Question

Two students,AandB,are both registered for a certain course. Assume that studentAattends class 80 percent of the time, studentBattends class 60 percent of the time, and the absences of the two students are independent. Consider the conditions of Exercise 7 of Sec. 2.2 again. If exactly one of the two students,AandB,is in class on a given day, what is the probability that it isA?

Short answer

Finally, \(21\) the percent of probability that it is \(A\).

Step-by-step solution

Given information

Two students\(A\)and\(B\)are both registered for a certain course. Given that\(A\)attends class\(80\)percent of that time and\(B\)attends class\(60\)percent of that time.

Also, given the condition absence of the two students are independent.

State the condition

\(A\) attends class \(80\) percent at that time, so \(A\) absent \(20\) percent of the time. Then the probability of \(A\) is \(p\left( A \right) = \frac{{20}}{{100}} = 0.2\).

\(B\)present class\(60\)percent of that time. Then\(B\)not present\(40\)percent of that time. Therefore, the probability of\(B\)given by\(p\left( B \right) = \frac{{40}}{{100}} = 0.4\).

Both are absent from the class on the same day. Then both absences are given by

\(\begin{aligned}{c}p\left( {A \cap B} \right)& = p\left( A \right) \times p\left( B \right)\\ &= 0.2 \times 0.4\\& = 0.08\end{aligned}\)

Therefore, the attends of both \(A\) \(B\) is find that

\(\begin{aligned}{c}p\left( {A \cap B} \right) &= 1 - p\left( {A \cup B} \right)\\& = 1 - 0.08\\ &= 0.92\end{aligned}\)

Compute the probability

To compute the probability by the method of the conditional theorem. The theorem is given by

\(p\left( {A\left| B \right.} \right) = \frac{{p\left( {A \cap B} \right)}}{{p\left( B \right)}}\) . In this condition, we use the theorem is

\(\begin{aligned}{c}p\left( {A\left| {A B} \right.} \right) &= \frac{{p\left( {A \cap \left( {A \cup B} \right)} \right)}}{{p\left( {A \cup B} \right)}}\\ &= \frac{{p\left( A \right)}}{{p\left( {A \cup B} \right)}}\end{aligned}\)

Hence the probability that it is \(A\) is given by

\(\begin{aligned}{c}\frac{{p\left( A \right)}}{{p\left( {A \cup B} \right)}} &= \frac{{0.2}}{{0.92}}\\ &= 0.21\end{aligned}\)

Finally, \(21\) the percent of probability that it is \(A\).