Question

In a class of 100 students, 60 students drink tea, 50 students drink coffee and 30 students drink both. A student from this class is selected at random, find the probability that student takes at least one of the two drinks (i.e. tea or coffee or both).

Answer 1

Total number of students in the class = 100
∴ n(S) = 100
Now,
Let A be the event that the students drink tea.
∴ n(A) = 60
$\Rightarrow \mathrm{P}\left(\mathrm{A}\right)=\frac{n\left(\mathrm{A}\right)}{n\left(\mathrm{S}\right)}=\frac{60}{100}=\frac{3}{5}$
Also,
Let B be the event that the students drink coffee.
∴ n(B) = 50
$\Rightarrow \mathrm{P}\left(\mathrm{B}\right)=\frac{n\left(\mathrm{B}\right)}{n\left(\mathrm{S}\right)}=\frac{50}{100}=\frac{1}{2}$
A$\cap$ B is the event that the students drink both.
We have:
n(A $\cap$ B) = 30
And,
$\mathrm{P}(\mathrm{A}\cap \mathrm{B})=\frac{n(\mathrm{A}\cap \mathrm{B})}{n\left(\mathrm{S}\right)}=\frac{30}{100}=\frac{3}{10}$
Probability that the student takes at least one of the two drinks (tea or coffee or both):
P(A$\cup$B) = P(A) + P(B) $-$ P(A $\cap$ B)

$$\begin{gathered} =\frac{3}{5}+\frac{1}{2}-\frac{3}{10} \\ =\frac{6+5-3}{10}=\frac{8}{10}=\frac{4}{5} \end{gathered}$$