Answer to Question #177733 in Statistics and Probability for Kojo

Question

1. Of the students in the college, 60% of the students reside in the hostel and 40% of the students are day scholars. Previous year result reports that 30% of all students who stay in the hostel scored A Grade and 20% of day scholars scored A grade. At the end of the year, one student is chosen at random and found that he/she has an A grade. What is the probability that the student is a hostlier? 

 

2. Two players A and B are competing at a trivia quiz gameinvolving a series of questions. On any individual question, the probabilities that A and B give the correct answer areαandβrespectively, for all questions, with outcomes for different questions being independent. The game finishes when a player wins by answering a question correctly. Compute the probability that A wins if

a) A answers the first question, 

b) B answers the first question.

Answer 1

1) Lets denote h for the students that reside in the hostel and d for the students that are day scholars.

Thus we have

P(h) = 0.6

P(d) = 0.4

P(A|h) = 0.3

P(A|d) = 0.2

The probability that the student is a hostler given that he/she has an A grade is calculated by the formula (using Bayes theorem)

$P(h|A) =\frac{P(A|h)\cdot P(h)}{P(A)}=\frac{P(A|h)\cdot P(h)}{P(A|h)\cdot P(h)+P(A|d)\cdot P(d)}=\frac{0.3\cdot 0.6}{0.3\cdot 0.6+0.2 \cdot 0.4}=0.69$

There is 69% chance that the student is a hostler given that he/she has an A grade.

2) Lets denote A for A answer the first question

F for game ends after the first question

W for A wins

need to find P(W|A') (B answers first question correctly)

P(F|A)=P [A answers first question correctly] = α

P(F′|A)=1−α and P(W|A∩F)=1 but P(W|A∩F′)=P(W|A′)

so that P(W|A)=P(W|A∩F)P(F|A)+P(W|A∩F′)P(F′|A)

P(W|A)=(1×α)+(P(W|A′)×(1−α))=α+P(W|A′)(1−α) (i)

We have P(F|A′)=P [B answers first question correctly] = β

P(F′|A)=1−β

but P(W|A′∩F)=0

Finally P(W|A′∩F′)=P(W|A)

thus P(W|A′)=(0×β)+(P(W|A)×(1−β))=P(W|A)(1−β) (ii)

Solving (i) and (ii) gives for B answers the first question:

$P(W|A')=\frac{\alpha(1-\beta)}{1-(1-\alpha)(1-\beta)}$

Following the same logic finally we have that for A answers the first question:

$P(W|A)=\frac{\beta(1-\alpha)}{1-(1-\alpha)(1-\beta)}$