Question

In a class of $100$ students, $60$ students drink tea, $50$ students drink coffee and $30$ students drink both. A student from 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).

• A. $\frac{1}{5}$
• B. $\frac{2}{5}$
• C. $\frac{3}{5}$
• D. $\frac{4}{5}$

Answer 1

Answer: (D)

$\frac{4}{5}$

It is given that,

$\left(1\right)$ The total students in a class $=100$.

$\left(2\right)$ Students drinking coffee $=50$.

$\left(3\right)$ Students drinking tea $=60$.

$\left(4\right)$ Student drinking both coffee and tea $=30$.

Using (1), we get

$n\left(S\right)=100$

Let $A$ be the event that the selected student drink tea.

Using $\left(3\right)$, $n\left(A\right)=60$.

$\therefore P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{60}{100}=\frac{3}{5}$.

Let $B$ be the event that the selected student drink coffee.

Using $\left(2\right)$, $n\left(B\right)=50$.

Therefore, $P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{50}{100}=\frac{1}{2}$.

Now $A\cap B$ is the event that selected student drinks both coffee and tea and $A\cup B$ is the event that the student selected drinks either tea or coffee or both.

Now using $\left(4\right)$, $n(A\cap B)=30$.

$\therefore P(A\cap B)=\frac{n(A\cap B)}{n\left(S\right)}=\frac{30}{100}=\frac{3}{10}$.

Now $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

$=\frac{3}{5}+\frac{1}{2}-\frac{3}{10}$

$=\frac{6+5-3}{10}$

$=\frac{8}{10}=\frac{4}{5}$

Thus $P(A\cup B)=\frac{4}{5}$

The probability that selected student takes at least one of the drinks is $\frac{4}{5}$.