I can't seem to make the decision tree and get the correct probability for this problem, can someone please tell me where my logic is flawed.
A bag contains 5 red balls and 5 blue balls. Three balls are drawn randomly without replacement. Find:
the conditional probability that we drew at least one blue ball given that we drew at least one red ball.
I keep wanting to write the branches as :
Red-Blue-Blue = 5/10 * 5/9 *4/8
Red-Blue-Red = 5/10 * 5/9 * 4/8
Red-Red-Blue = 5/10 * 4/9 * 5/8
Adding those together and I am way off. Can someone tell me the correct setup for this? and where my logic is skewed.
Best Answer
Let $E$ be the event that at least one red ball is drawn; let $F$ be the event that at least one blue ball and at least one red ball are drawn. We wish to find the conditional probability that at least one blue ball is drawn given that at least one red ball is drawn, which is $$\Pr(F \mid E) = \frac{\Pr(E \cap F)}{\Pr(E)}$$
In calculating the conditional probability, it is important to remember that your sample space is the set of all drawings of three balls in which at least one red ball is drawn. These are the drawings that do not contain at least one blue ball. $$\Pr(E) = 1 - \frac{5}{10} \cdot \frac{4}{9} \cdot \frac{3}{8}$$ If you insist on calculating the probability that at least one red ball is drawn directly, you need to add the probabilities of the following seven sequences of events:
blue, blue, red
blue, red, blue
red, blue, blue
blue, red, red
red, blue, red
red, red, blue
red, red, red
See Satish Ramanathan's answer for an efficient way of doing this.
The probability that at least one red ball is selected is your denominator.
Your numerator is the probability that at least one red and at least one blue ball are selected.
You can find this by adding the probabilities of the first six events listed above or by subtracting the probability that three red balls were drawn from the probability that at least one red ball is drawn. $$\Pr(E \cap F) = 1 - \frac{5}{10} \cdot \frac{4}{9} \cdot \frac{3}{8} - \frac{5}{10} \cdot \frac{4}{9} \cdot \frac{3}{8}$$
A combinatorial approach
Our sample space consists of all drawings of three balls that contain at least one red ball, which can be found by subtracting the number of selections that contain only blue balls from the total number of selections, which is $$|E| = \binom{10}{3} - \binom{5}{3}$$ For a drawing to be successful, it must contain at least one red ball and at least one blue ball. The number of such selections can be found by adding the number of ways of selecting one red ball and two blue balls or selecting two red balls and one blue ball, which is $$|E \cap F| = \binom{5}{1}\binom{5}{2} + \binom{5}{2}\binom{5}{1}$$ Alternatively, it can be found by subtracting the number of drawings that contain only red balls from the number that contain at least one red ball, which is $$|E \cap F| = \binom{10}{3} - \binom{5}{3} - \binom{5}{3}$$ Thus, the conditional probability that at least one blue ball is drawn given that at least one red ball is drawn is $$\Pr(F \mid E) = \frac{\Pr(E \cap F)}{\Pr(E)} = \frac{|E \cap F|}{|E|} = \frac{\dbinom{5}{1}\dbinom{5}{2} + \dbinom{5}{2}\dbinom{5}{1}}{\dbinom{10}{3} - \dbinom{5}{3}} = \frac{\dbinom{10}{3} - \dbinom{5}{3} - \dbinom{5}{3}}{\dbinom{10}{3} - \dbinom{5}{3}}$$