A bag contains 5 red and 5 blue identical balls. Three balls are selected at random without replacement. Determine the probability of selecting balls alternating in color.
\(\frac{7}{18}\)
\(\frac{5}{18}\)
\(\frac{5}{36}\)
\(\frac{1}{36}\)
Correct answer is C
probability of selecting three balls with alternating colours = probability of selecting a blue ball (without replacement) x probability of selecting a red ball (without replacement) x probability of selecting a blue ball (without replacement)
In this case, we will be using the first instance (both would give the same answer since the number of blue and red balls are equal)
first: probability of selecting a red ball (without replacement) = \(\frac{5}{10}\) = \(\frac{1}{2}\)
second: the probability of selecting a blue ball (without replacement) = \(\frac{5}{9}\)
third: probability of selecting a red ball (without replacement) = \(\frac{4}{8}\) = \(\frac{1}{2}\)
Therefore, the probability of selecting three balls with alternating colours = \(\frac{1}{2}\) x \(\frac{5}{9}\) x \(\frac{1}{2}\)
= \(\frac{5}{36}\)