Given that the chosen ticket is a first class ticket, what is the probability that it is a long distance ticket

conditional probabilityprobability

The question is:

Train tickets are distributed into three boxes. Each ticket can be classified as either First Class (FC) or Third Class (TC) and as either Long Distance (LD) or Short Distance (SD). Each box contains the following proportions tickets of each type:

\begin{array}{|c|c|c|c|}
\hline
\text{Box}& \text{FC and SD} & \text{FC and LD} & \text{TC and SD} & \text{TC and LD} \\ \hline
\text{Box 1}& 10\% & 25\% & 15\% & 50\%\\ \hline
\text{Box 2}& 24\% & 22\% & 33\% & 21\%\\ \hline
\text{Box 3}& 40\% & 40\% & 10\% & 10\%\\ \hline
\end{array}

A box is randomly selected and a ticket is randomly chosen from that box. Given that the chosen ticket is a first class ticket, what is the probability that it is a long distance ticket?

So far I have the following:

Let $C_1$ and $C_2$ denote the events of choosing either a first class ticket or a third class ticket, respectively. Further, let $D_1$ and $D_2$ denote the events of choosing a short distance or long distance ticket, respectively. Also, let $B_1,B_2,B_3$ denote the events of choosing the first, second or third box. We are asked what is the probability that it is a long distance ticket, given that the chosen ticket is a first class ticket. I have:

$$\Pr \left( {{D_2}\left| {{C_1}} \right.} \right) = \frac{{\Pr \left( {{D_2} \cap {C_1}} \right)}}{{\Pr \left( {{C_1}} \right)}} = \frac{{\sum\limits_{n = 1}^3 {\Pr \left( {{D_2} \cap {C_1} \cap {B_n}} \right)} }}{{\sum\limits_{n = 1}^3 {\Pr \left( {{C_1} \cap {B_n}} \right)} }}$$

Is it correct that I'm applying the law of total probability? If so, the numerator is easy to find from the table, but how would I find ${\sum\limits_{n = 1}^3 {\Pr \left( {{C_1} \cap {B_n}} \right)} }$?

Best Answer

Simply looking at your table, the probability to get a First class ticket is

$$\mathbb{P}[FC]=\frac{1}{3}\cdot0.35+\frac{1}{3}\cdot0.46+\frac{1}{3}\cdot0.80$$

similarly, the probability to get a ticket that is FC AND LD is

$$\mathbb{P}[FC \cap LD]=\frac{1}{3}\cdot 0.25+\frac{1}{3}\cdot0.22+\frac{1}{3}\cdot0.40$$

your answer is the ratio of the two above probabilities.

But before calculating it observe that, being the probability to choose a single box uniformly, this can be written off.

thus

$$\mathbb{P}[LD|FC]=\frac{0.25+0.22+0.4}{0.35+0.46+0.8}=\frac{87}{161}$$

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