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.
help: 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: Find \( P(\text{Red}|\text{Green}) \)

Conditional probability \( P(A|B)=\frac{P(A\cap B)}{P(B)} \). In the context of the tree - diagram, when we want \( P(\text{Red}|\text{Green}) \), after picking a green object first (probability of picking green first is \( \frac{4}{9} \)), the number of remaining objects is \( 8 \). The conditional probability \( P(\text{Red}|\text{Green}) \) is given directly from the tree - diagram as \( \frac{5}{8} \).

Step2: Find the probability of two green objects

The probability of picking a green object first is \( \frac{4}{9} \). After picking one green object, the number of green objects left is \( 3 \) and the total number of objects left is \( 8 \). So the probability of picking a second green object given that the first one was green is \( \frac{3}{8} \).
By the multiplication rule of probability for dependent events \( P(\text{Green}_1\cap\text{Green}_2)=P(\text{Green}_1)\times P(\text{Green}_2|\text{Green}_1) \).

$$P(\text{Green}_1\cap\text{Green}_2)=\frac{4}{9}\times\frac{3}{8}=\frac{12}{72}=\frac{1}{6}$$

Answer:

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