Answer
No. The first ball and the second ball must have the same color. So, order is not important here.
Work Step by Step
- First method:
Sample space:
First ball: 9 choices.
Second ball: 8 choices.
$n(S)=9\times8=72$
Event: two red balls:
First ball: 5 choices.
Second ball: 4 choices.
$n(E)=5\times4=20$
$P(E)=\frac{n(E)}{n(S)}=\frac{20}{72}=\frac{5}{18}$
- Second method:
Sample space: all possible combinations of 2 balls from a bag with 9 balls:
$n(S)=C(9,2)=\frac{9!}{(9-2)!\times2!}=36$
Choose two from five red balls:
$n(E)=C(5,2)=\frac{5!}{(5-2)!\times2!}=10$
$P(E)=\frac{n(E)}{n(S)}=\frac{10}{36}=\frac{5}{18}$
