91Ó°ÊÓ

A and B, two individuals, are both enrolled in the same course.

Student A attends class 80% of the time, and student B attends class 60% of the time, and the two individuals' absences are independent.

Let:

\(A = \)The event that student A attends class on a given day

\(B = \)The event that student B attends class on a given day

Thus, the given information can be summarized as follows:

\(\begin{aligned}{}P\left( A \right) = 0.80\\P\left( B \right) = 0.60\end{aligned}\)

The Additional rule of probability results in the probability of appearance of either of the events A or B. Mathematically, it is given by:

\({\bf{P}}\left( {{\bf{A}} \cup {\bf{B}}} \right){\bf{ = P}}\left( {\bf{A}} \right){\bf{ + P}}\left( {\bf{B}} \right) - {\bf{P}}\left( {{\bf{A}} \cap {\bf{B}}} \right)\)

It is given that the absences of the two students are independent. This indicates that the presences of the two students are also independent.

Thus,

\(P\left( {A \cap B} \right) = P\left( A \right) \times P\left( B \right)\)

So, using the Additional Rule of Probability to obtain the probability that at least one of two students will be in class on a given day as:

\(\begin{aligned}{}P\left( {{\rm{at least one of the two students will be in class}}} \right) &= P\left( {A \cup B} \right)\\ &= P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\\& = P\left( A \right) + P\left( B \right) - P\left( A \right) \times P\left( B \right)\end{aligned}\)

Substituting the values,

\(\begin{aligned}{}P\left( {{\rm{at least one of the two students will be in class}}} \right) &= 0.80 + 0.60 - 0.80 \times 0.60\\& = 1.4 - 0.48\\ &= 0.92\end{aligned}\)

Therefore, the required probability is 0.92.

The Conditional probability of an event A given some other event B has appeared is given by:

\({\bf{P}}\left( {{\bf{A|B}}} \right){\bf{ = }}\frac{{{\bf{P}}\left( {{\bf{A}} \cap {\bf{B}}} \right)}}{{{\bf{P}}\left( {\bf{B}} \right)}}\)

By the definition of conditional probability,

If at least one of the two students will be in class on a given day, the probability that A is in class that day is obtained as:

\(\begin{array}{}P\left[ {A|\left( {A \cup B} \right)} \right] = \frac{{P\left[ {A \cap \left( {A \cup B} \right)} \right]}}{{P\left( {A \cup B} \right)}}\\ = \frac{{P\left( A \right)}}{{P\left( {A \cup B} \right)}}\;\;\;\;\left\{ {A \subseteq \left( {A \cup B} \right)} \right\}\\ = \frac{{0.80}}{{0.92}}\\ \approx 0.8696\end{array}\)

Therefore, the required probability is approximately 0.8696.