solve for $\hat{\theta}$. Specifically,

$\frac{\partial \ell}{\partial \theta}=\sum_{i=1}^n\left(x_i-\theta\right)=n \bar{x}-n \theta=0$

where $\bar{x} = \frac{1}{n} \sum_{i=1}^n x_i$ is the sample mean. Thus, the maximum likelihood estimate of $\theta$ is $\hat{\theta} = \bar{x}$.

Let $X$ and $Y$ be the numbers of wins in 4 and 8 games respectively. For 4 games there are $2^4=16$ equally likely outcomes e.g. $W L W W$ which has 3 wins so $X=3$. Using our basic counting principles there will be $\left(\begin{array}{l}4 \ j\end{array}\right)$ outcomes containing $j$ wins and so $P(X=3)=4 \times 0.5^4=0.25$.
Similarly with 8 games there are $2^8=256$ equally likely outcomes and this time $P(Y=5)=56 \times 0.5^8=$ $0.2188$ so the former is larger.
For part (b) remember that $X \geq 3$ means all the outcomes with at least 3 wins out of 4 etc and that we sum probabilities over mutually exclusive outcomes. Doing the calculations, $P(X \geq 3)=$ $0.25+0.0625=0.3125$ is less than $P(Y \geq 5)=0.2188+0.1094+0.0313+0.0039=0.3633-$ we deduce from this that the chance of a drawn series falls as the series gets longer.

There are $\left(\begin{array}{c}100 \ 3\end{array}\right)$ choices of three packages (in any ordering). There are $\left(\begin{array}{c}60 \ 3\end{array}\right)$ choices of three packages without prizes. Hence $P(X=0)=\left(\begin{array}{c}60 \ 3\end{array}\right) /\left(\begin{array}{c}100 \ 3\end{array}\right) \approx 0.2116$. If a single prize is won this can happen in $\left(\begin{array}{c}40 \ 1\end{array}\right) \cdot\left(\begin{array}{c}60 \ 2\end{array}\right)$ ways. Hence $P(X=1)=\left(\begin{array}{c}40 \ 1\end{array}\right) \cdot\left(\begin{array}{c}60 \ 2\end{array}\right) /\left(\begin{array}{c}100 \ 3\end{array}\right) \approx 0.4378$ and similarly $P(X=2)=\left(\begin{array}{c}40 \ 2\end{array}\right)$. $\left(\begin{array}{c}60 \ 1\end{array}\right) /\left(\begin{array}{c}100 \ 3\end{array}\right) \approx 0.2894$ and $P(X=3)=\left(\begin{array}{c}40 \ 3\end{array}\right) /\left(\begin{array}{c}100 \ 3\end{array}\right) \approx 0.0611$ (there is some small rounding error in the given values).