Question

Two students, X and Y, forgot to put their names on their exam papers. The professor knows that these two students do well on the exam with probabilities 0.8 and 0.4, respectively. After grading, the professor notices that X and Y forgot to put their names on their exams. One of their exams was done well and the other was done poorly. Given this information, and assuming that students worked independently of each other, what is the probability that the good exam belongs to student X

Answer 1

Answer:

The probability that the good exam belongs to student X is 0.8571.

Step-by-step explanation:

It is provided that the probability that X did well in the exam is, P (X) = 0.90 and the probability that X did well in the exam is, P (Y) = 0.40,

Compute the probability that exactly one student does well in the exam as follows:

$P(Either\ X\ or\ Y\ did\ well)=P(X\cap Y^{c})+P(X^{c}\cap Y)\\=P(X)P(Y^{c})+P(X^{c})P(Y)\\=P(X)[1-P(Y)]+[1-P(X)]P(Y)\\=(0.80\times0.60)+(0.20\times0.40)\\=0.56$

Then the probability that X is the one who did well in the exam is:

$P(X\ did\ well\ in\ the\ exam)=\frac{P(X\cap Y^{c})}{P(X\cap Y^{c})+P(X^{c}\cap Y)}\\ =\frac{P(X)[1-P(Y)]}{P(X\cap Y^{c})+P(X^{c}\cap Y)} \\=\frac{0.80\times0.60}{0.56}\\=0.857143\\\approx0.8571$

Thus, the probability that the good exam belongs to student X is 0.8571.