Consider the following experiment that Kahneman and Tversky ran to study base-ra

Question

Consider the following experiment that Kahneman and Tversky ran to study base-rate neglect. In Problem A, subjects are told that Jack has been drawn from a population of 30% engineers and 70% lawyers and that Jack wears a pocket protector. (a) Let p_A denote the probability that Jack is an engineer, given that he wears a pocket protector. Using Bayes' rule, show that the odds that Jack is an engineer as opposed to a lawyer are given by: p_A/1 - p_A = Pr (pocket protector | Jack is an engineer)/Pr (pocket protector | Jack is a lawyer) times 0.3/0.7 In Problem B, subjects are told that Jack has been drawn from a population of 70% engineers and 30% lawyers and that Jack wears a pocket protector. (b) Let p_B denote the probability that Jack is an engineer, given that he wears a pocket protector. Show that p_B/1 - p_B = Pr (pocket protector | Jack is an engineer)/Pr(pocket protector | Jack is a lawyer) times 0.7/0.3 And conclude that, if subjects form beliefs according to the laws of probability, it must be the case that: p_A/(1 - p_A)/p_B/(1 - p_B) = 9/49. (c) Explain intuitively why the odds ratio in (*) does not depend on Pr (pocket protector | Jack is an engineer) and why Kahneman and Tversky set up the experiment in this way. (d) What values of p_A/(1 - p_A)/p_B/(1 - p_B) imply that subjects exhibit base-rate neglect?

Explanation / Answer

Alway remeber the formula of Bayes' theorm

P(A/B)=P(B/A)*P(A)/P(B)

let's denote ; Wears pocket protector=Po

Engineer =E and Lawer= L for the future calculations

and P(E)=0.3 and P(L)=0.7 is given

A)Pa=P(E/Po) = P(Po/E)*P(E)/P(Po)----------------------------(1)

so 1-Pa= jack is a lawer given that he wears a pocket protector = P(L/Po)= P(Po/L)*P(L)/P(Po) -------(2)

So., dividing eqn 1 & 2 we get Pa/(1-Pa)=[P(Po/E)*P(E)]/[P(Po/L)*P(L)] =[P(Po/E)/P(Po/L)]*[0.3/0.7] -------(a)

which is the given solution to be proved

B) same as above let's substitute Pb/(1-Pb) =[P(Po/E)*P(L)]/[P(Po/L)*P(E)] =[P(Po/E)/P(Po/L)]*[0.7/0.3]---(b)

if you observe the ans of A you can notice here that E & L only have been iterchanged and you got the correct ans here as well.

now take the ratio of a &b which is [Pa/(1-Pa)]/[Pb/(1-Pb)]=0.3*0.3/0.7*0/7 = 9/49 rest were cancelled out

C) As you can observe the rqn a and b both have the same terms attached with 0.3/0.7 and 0.7/0.3 respectively which gets cancelled out eventually and so the ratio don't depend on Pr(Po/E)

D)any rate below 9/49 reflects the base rate Fallacy or what you have said here is base - rate neglect.

Pls rate the ans if it has helped you enrich your knowledge. Good Luck!!

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.