Conditional probability, In studying the causes of power failures, the following data have been gathered:

Question

In studying the causes of power failures, the following data have been gathered: 10% are due to a transformer damage, 75% are due to line damage, 5% involve both problems. Based on these percentages, find the probability that a given power failure involves:

a) line damage given that there is transformer damage

b) transformer damage given that there is line damage

c) transformer damage but not line damage

d) transformer damage given that there is no line damage

e) transformer damage or line damage.

Now, I am familiar with conditional probabilities ( to some degree at least ) and the first thing that came to my mind for point a) was Bayes

so

$ T $ - it involves transformer damage

$ L $ - it involves line damage

$ B $ - it involves both

$$ P(L/T)=\frac{P(L)*P(T/L)}{P(T)} $$

but the problem is that I am stuck at $P(T/L)$ I have no idea from were to start and maybe this approach is not even the correct one so I would appreciate some help, maybe a hint on how to proceed...

Answer 1

According to your notation, we have $\textbf{P}(T) = 0.1$, $\textbf{P}(L) = 0.75$ and $\textbf{P}(T\cap L) = 0.05$. Therefore:

(a) The sought probability is given by $\textbf{P}(L|T)$, which can be rewritten as \begin{align*} \textbf{P}(L|T) = \frac{\textbf{P}(L\cap T)}{\textbf{P}(T)} = \frac{0.05}{0.1} = 0.5 \end{align*}

(b) Analogously, we have \begin{align*} \textbf{P}(T|L) = \frac{\textbf{P}(T\cap L)}{\textbf{P}(L)} = \frac{0.05}{0.75} = \frac{1}{15} \end{align*}

(c) Here, the event in which we are interested in is described by $\textbf{P}(T\cap L^{c})$: \begin{align*} \textbf{P}(T\cap L^{c}) = \textbf{P}(T) - \textbf{P}(T\cap L) = 0.1 - 0.05 = 0.05 \end{align*}

(d) In this case, the target event is described by $\textbf{P}(T|L^{c})$: \begin{align*} \textbf{P}(T|L^{c}) = \frac{\textbf{P}(T\cap L^{c})}{\textbf{P}(L^{c})} = \frac{\textbf{P}(T\cap L^{c})}{1 - \textbf{P}(L)} = \frac{0.05}{1 - 0.75} = \frac{0.05}{0.25} = 0.2 \end{align*}

(e) Finally, the last event is given by \begin{align*} \textbf{P}(T\cup L) = \textbf{P}(T) + \textbf{P}(L) - \textbf{P}(T\cap L) = 0.1 + 0.75 - 0.05 = 0.8 \end{align*}

Hope this helps.

Answer 2

Going off my hint: recall that for two events $X$ and $Y$ (with $P(Y) \neq 0$) we have

$$ P(X|Y) = \frac{P(X \cap Y)}{P(Y)} $$

So in your case, you want

$$ P(L|T) = \frac{P(L \cap T)}{P(T)} $$

Note that the event $L \cap T$ is exactly $B$. So

$$P(L|T) = \frac{P(B)}{P(T)}$$