Answer
(a)
$\text{This is the probability that $E$ occurs if we know that $F$ has occurred.}$
$$
P(E | F)=\frac{P(E \cap F)}{P(F)}
$$
(b)
$\text{Two events will be independent if the occurrence of one of this events}$ $\text{does not affect probability of occurrence of the other event.}$
(c)
$P(E \cap F)=P(E) P(F | E)$
If $E$ and $F$ are independent, then
$$
P(E \cap F)=P(E) P(F)
$$
(d)
$\quad (i) $
$\quad$In this case $E$ and $F$ are independent, so
$\quad$$P(E \cap F)=P(E) P(F)=\frac{7}{10} \cdot \frac{7}{10}=\frac{49}{100}$
$\quad (ii)$
$\quad$In this case $E$ and $F$ are not independent, so
$\quad$$P(E \cap F)=P(E) P(F | E)=\frac{7}{10} \cdot \frac{6}{9}=\frac{7}{15}$
Work Step by Step
(a)
$\text{This is the probability that $E$ occurs if we know that $F$ has occurred.}$
$$
P(E | F)=\frac{P(E \cap F)}{P(F)}
$$
(b)
$\text{Two events will be independent if the occurrence of one of this events}$ $\text{does not affect probability of occurrence of the other event.}$
(c)
$P(E \cap F)=P(E) P(F | E)$
If $E$ and $F$ are independent, then
$$
P(E \cap F)=P(E) P(F)
$$
(d)
$\quad (i) $
$\quad$In this case $E$ and $F$ are independent, so
$\quad$$P(E \cap F)=P(E) P(F)=\frac{7}{10} \cdot \frac{7}{10}=\frac{49}{100}$
$\quad (ii)$
$\quad$In this case $E$ and $F$ are not independent, so
$\quad$$P(E \cap F)=P(E) P(F | E)=\frac{7}{10} \cdot \frac{6}{9}=\frac{7}{15}$
