Question

In a box, there are $3$ red marbles, $2$ green marbles and $7$ blue marbles. Oliver picks a marble blindfolded. What is the probability of picking a red marble or a green marble?

• A. $\frac{3}{12}$
• B. $\frac{5}{12}$
• C. $\frac{7}{12}$
• D. $\frac{4}{12}$

Answer 1

The correct option is B $\frac{5}{12}$

$\underset{–}{\text{Given}:}$
In a box,
$\bullet$ Number of red marbles $=3$
$\bullet$ Number of green marbles $=2$
$\bullet$ Number of blue marbles $=7$

$\therefore$ Total number of marbles in the box $=3+2+7=12$

Here,
$\text{P(picking a red marble)}=\frac{\text{Number of red marbles}}{\text{Total number of marbles}}$

$\Rightarrow \text{P(picking a red marble)}=\frac{3}{12}$

Next,
$\text{P(picking a green marble)}=\frac{\text{Number of green marbles}}{\text{Total number of marbles}}$

$\Rightarrow \text{P(picking a green marble)}=\frac{2}{12}$

$\underset{–}{\text{Need to Find}:}$
Probability that the blindfoldly picked marble by Oliver is either red or green, i.e., $\text{P(picking a red marbleor a green marble)}$

$\text{P(picking a red marble or a green marble) = P(picking a red marble) + P(picking a green marble)}$

$\Rightarrow \text{P(picking a red marble or a green marble)}=\frac{3}{12}+\frac{2}{12}$

$\Rightarrow \text{P(picking a red marble or a green marble)}=\frac{3+2}{12}=\frac{5}{12}$

$\therefore$ The probability of picking a red marble or a green marble is $\frac{5}{12}$.

Hence, option (b.) is the correct choice.