Question

a bowl contains slips of paper numbered 1-9. two slips are drawn, one after the other and without replacement. find the probability of the first number being drawn is greater than 8 and the second number is less than 3

Answer 1

\[\frac{1}{9}\times \frac{2}{8}=\frac{2}{72}=\frac{1}{36}\]

Answer 2

First time you picked, you get to choose from 9 slips and there is only one slip that is greater than 8.
1/9
Second time , you have 8 slips left, and there are 2 slips that is lower than 3
2/8

1/9 * 2/8=

Answer 3

But if the first number you pick is less than three, than the probability of drawing a number less than three the second time is cut in half

Answer 4

That's not the situation the question's asking about.

Answer 5

it has to be taken into account

Answer 6

Are you disputing that the probability of initially picking a number greater than 8 is 1/9 or that the probability of picking a number less than 3 is 2/8?

Answer 7

i am disputing the fact that the second number drawn may or may not be affected by the first draw. if the first draw were to be a 2, then the probability of drawing a number less than three on the second draw would indeed be 1/8 not 2/8

Answer 8

You are absolutely correct, but if the first number picked was a 2 then the probability of the first number being greater than 8 is zero, so this doesn't affect the question.

Answer 9

ah i see now thank you. wanna take a shot another probability question?

Answer 10

Go for it, quick though, I need to go out in a min.