# Updating Probability Using Bayes’ Formula

Bayes’ formula is used to calculate an updated or posterior probability given a set of prior probabilities for a given event. It is a theorem named after the Reverend T Bayes and is used widely in Bayesian methods of statistical inference.

It is the logic used to come up with the formula:

Let $$E_1, E_2, E_3, …, E_n$$ be a set of mutually exclusive and exhaustive events.

Using the conditional probability:

$$P(E_i | A) =\cfrac {P(E_i A)}{P(A)}$$

And also the relationship:

$$P(E_iA) = P(AE_i) = P(E_i)P(A | E_i)$$

And the total probability rule:

$$P(A) = \sum {P(AE_j)} \quad \text { for all j} = 1, 2,…,n$$

We can finally substitute for $$P(E_iA)$$ and $$P(A)$$ in equation 1.

This gives:

$$P(E_{ i }|A)=\cfrac { P(E_{ i })P(A|E_{ i }) }{ \sum _{ j=1 }^{ n }{ P(E_{ i })P(A|E_{ i }) } }$$

This is the Bayes’ formula, and it allows us to ‘turnaround’ conditional probabilities, i.e., we can calculate $$P(E_i|A)$$  if given information only about $$P(A|E_i)$$.

Take note of the explanations given below.

1. $$P(E_j)$$ are known as prior probabilities.
2. Event $$A$$ is some event known to have occurred.
3. $$P(E_i|A)$$ is the posterior probability.

#### Example: Bayes’ Formula

A Civil Engineer wishes to investigate the punctuality of electric trains by considering the number of train journeys. In the sample, 50% of trains were destined for New York, 30% for Vegas, and 20% for Washington, DC. The probabilities of a train arriving late in New York, Vegas, and Washington, DC, are 40%, 35%, and 25%, respectively. If the Engineer picks a train at random from this group, what is the probability that it would be one destined for New York?

Solution:

We are looking for $$P\text{(New York | Late)}$$.

Let us define the events that are critical in our calculation.

First, $$N$$ is the event “A train chosen at random would be one destined for New York.”

Secondly, $$V$$ is the event “A train chosen at random would be destined for Vegas.”

And $$W$$ is the event “A train chosen at random would be destined for Washington DC.”

Finally, let $$L$$ be the event “A randomly chosen would arrive late.”

\begin{align*} P(N|L) & =\cfrac { P(N)P(L|N) }{ P(N)P(L|N)+P(V)P(L|V)+P(W)P(L|W) } \\ & =\cfrac { 0.5×0.4 }{ 0.5× 0.4+0.3×0.35+0.2× 0.25 } \\ & =\cfrac { 0.2 }{ 0.355 } \\ & =0.5634 \\ & =56.3\% \\ \end{align*}

We have computed $$P(N | L)$$ given only $$P(L | N)$$, hence the phrase ‘turnaround conditional probability’.

## Question

A chartered analyst can choose any one of three routes, A, B, or C, to get to work. The probabilities that she arrives on time using routes A, B, and C are 50%, 52%, and 60%, in that order. If she is equally likely to choose any one of the routes and arrive on time, the probability that she chose route A is closest to:

A. 30.9%.

B. 16.67%.

C. 25%.

Solution

First, you should define the relevant events.

Let $$A$$ be the event “Chooses route A.”

Let $$B$$ be the event “Chooses route B.”

And let $$C$$ be the event “Chooses route C.”

Lastly, define event T as “Arrives to work on time.”

Now, what we have is $$P(T | A)$$, i.e., the probability that the analyst arrives on time, given that she chooses route A.

However, we want to find the turnaround probability $$P(A | T)$$, i.e., the probability that the analyst chooses route A, given that she arrives on time.

This is what calls for the application of Bayes’ formula:

\begin{align*} P(A|T) & =\cfrac { P(A)P(T|A) }{ P(A)P(T|A)+P(B)P(T|B)+P(C)P(T|C) } \\ & =\cfrac { \frac {1}{3} ×0.5 }{ \frac {1}{3} ×0.5+\frac {1}{3} × 0.52+\frac {1}{3} × 0.6 } \\ & =\cfrac { 0.16667 }{ 0.54 } \\ & =0.30865 \\ & =30.9\% \\ \end{align*}

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