Consider a die that has been designed such that all even numbers are equally likely, all odd numbers are equally likely, but an even number is twice as likely as an odd number. what is the probability that this die lands 2?

Accepted Solution

Answer: Β  2/n where n is the number of faces on the dieStep-by-step explanation:A 6-sided die could be numbered 1, 2, 2, 3, 4, 4 to meet the probability requirements 2/6 = 1/3 of the faces have the number 2, so the probability of 2 would be 1/3.___To meet the probability requirements nicely requires a die with a number of faces that is a multiple of 3. The regular solids meeting this requirement arecube β€” 6 facesdodecahedron β€” 12 facesA dodecahedron could be numbered 1, 1, 2, 2, 2, 2, 3, 3, 4, 4, 4, 4 and the probability of a 2 would be the same as for a cube. It could more reasonably be numbered 1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 8, 8, in which case, the probability of 2 would be 1/6.___If you use some other shape or put blank faces on the die (say a 20-faced die with 2 blanks), then the probability will be different from the numbers above. If you still use 2 faces for each even number, then the probability of 2 using an n-faced die is 2/n.