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Question: enter image description here

What I tried. Well for part i the probability of any face being landed on is 1/4. I'm not sure if the way I explanation for the approach I took is correct because it gets the same answer but... $$ {\binom {3}{3}}*({\frac 1 4} )^3$$

My reasoning is that out of the you choose all 3 tetrahedrons to land on black and then 1/4 is to actually land it on the black?

D.Ronald
  • 540
  • The problem whose image you post describes the tetrahedra as "uniform". Another way to describe these would be "regular tetrahedra", meaning all the faces are congruent (equilateral) triangles. This is crucial to having all the sides having equal chances of "being landed on". – hardmath Oct 31 '17 at 12:21
  • Please take the time to enter important parts of your question as text instead of pasting a picture. Images are neither searchable nor accessible to people using screen readers. – amd Oct 31 '17 at 20:13

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