### Video Transcript

A bag contains an unknown number of
balls. Given that one-sixth of the balls
are white, one-fifth of them are green, and the rest are blue, what is the
probability that a ball drawn at random from the bag is blue?

In this question, we are told that
a bag contains three different colored balls. They are white, green, and
blue. We are not told the number of balls
that are each color or the total number of balls in the bag. We are told, however, that
one-sixth of the balls are white. And this means that the probability
of selecting a white ball at random is one-sixth. Likewise, we are told that
one-fifth of the balls are green, so the probability of selecting a green ball is
one-fifth. It is the probability of selecting
a blue ball, that we will call 𝑥, that we are trying to calculate.

We recall that the sum of the
probabilities of all outcomes in a sample space must equal one. This means that in this question
the probability of selecting a white ball plus the probability of selecting a green
ball plus the probability of selecting a blue ball must equal one. Substituting in the values we know,
we have one-sixth plus one-fifth plus 𝑥 is equal to one. On the left-hand side of our
equation, we can begin by adding one-sixth and one-fifth by finding the lowest
common denominator, which is equal to 30. One-sixth is equivalent to five
thirtieths and one-fifth is equivalent to six thirtieths.

Adding the numerators, we see that
one-sixth plus one-fifth is equal to eleven thirtieths. Our equation simplifies to eleven
thirtieths plus 𝑥 is equal to one. Noting that one whole one is equal
to thirty thertieths. We can subtract eleven thirtieths
from both sides. This gives us 𝑥 is equal to
nineteen thirtieths. And we can therefore conclude that
the probability of selecting a blue ball from the bag is nineteen thirtieths.