Question

Inside a box, there are $8$ tickets of $3$ colors; $4$ are purple, $3$ are blue and $1$ gold. Two tickets are picked from the box at random without replacement. What is the probability that both the tickets are not purple?

• A. $\frac{2}{7}$
• B. $\frac{3}{4}$
• C. $\frac{3}{14}$
• D. $\frac{11}{14}$

Answer 1

The correct option is D $\frac{11}{14}$

In the given box,
total number of tickets= $8$
number of purple tickets= $4$
number of blue tickets= $3$
number of gold ticket= $1$

Now, since two tickets are picked from the box without replacement,
$\text{P(both purple)=P(purple)}\times \text{P(purple, when first picked ticket is purple})$

$\Rightarrow \text{P(both purple)}=\frac{4}{8}\times \frac{4-1}{8-1}$

$\Rightarrow \text{P(both purple)}=\frac{1}{2}\times \frac{3}{7}$

$\Rightarrow \text{P(both purple)}=\frac{1\times 3}{2\times 7}=\frac{3}{14}$

Now,
$\text{P(both not purple)}=1-\text{P(both purple)}$

$\Rightarrow \text{P(both not purple)}=1-\frac{3}{14}$

$\Rightarrow \text{P(both not purple)}=\frac{14}{14}-\frac{3}{14}$

$\Rightarrow \text{P(both not purple)}=\frac{14-3}{14}=\frac{11}{14}$

$\therefore$ The probability of both the picked tickets are not purple is $\frac{11}{14}$.