Question

In a box containing only purple and green marshmallows, $6$ marshmallows are purple. If the probability of choosing a purple marshmallow from the box is $\dfrac{1}{3}$, calculate the number of green marshmallows in the box.
A)$2$
B)$6$
C)$9$
D)$12$
E)$18$

Answer 1

Hint: We know probability $ = $ Number of favourable outcomes$ \div $Total number of outcomes.
Use this formula, you will get the answers.

Complete step-by-step answer:
So, we are provided with a box containing purple and green marshmallows. Now if we randomly pick any marshmallow, it may be a purple or green. So we can define the probability of taking out green or purple marshmallows.
So, basically, given that there are $6$ purple marshmallow and probability of taking purple marshmallow from the box is $\dfrac{1}{3}$
And we have to calculate the number of green marshmallows. So, let's assume the number of green marshmallows is $x$.
So, total marshmallow in the box $ = $Green marshmallow $ + $Purple marshmallow
Total $ = x + 6$
We know that probability $ = $ Number of favourable outcomes$ \div $Total number of outcomes
Here in the question, it is given that
Probability of taking out purple marshmallow is $\dfrac{1}{3}$.
This means, Number of purple marshmallow $ \div $Total marshmallow $ = \dfrac{1}{3}$
$\dfrac{6}{{x + 6}} = \dfrac{1}{3}$
Now, simply this
$
  18 = x + 6 \\
  x = 12 \\
 $
Here $x$ is defined as the number of green marshmallows.
So, number of green marshmallows $ = 12$

So, the correct answer is “Option D”.

Note: You can also consider that total number of marshmallows be $N$. So that the probability of taking out purple marshmallows be
$
  \dfrac{6}{N} = \dfrac{1}{3} \\
  N = 18 \\
 $
Probability of taking out green marshmallows $ = 1 - \dfrac{6}{N}$
No of green marshmallows $ \div $Total no. of marshmallows $ = \dfrac{{N - 6}}{N}$
$\dfrac{x}{N} = \dfrac{{N - 6}}{N}$ (Let X be the number of green marshmallows)
$
  x = N - 6 \\
   = 18 - 6 \\
   = 12 \\
 $