A bag contains 1 white and 8 red balls. The probability for randonly drawing a red ball 8 successive time is 11.11%= $\frac{1}{9} (p = \frac{8}{9}\cdot \frac{7}{8}\cdot \frac{6}{7}\cdot \frac{5}{6}\cdot \frac{4}{5}\cdot \frac{3}{4}\cdot \frac{2}{3} \cdot \frac{1}{2})$. What is the 88.89% = $\frac{8}{9}$ probability against this signifies? Does it mean that there is 88.89% probability that the 1 white ball would be randomnly chosen before 8 successive red balls are drawn?
$\begingroup$
$\endgroup$
1
1 Answer
$\begingroup$
$\endgroup$
I think you are assuming, when you say "drawing a red ball $8$ successive times", that "drawing a red ball in first $8$ draws". You are not considering the case when you get the white ball in the first draw and then 8 red balls successively.
Assuming that you want to say "drawing a red ball in first 8 draws", the answer should be: $8/9$ signifies that "you have not drawn only red balls in first 8 draws".
[self-study]tag & read its wiki. $\endgroup$