Answer:
a. 9
b. [tex]\mathbf{\frac{1}{3}}[/tex]
c. [tex]\mathbf{\frac{1}{4}}[/tex]
d. [tex]\mathbf{\frac{1}{12}}[/tex]
Step-by-step explanation:
a. The total number of balls in the bag is
[tex] 6 + 3 = 9[/tex]
b. Let A denote the event that the 1st chosen ball is orange. Let B denote the event that the 2nd chosen ball is orange.
[tex]P(A) = \frac{\text{no. of orange balls in the bag}}{\text{Total no. of balls}} = \frac{3}{9} = \mathbf{\frac{1}{3}}[/tex]
c. If the first choice is orange, (i.e A occurs) then there are 9 - 1 = 8 balls left in the bag with 3 - 1 = 2 orange balls.
[tex]P(B|A) = \frac{2}{8} = \mathbf{\frac{1}{4}}[/tex]
d. Here we are required to compute [tex]P(B \cap A)[/tex]. We use the fact that
[tex]P(B|A) = \frac{P(B \cap A)}{P(A)} \implies P(B \cap A) = P(B | A) \times P(A) = \frac{1}{4} \times \frac{1}{3} = \mathbf{\frac{1}{12}}[/tex]
