Statistics Question, Bayes Rule, Barometer Question

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Bayes rule states that P(A|B) = [P(B|A)P(A)]/P(B).

Let's suppose rain falls one day in three, and when it does a barometer shows rain with a probability of 0.7, otherwise it has a 0.1 probability of showing rain. Find the probability, that on a random day, the barometer shows "rain".

Does anyone know how to find the required probability by using Bayes Rule? Thank you in advance.

EDIT found very useful link as to what solution to use and when When to use Total Probability Rule and Bayes' Theorem.

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Bayes' Rule is unnecessary for this question. What you need is the Law of Total Probability:

$$ P(A) = \sum_{i} P(A|B_i)P(B_i). $$

In this case, where we have two possibilities: $B = \text{rain}$ and $\neg B = \text{no rain}$,you can write:

$$ P(A) = P(A|B) P(B) + P(A|\neg B)P(\neg B). $$

therefore:

$$ P(\text{barometer shows rain}) = P(\text{barometer shows rain}| \text{rain}) P(\text{rain}) + P(\text{barometer shows rain}|\text{no rain})P(\text{no rain}). $$

Plugging in the values one gets $P(\text{barometer shows rain}) = 0.7 \times \left(\frac{1} {3}\right) + 0.1 \times \left(1 - \frac{1}{3}\right) = 0.3.$