Question

In a class of $100$ students, $60$ students drink tea, $50$ students drink coffee and $30$ students drink both tea and coffee. A student from this class in selected at random. Find the probability that the student takes at least one of the drinks.

Answer 1

There are $100$ students in the class
$\therefore n\left(S\right)=100$
Let $A$ is the event that the students drink tea.
$\therefore n\left(A\right)=60$
$\therefore P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{60}{100}$
Let $B$ is the event that the students drink coffee.
$\therefore n\left(B\right)=50$
$\therefore P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{50}{100}$
Now the number of students drinking both tea and coffee is 30
$\therefore n(A\cap B)=30$
$\therefore P(A\cap B)=\frac{n(A\cap B)}{n\left(S\right)}=\frac{30}{100}$
$\therefore$ Probability that selected students who take at least one of the two drinks is
$\therefore P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$
$=\frac{60}{100}+\frac{50}{100}-\frac{30}{100}$
$=\frac{60+50-30}{100}$
$=\frac{110-30}{100}$
$=\frac{80}{100}$
$=\frac{4}{5}$