m00li

this answer assumes that we know the number of bad computers, but we don't. what is the actual answer?

No plasmid. What you are stating will work only if condition 3 is relaxed

wouldn't the area be infinitesimally off yup, sowwee

Everything in your post is correct. But remember what the question is. Whatever value a has, how close does the ray come to the point O? The phrase "before it exits to the left" is appropriate to a being acute, as in the figure. If you want to include a being obtuse, so that the ray would not exit to the left, that just gives you "forever" as the time frame to consider the ray's point of closest approach. Although of course you wouldn't need "forever" to determine it. In other words, the wording is not meant to be tricky: assume a to be acute if you like. Hi Bonanova, Sorry, I do not understand. Is my answer wrong? or incomplete?

Yes, bonanova. I think the answer depends on a (e.g. if a is obtuse the ray doesnt exit on left and never hits the top mirror). It coinicides with the icident ray if a is a factor of pi/2 radian

hi bonanova, i have given it for any arbitrary a (both when its sub multiple and when its not). only case to be added is, if a is integer multiple of pi, then the ray doesnt return at all

Had bmad doubled up her pace, you would have died waiting for her (note) call.

Isnt it just 1/n assuming the casino deck is made up of n distinct cards?

You are not alone Plasmid. I am having trouble too I must have been high on something that day My solution is for cases where a 'selection' is being made out of n different colors. I should have calculated permutations and not selections.

Awesome Rob!

hi bmad, Is it purely mathematical or does it require some general knowledge too?

how do we find the mean distance for a square? and for a regular unit pentagon? seems very very tough to me

2 items from a group of 4 can be selected in 10 distinct ways if repetitions are allowed. e.g. {(1,1),(1,2),(1,3),(1,4),(2,2),(2,3),(2,4),(3,3),(3,4),(4,4)} Are each of these selections equally likely? Why or why not?

At 9:05:25 a regular wrist watch hands are at: hr @ -87.291667 degrees min @ 32.5 degrees sec @ 150 degrees therefore, a regular wristwatch cannot have its hands separated equally at 9:05:25. Hence, if a wristwatch has it hands separated equally at 9:05:25, it cannot be working properly. The question doesnt seem to be correct in stating BOTH "A strange wrist watch have it hands separated equally on 9:05:25 am." AND ".and seem to be working properly."

continuing...

Yes, the solution is absolutely incorrect