Prepared student and correct answer

Problem

$40\%$ of the students in a class are prepared. A prepared student answers correctly with probability $1$, an unprepared one with probability $1/5$. Construct a contingency table, tree diagram and Venn diagram for the events $A=$ «prepared» and $B=$ «answers correctly». Finally compute $P(A\mid B)$.

Solution

The data are $P(A)=0,4$, $P(B\mid A)=1$, $P(B\mid \overline{A})=1/5=0,2$.

Contingency table (over $100$ imaginary students):

	answers correctly ($B$)	answers wrongly ($\overline{B}$)	total
prepared ($A$)	$40$	$0$	$40$
unprepared ($\overline{A}$)	$12$	$48$	$60$
total	$52$	$48$	$100$

Tree diagram: from the first fork $A$ ($0,4$) / $\overline{A}$ ($0,6$); from branch $A$ one reaches $B$ with probability $1$, from branch $\overline{A}$ one reaches $B$ with probability $0,2$.

The total probability of answering correctly is $P(B) = 0,4\cdot 1 + 0,6\cdot 0,2 = 0,52.$ By Bayes’ theorem: $P(A\mid B) = \frac{P(B\mid A)\,P(A)}{P(B)} = \frac{0,4}{0,52} = \boxed{\dfrac{10}{13} \approx 0,77}$

Links

Topics: Probabilita
Concepts: Diagramma albero · Tabella di contingenza · Teorema di bayes
Methods: Bayes · Prob condizionata
Skills: Calcolo probabilita · Modellizzare
Exercise type: Problema probabilita