Question

suppose 6 green and 4 orange 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 c =?
b) what is the value of the branch probability d =?
c) what is the value of the branch probability e =?

Explanation:

Step1: Calculate the total number of objects initially

There are \(6\) green (\(G\)) and \(4\) orange (\(O\)) objects. So the total number of objects initially \(n = 6 + 4=10\).

Step2: Find the value of \(a\) (probability of first - draw being green)

The probability of drawing a green object first, \(a=\frac{6}{10}=\frac{3}{5}\).

Step3: Find the value of \(c\) (probability of second - draw being green given first - draw is green)

After drawing one green object, the number of green objects left is \(6 - 1=5\) and the total number of objects left is \(10-1 = 9\). So \(c=\frac{5}{9}\).

Step4: Find the value of \(d\) (probability of second - draw being orange given first - draw is green)

After drawing one green object, the number of orange objects is still \(4\) and the total number of objects left is \(9\). So \(d=\frac{4}{9}\).

Step5: Find the value of \(b\) (probability of first - draw being orange)

The probability of drawing an orange object first, \(b = \frac{4}{10}=\frac{2}{5}\).

Step6: Find the value of \(e\) (probability of second - draw being green given first - draw is orange)

After drawing one orange object, the number of green objects is \(6\) and the total number of objects left is \(9\). So \(e=\frac{6}{9}=\frac{2}{3}\).

Answer:

a) \(c=\frac{5}{9}\)
b) \(d=\frac{4}{9}\)
c) \(e=\frac{2}{3}\)