# Conditional probability

A girl goes to school by bus every day. If it doesn't rain, probability that she will be late for bus is 1/5. If it rains probability that she will be late is 2/3. Probability that it is raining is 1/4.

Girl forgot to pick up a bus. Define the probability that it was raining.

In this task I have done following using my logic, but I am not sure is it correct: P(late for bus and rained that day) = P(rains)/P(late for school and is raining )= 1/4/2/3 = 3/8.

I am not sure if it is correct, so I plead for your help.

• What do you mean by "forgot to pick up a bus"? Can we consider "forgetting" as a reason for "being late for a bus"? Jul 2, 2012 at 19:35
• I am sorry I didn't properly explained this situation. When I meant "forgot to pick up a bus" I wanted to say " she was late for the bus that day ". Jul 2, 2012 at 19:38

It may be useful to introduce some symbols. Let $L$ be the event she is late for the bus, and let $R$ be the event it rains. We want the probability that it rains, given that she is late. In symbols, we want $\Pr(R|L)$.

There are some useful formulas that come into play. The most basic one comes from the definition of conditional probability. We have $$\Pr(R|L)\Pr(L)=\Pr(R\cap L).$$ If we can calculate $\Pr(L)$ and $\Pr(R\cap L)$ we will be finished.

We first compute $\Pr(R\cap L)$. The probability it rains is $\frac{1}{4}$. Given that it rains, the probability she is late is $\frac{2}{3}$. So the probability it rains and she is late is $\frac{1}{4}\cdot\frac{2}{3}$.

Next we compute $\Pr(L)$. She can be late in two ways: (i) It rains and she is late or (ii) It doesn't rain and she is late. We already computed the probability of (i). In the same way, we find that the probability it doesn't rain and she is late is $\frac{3}{4}\cdot\frac{1}{5}$. To find $\Pr(L)$, add together the probabilities of (i) and of (ii).

Remark: Here is an imprecise but useful way to view the calculation. Imagine we track what happens over $300$ days. It will rain about $75$ days. On two-thirds of these days she will be late, so she will be late on about $50$ days because of rain. It will not rain about $225$ days, and she will be late on about one-fifth of these days, so $45$ days.

So she is late a total of $50+45$ days. Let's confine attention (cut down the sample space) to days she was late. On $50$ of those days it was raining. So the probability it is raining given that she is late should be about $\frac{50}{95}$.

• This is truly extraordinary and explained me this task in a most great way! Now I understand it fully! Jul 2, 2012 at 19:59
• I can't give you +1 I need 15 rep. :( Jul 2, 2012 at 20:01
• @Takarakaka: It is not a problem, knowing it was useful is worth much more. Jul 2, 2012 at 20:09
• It explained me a lot of things here! Thank you, @Andre :) Jul 2, 2012 at 20:12

This is an application of the Bayes' theorem: $$\mathbb{P}(\text{rain}| \text{on time}) = \frac{\mathbb{P}(\text{rain} \land \text{on time} )}{\mathbb{P}(\text{on time} )} = \frac{\mathbb{P}( \text{on time} | \text{rain} ) \mathbb{P}(\text{rain})}{\mathbb{P}(\text{on time} | \text{rain} ) \mathbb{P}(\text{rain}) + \mathbb{P}(\text{on time} | \text{no rain} ) \mathbb{P}(\text{no rain})} = \frac{\frac{1}{3} \cdot \frac{1}{4} }{ \frac{1}{3} \cdot \frac{1}{4} + \frac{4}{5} \cdot \frac{3}{4} } = \frac{5}{41}$$

• Thank you on the formula! This looks a bit complicated, but will figure it out! Jul 2, 2012 at 19:42

Given that it rains, the probability she is late is 2/3 is P(L|R) not P(R|L), so your answer must be corrected.