Question

suppose 5 red and 2 green 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. note: please enter your answers below in simplified fraction form. a) what is the conditional probability ( p(red|red)=? ) b) what is the probability of a final outcome such that first a green object and then a red object was selected?

Explanation:

Step1: Find \(P(\text{Red}|\text{Red})\)

Conditional probability formula: \(P(A|B)=\frac{P(A\cap B)}{P(B)}\). Here, \(A\) and \(B\) are both "selecting a red object". From the tree - diagram, when the first object is red (probability \(P(\text{Red}_1)=\frac{5}{7}\)), and then the second object is red (probability \(P(\text{Red}_2|\text{Red}_1)=\frac{4}{6}\)). Also, \(P(\text{Red}|\text{Red})\) can be directly read from the tree - diagram. The value on the branch from the first "Red" to the second "Red" is \(\frac{4}{6}=\frac{2}{3}\).

Step2: Find \(P(\text{Green then Red})\)

Use the multiplication rule for dependent events \(P(A\cap B)=P(A)\times P(B|A)\). Let \(A\) be "selecting a green object first" and \(B\) be "selecting a red object second". \(P(\text{Green}_1)=\frac{2}{7}\), and \(P(\text{Red}_2|\text{Green}_1)=\frac{5}{6}\). Then \(P(\text{Green then Red})=\frac{2}{7}\times\frac{5}{6}=\frac{2\times5}{7\times6}=\frac{10}{42}=\frac{5}{21}\).

Answer:

a) \(\frac{2}{3}\)
b) \(\frac{5}{21}\)