How do I solve this last problem? I don't know how to find

>>210664
If you're looking for an "or" probability, you take the sum of the probabilities of both outcomes.

So add the .1147 you got for 16 attending to the probability you get for 15 attending, and voila.

>>210666
this

when in doubt about mechanics, look at a simpler version: if you do one throw, what are the odds of either "2" or "3" coming up on a 6-sided die?

File: 245e74db02480ec392da362b043e6775.png (37KB, 1573x777px) Image search: [Google]

37KB, 1573x777px

>>210666
>>210678
Alright, I got it. Thanks. One more question, I know I can't fix this now but I still wanna figure out how to do this. The "at least" threw me off, and I didn't know what to do.

>>210679
Ok, there's a couple different ways you can get the correct solution for this one.

The easy way, given your chart above, is to subtract the odds for both cards coming up yellow from 1. Because 1 - (YY) = odds of not both yellow = odds of at least one green.
So 1 - 6/56 = 50/56 = 25/28

The other way you could do it, is to calculate the odds of all the possible green combinations and add them together (as each outcome is independent). Now, if the first card you draw is green, you don't need to break it down further - you've already satisfied "at least one green", so you'd take:

Odds first card green + Odds first card yellow & second card green

Which is:

5/8 + (3/8)(5/7) = 50/56

Finally, since you filled out that chart, you could have used the complete breakdown that encompasses all possible combinations:

GG + GY + YG = 20/56 + 15/56 + 15/56

>>210692
>The easy way, given your chart above, is to subtract the odds for both cards coming up yellow from 1. Because 1 - (YY) = odds of not both yellow = odds of at least one green.
>So 1 - 6/56 = 50/56 = 25/28

I feel dumb for not remember this. Thank you so much.