Question

suppose 9 orange and 3 blue 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, but the branch probabilities are missing. note: for all questions below, please enter your answers in fraction form. a) what is the value of the branch probability ( b =? ) b) what is the value of the branch probability ( c =? ) c) what is the value of the branch probability ( e =? )

Explanation:

Step1: Calculate the value of \(b\)

The total number of objects initially is \(9 + 3=12\). The probability of selecting a blue object first (\(b\)) is the number of blue objects divided by the total number of objects. So, \(b=\frac{3}{12}=\frac{1}{4}\).

Step2: Calculate the value of \(c\)

After selecting an orange object first (probability \(a=\frac{9}{12}\)), the number of orange objects left is \(9 - 1 = 8\) and the total number of objects left is \(12- 1=11\). So, \(c=\frac{8}{11}\).

Step3: Calculate the value of \(e\)

After selecting a blue object first (probability \(b = \frac{3}{12}\)), the number of orange objects is still \(9\) and the total number of objects left is \(12 - 1=11\). So, \(e=\frac{9}{11}\).

Answer:

a) \(\frac{1}{4}\)
b) \(\frac{8}{11}\)
c) \(\frac{9}{11}\)