When events are not independent we must use conditional probability. The conditional probability of A occurring given B is the probability of event A occurring given that event B has already taken place and is denoted \[\mathrm{P}(B|A).\]
We can use this conditional probability, with the multiplication law, to give:
\begin{equation} \mathrm{P}(A \; \text{and} \; B) = \mathrm{P}(A) \times \mathrm{P}(B|A) \end{equation}
or
\begin{equation} \mathrm{P}(A \; \text{and} \; B) = \mathrm{P}(B) \times \mathrm{P}(A|B) \end{equation}
$70\%$ of households in Newcastle watch the news on an evening and $50\%$ of households watch the news on both the evening and in the morning. What is the probability that a household, which watches the news in the evening, will also watch the morning news?
First we denote by $E$ the event that the household watches the news on the evening and $M$ the event that the household watches the morning news.
From the question we have $\mathrm{P}(E) = 0.7$ and $\mathrm{P}(E \; \text{and} \; M) = 0.5$.
We want to calculate $\mathrm{P} (M \vert E)$.
We can use and rearrange the equation of the multiplication law in the blue box above as follows:
\begin{equation} \begin{split} \mathrm{P}(E \; \text{and} \; M) &&= \mathrm{P}(E) \times \mathrm{P}(M|E)\\ \mathrm{P}(M \vert E) &&= \dfrac{\mathrm{P}(E \; \text{and} \; M)}{\mathrm{P}(E)}\\ \end{split} \end{equation} Inserting the values of these probabilities given above into this equation gives:
\begin{align} \mathrm{P}(M \vert E) &= \dfrac{0.5}{0.7}\\ &=\dfrac{5}{7}\\ &= 0.714\text{ (to 3 d.p).}\\ \end{align}
The probability that a household that watches the news on the evening also watches the morning news is $\dfrac{5}{7} = 0.714\text{ (to 3 d.p).}$
To develop these ideas further see To develop these ideas further see the pages on discrete probability distributions and continuous probability distributions.