Question

suppose 4 green and 5 red identical objects are in a jar. a blindfolded person randomly selects two of these objects from the jar one after another, without replacing them back into the jar. the following tree diagram depicts all the possible random outcomes and their probabilities. hint: please enter your answers below in simplified fraction form. a) what is the conditional probability ( p(red|green)=? ) b) what is the probability of obtaining a final outcome with two green objects?

Explanation:

Step1: Recall the formula for conditional probability

The formula for conditional probability is \(P(A|B)=\frac{P(A\cap B)}{P(B)}\). In the context of the tree - diagram, for \(P(\text{Red}|\text{Green})\), we can directly read the value from the tree - diagram. After selecting a green object first (probability of selecting a green object first is \(\frac{4}{9}\)), the probability of selecting a red object second (given that the first was green) is shown in the tree - diagram.

Step2: Calculate the probability of two green objects

The probability of getting a green object on the first draw is \(\frac{4}{9}\). After drawing one green object, there are \(4 - 1=3\) green objects left and a total of \(9-1 = 8\) objects left. So the probability of getting a green object on the second draw given that the first was green is \(\frac{3}{8}\). Using the multiplication rule for independent (in the sense of sequential, non - replacement) events \(P(A\cap B)=P(A)\times P(B|A)\), we have \(P(\text{two green})=\frac{4}{9}\times\frac{3}{8}\).

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Answer:

a) \(\frac{5}{8}\)
b) \(\frac{1}{6}\)