from the elementary sum rule $p(A) + p(\bar A) = 1$ and the product rule. Here, you may make use of the (Boolean logic) fact that $A + B = \overline {\bar A \bar B }$.
$$\begin{align*}
p\left( {A + B} \right) &\underset{\mathrm{bool}}{=} p\left( {\overline {\bar A \bar B } } \right) \ &\underset{\mathrm{sum}}{=} 1 - p\left( {\bar A \bar B } \right) \ &\underset{\mathrm{prod}}{=} 1 - p\left( {\bar A |\bar B } \right)p\left( {\bar B } \right) \ &\underset{\mathrm{sum}}{=} 1 - \left( {1 - p\left( {A|\bar B } \right)} \right)\left( {1 - p\left( B \right)} \right) \ &= p(B) + \left( {1 - p\left( B \right)} \right)p\left( {A|\bar B } \right) \ &\underset{\mathrm{prod}}{=} p(B) + \left( {1 - p\left( B \right)} \right)p\left( {\bar B |A} \right)\frac{{p\left( A \right)}} {{p\left( {\bar B } \right)}} \ &\underset{\mathrm{sum}}{=} p(B) + p\left( {\bar B |A} \right)p\left( A \right) \ &\underset{\mathrm{sum}}{=} p(B) + \left( {1 - p\left( {B|A} \right)} \right)p\left( A \right) \ &\underset{\mathrm{sum}}{=} p\left( A \right) + p(B) - p\left( {A,B} \right) \end{align*}$$ Note that, aside from the first boolean rewrite, everything follows straight application of sum and product rules.
$$\begin{align*} p(b_1) &= p(b_2) = 1/2\\ p(a|b_1) &= 8/12, \quad p(a|b_2)=10/12\\ p(o|b_1) &= 4/12, \quad p(o|b_2)=2/12 \end{align*}$$The following probabilities are given in the problem statement,
(a) $p(a) = \sum_i p(a,b_i) = \sum_i p(a|b_i)p(b_i)=\frac{8}{12}\cdot\frac{1}{2} + \frac{10}{12}\cdot\frac{1}{2} = \frac{3}{4}$
(b) $p(b_1|a) = \frac{p(a,b_1)}{p(a)} = \frac{p(a|b_1)p(b_1)}{p(a)} = \frac{\frac{8}{12}\cdot\frac{1}{2}}{\frac{3}{4}} = \frac{4}{9}$
$$\begin{align*} p(S_1=\text{t})&= 1/3\\ p(S_1=\text{f})&= 1 - p(S_1=\text{t})= 2/3\\ p(S_2=\text{y}|S_1=\text{t})&= 1/3 \\ p(S_2=\text{y}|S_1=\text{f})&= 2/3 \end{align*}$$We use variables $S_1 \in \{\text{t},\text{f}\}$ and $S_2 \in \{\text{y},\text{n}\}$ for statements 1 and 2 and shorthand "y", "n", "t" and "f" for "yes", "no", "true and "false", respectively. The problem statement provides us with the following probabilities,
We are asked to compute $p(S_1=\text{t}|S_2=\text{y})$. Use Bayes rule, $$\begin{align*} p(S_1=\text{t}|S_2=\text{y}) &= \frac{p(S_1=\text{t},S_2=\text{y})}{p(S_2=\text{y})}\\ &=\frac{\overbrace{p(S_2=\text{y}|S_1=\text{t})p(S_1=\text{t})}^{\text{both speak the truth}}}{\underbrace{p(S_2=\text{y}|S_1=\text{t})p(S_1=\text{t})}_{\text{both speak the truth}}+\underbrace{p(S_2=\text{y}|S_1=\text{f})p(S_1=\text{f})}_{\text{both lie}}}\\ &= \frac{\frac{1}{3}\cdot\frac{1}{3}}{\frac{1}{3}\cdot\frac{1}{3}+\frac{2}{3}\cdot\frac{2}{3}} = \frac{1}{5} \end{align*}$$
There are two hypotheses: let $H = 0$ mean that the original ball in the bag was white and $H = 1$ that is was black.
Assume the prior probabilities are equal. The data is that when a randomly selected ball was drawn from the bag, which contained a white one and the unknown one, it turned out to be white. The probability of this result according to each hypothesis is: $$ P(D|H =0) = 1,\quad P(D|H =1) = 1/2$$ So by Bayes theorem, the posterior probability of $H$ is $$P(H =0|D) = 2/3,\quad P(H =1|D) = 1/3$$
$$\begin{align*} p(S_1=\text{R}|S_2=\text{G}) &=\frac{p(S_2=\text{G}|S_1=\text{R})p(S_1=\text{R})}{p(S_2=\text{G}|S_1=\text{R})p(S_1=\text{R})+p(S_2=\text{G}|S_1=\text{G})p(S_1=\text{G})}\\ &= \frac{\frac{7}{11}\cdot\frac{5}{12}}{\frac{7}{11}\cdot\frac{5}{12}+\frac{6}{11}\cdot\frac{7}{12}} = \frac{5}{11} \end{align*}$$(a) $p(S_1=R) = \frac{N_R}{N_R+N_G}= \frac{5}{12}$
(b) The outcome of the $n$th draw is referred to by variable $S_n$. Use Bayes rule to get
When a data generating distribution is considered as a function of the model parameters for given data, i.e. $L(\theta) \triangleq \log p(D|\theta)$, it is called a likelihood. It is more correct to speak about the likelihood of a model (or of the likelihood of the parameters).
That depends. The term 'probabilistic' refers to a state-of-knowledge (or beliefs) about something (in this case, about the values of a speech signal). The fundamental issue here is to realize that the signal itself is not probabilistic (nor deterministic), but rather that these attributes reflect a state-of-knowledge. If you had a perfect microphone and recorded a speech signal perfectly at its source, then you would know all the signal values perfectly. You could say that the signal is deterministic since there is no uncertainty. However, before you would record the signal, how would you describe your state-of-knowledge about the signal values that your are going to record? There is uncertainty, so you would need to describe that speech signal by a probability distribution over all possible values.
where $\mathbb{E}[\cdot]$, $\mathbb{V}[\cdot]$ and $\mathbb{V}[\cdot,\cdot]$ refer to the expectation (mean), variance and covariance operators respectively. You may make use of the more general theorem that the mean and variance of any distribution $p(x)$ is processed by a linear tranformation as $$\begin{align*} \mathbb{E}[Ax +b] &= A\mathbb{E}[x] + b \\ \mathbb{V}[Ax +b] &= A\,\mathbb{V}[x]\,A^T \end{align*}$$
$$\begin{align*} \mathbb{E}[z] &= \mathbb{E}[Aw] \\ &= \mathbb{E}[x+y] \\ &= \mathbb{E}[x]+\mathbb{E}[y]\\ \mathbb{V}[z] &= \mathbb{V}[Aw] \\ &= A\mathbb{V}[w]A^T \\ &= \begin{bmatrix}I & I \end{bmatrix}\begin{bmatrix} \mathbb{V}[x] & \mathbb{V}[x,y] \\ \mathbb{V}[x,y] & \mathbb{V}[y]\end{bmatrix}\begin{bmatrix}I \\ I \end{bmatrix} \\ &= \mathbb{V}[x]+\mathbb{V}[y]+2\mathbb{V}[x,y] \end{align*}$$Define $A = [I, I]$, $w = [x;y]$ (where the notation ";" stacks the columns of $x$ and $y$ and $I$ is the identity matrix). Then $z = A w$. Now apply the formula for the mean and variance of a RV after a linear transformation.