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Mathematics 11 Online
OpenStudy (anonymous):

Can someone please help me answer this question A tetrahedral die has 4 faces, number 1-4. If the die is weighted in such a way that each number is twice as likely to land facing down as the next number (1 twice as likely as 2, 2 twice a likely as 3, and so on. What is the probability distribution for the face landing down?

OpenStudy (zarkon):

Let D= die roll start with \[P(D=4)=p\] work your way backwards to D=1....writing the probability in terms of p then use the law of total probability to determine the value of p.

OpenStudy (anonymous):

weel i got 2 answer i did 1680/8=210 and 6*12*4=24/4=6

OpenStudy (zarkon):

those are not probabilities

OpenStudy (zarkon):

the probability D is 3 is twice the value of prob D=4 so \[P(D=3)=2p\]

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