Last question in the exercise file
In the last question, "What is the likelihood of getting admitted, having been offered a place on the waitlist?", the way it's calculated in the solutions is 33/1299. As I understand, that's because the numerator is 33/1299 and the denominator 1299/1299 right?
However, the way I calculated is (33/1299)/(1299/5678), so I divided intersection over the probability of being offered a place on the waitlist for all the students. Is this approach wrong?
Thank you for your question!
As the question asks, we would like to find the likelihood of being admitted to the college given that you have been offered a place on the waiting list. The number of students who have been offered a place on the waiting list is 1299. Out of these 1299, 33 students have been admitted. Therefore, there is a 33/1299 chance of being admitted given that you have been offered a place on the waiting list.
To answer your question: for this exercise, we don't actually need the probability of being offered a place on the waiting list (1299/5678). We start with the assumption that we are already on the waiting list and what we need is the probability of being admitted :)
Hope this helps!
i have also solved this using simple probability formula of number of favorable outcomes/total no of outcomes using 33/1299.
But can this also be solved using conditional probability formula? P (Being Admitted | offered a placed on the waitlist).
As this also involved condition to find probability, why cannot we use the Bayes rule here?
Thank you for your comment!
Could you share a solution using Bayes' theorem? I would be happy to see it.