Answered

(c) Is it more likely that a person who takes a train to work lives in Chicago or more likely that a person
who lives in Chicago will take a train to work. Support your work using conditional probabilities.



Answer :

chibabykalu16

Step-by-step explanation:

[tex]\boxed{\begin{array}{l}\underline{\textsf{Probability}}\\\sf \text{Given:} \\\phantom{ww}\bullet\;\textsf{$P(\text{Chicago})$: Probability of living in Chicago.} \\\phantom{ww}\bullet\;\textsf{$P(\text{Train})$: Probability of taking a train.} \\\phantom{ww}\bullet\;\textsf{$P(\text{Train} \mid \text{Chicago})$: Probability of taking a train given living in Chicago.} \\\phantom{ww}\bullet\;\textsf{$P(\text{Chicago} \mid \text{Train})$: Probability of living in Chicago given taking a train.} \end{array}}[/tex]

To address this question, we need to understand two key concepts in probability: the conditional probability of taking a train given living in Chicago, and the conditional probability of living in Chicago given taking a train. These can be represented as follows:

1. [tex]\sf  P(\text{Train} \mid \text{Chicago}) [/tex] - Probability of taking a train given that a person lives in Chicago.

2. [tex]\sf  P(\text{Chicago} \mid \text{Train}) [/tex] - Probability of living in Chicago given that a person takes a train.

We also need to consider the overall probability of living in Chicago ([tex]\sf  P(\text{Chicago}) [/tex]) and the likelihood of taking a train ([tex]\sf  P(\text{Train}) [/tex]).

  1. [tex]\sf  P(\text{Chicago}) [/tex]: Probability a person lives in Chicago. This probability is relatively low given that Chicago is one city among many in the U.S..
  2. [tex]\sf  P(\text{Train}) [/tex]: Probability a person takes a train. This will depend on whether train travel is common in their region.
  3. [tex]\sf  P(\text{Train} \mid \text{Chicago}) [/tex]: This is likely higher because Chicago has a well-developed public transportation system including trains (like the 'L').
  4. [tex]\sf  P(\text{Chicago} \mid \text{Train}) [/tex]: This is likely lower since many cities in the U.S. and around the world have train systems, not just Chicago.

To find these probabilities, we use the formula for conditional probability:

[tex]\sf  P(A \mid B) = \frac{P(A \cap B)}{P(B)} [/tex]

Where:

  • [tex]\sf  A [/tex] and [tex]\sf  B [/tex] are events,
  • [tex]\sf  P(A \cap B) [/tex] is the probability of both [tex]\sf  A [/tex] and [tex]\sf  B [/tex] occurring,
  • [tex]\sf  P(B) [/tex] is the probability of [tex]\sf  B [/tex] occurring.

Based on the general data:

  1. [tex]\sf  P(\text{Chicago}) [/tex] is the proportion of the population living in Chicago compared to the total population.
  2. [tex]\sf  P(\text{Train}) [/tex] is the proportion of the population that uses the train.
  3. [tex]\sf  P(\text{Train} \cap \text{Chicago}) [/tex] would be higher as a proportion of Chicago's population uses the train.

Assuming:

Chicago's population is less compared to the total population using trains across various cities, [tex]\sf  P(\text{Chicago}) [/tex] is lower than [tex]\sf  P(\text{Train}) [/tex].

Therefore:

  • [tex]\sf  P(\text{Train} \mid \text{Chicago}) [/tex] is high because a significant portion of Chicago's population uses the train due to its availability and convenience.
  • [tex]\sf  P(\text{Chicago} \mid \text{Train}) [/tex] is low because many train users live in different cities.

It is more likely that a person who takes a train to work lives in a city other than Chicago (given the global use of trains) than a person who lives in Chicago will take a train (even though many do). Hence, [tex] P(\text{Train} \mid \text{Chicago}) [/tex] is more significant relative to [tex]\sf P(\text{Chicago} \mid \text{Train})[/tex].