A bag contains white, red, and
black balls. The probability of drawing a white
ball at random is 11 out of 20 and a red ball at random is three out of 10. What is the smallest number of red
balls and black balls that could be in the bag?
We are told in the question that
there are three different color balls in the bag. The probability of selecting a
white ball is 11 out of 20 or eleven twentieths. The probability of selecting a red
ball is three out of 10 or three-tenths. We are not given the probability of
selecting a black ball.
In order to compare fractions, we
need to ensure that the denominators are the same. The lowest common multiple of 10
and 20 is 20, so we need to multiply the denominator of the second fraction by
two. Whatever we do to the denominator,
we must do to the numerator. Three multiplied by two is equal to
six, and 10 multiplied by two is 20. Therefore, the fractions
three-tenths and six twentieths are equivalent.
We know that the sum of all
probabilities is one. In this question, as our
denominator is 20, this is equal to twenty twentieths. Eleven twentieths plus six
twentieths is equal to seventeen twentieths. When the denominators are the same,
we just add the numerators. Subtracting this from one or twenty
twentieths gives us three twentieths. This means that the probability of
selecting a black ball is three twentieths.
We now have three probabilities
that we can compare as the denominators are all the same. The ratio of white to red to black
balls is 11 to 6 to three. This means that the smallest number
of balls in total is 20, where 11 would be white, six red, and three black. The smallest number of red balls
and black balls that could be in the bag are six and three, respectively.
As we’re only given the
probabilities, the total number of balls could be any multiple of 20. For example, we could have 40 balls
in total where 22 are white, 12 are red, and six are black. However, as we were looking for the
smallest number of red and black balls, the correct answer is six red and three