Probability Bayes Rule

Introduction

In practice, Bayes's formula is very useful when it comes to conditional probabilities. The idea behind the rule is that, when having two events $A$ and $B$ and a conditional event $A\mid B$, you can compute the probability of event $B\mid A$.

Learning objectives

In this lab, you'll

• learn how to apply Bayes' rule, both in the simple form and the long form
• practice creating two-way tables

Exercise 1

Each year, 2000 students apply for undergrad programs at Stanford University. Among the applicants, 20% have an SAT score above 1550, 55% have an SAT score between 1450 and 1550 and 25% have an SAT below 1450.

Obviously, the probability of being admitted grows with the SAT score. People that have an SAT score of 1550 or higher have a probability of 80% of being admitted. For students that have scores between 1450 and 1550 this probability drops to 15%, for students that have an SAT score below 1450 this probability drops to 5%.

• what is the probability of being admitted?
• what is the probability of having an SAT of over 1550 if you were admitted?

Solution

what is the probability of being admitted?

According to the decomposition formula:

P_admitted = None
P_admitted # answer : 0.255

what is the probability of having an SAT of over 1550 if you were admitted?

P_1550_given_admitted = None
P_1550_given_admitted  # answer: 0.62745

Recall that the probability of being admitted given an SAT of > 1550 is different! So this is a good illustration to show that generally $P(A\mid B) \neq P(B\mid A)$!

Exercise 2

There is a disease that affects a proportion of $\gamma$ of the human population.

Having the disease is affected by a genetic factor, and we know that the joint probability of you and your father having the disease is equal to $\delta$.

Now our question is: given that your father doesn't have the disease, what is the probability that you won't get the disease? Create a two-way table to solve for this in terms of $\gamma$ and $\delta$:

P(A) = You have the disease

P(B) = Your dad has the disease

Exercise 3

Now, let the probability of having the disease be 0.065 and the joint probability of a person and their father having the disease be 0.01, what is the probability that you don't have the disease given that your father doesn't have it?

prob_no_disease = None
prob_no_disease # correct answer: 0.94118

Exercise 4

Two boxes are filled with black and white balls. In the first box, there are 3 white balls and 7 black ones, in the second box, there are 6 white balls and 4 black ones. Jenna takes one ball out of box 1 and puts it in the second box without looking at the color. then he takes a ball out of the second box and notices that it's white. Now, it's your turn to compute the probability that Jenna put a white ball from the first box into the second. Use bayes formula to do this! It will help if you define events $A, A'$ and $B$.

Solution

Denote event A the event that Jenna put a white ball from the first box to the second.

Denote event B the event that Jenna draws a white ball from box 2.

$P(A \mid B)= \dfrac{P(A)P(B\mid A)}{P(A)P(B\mid A)+ P(A')P(B\mid A')}$

None
None
None
None

p_a_given_b = None
p_a_given_b # correct answer: 0.33333