Question

Two students, A and B, are both registered for a certain course. Assume that student A attends class 80 percent of the time, student B attends class 60 percent of the time, and the absences of the two students are independent.

a. What is the probability that at least one of the two students will be in class on a given day?

b. If at least one of the two students is in class on a given day, what is the probability that A is in class that day?

Short answer

a. The probability that at least one of two students will be in class on a given day is 0.92.

b. 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 0.8696.

Step-by-step solution

Given information

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.

Defining events

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:

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

(a) Computing the probability in part a

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.

(b) Computing the probability in part b

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.