There are many ways to produce crooked dice. To load a die so that 6 comes up to

Question

There are many ways to produce crooked dice. To load a die so that 6 comes up too often and 1 (which is opposite 6) comes up too seldom, add a bit of lead to the filling of the spot on the 1 face. Because the spot is solid plastic, this works even with transparent dice. If a die is loaded so that 6 comes up with probability 0.2 and the probabilities of the 2, 3, 4, and 5 faces are not affected, what is the assignment of probabilities to the six faces? (Round your answers to three decimal places.)

Explanation / Answer

From the given problem,

P(6)=0.2

P(2)=P(3)=P(4)=P(5)=1/6=0.167=k(say)................(Since probabilities of getting 2,3,4 & 5 are unaffected)

Since we know,

P(1)+P(2)+P(3)+P(4)+P(5)+P(6)=1

=> P(1)+4k+P(6)=1

=> P(1)=1-4k-P(6)

=> P(1)=1-4*(1/6)-0.2

=> P(1)=0.133

Therefore we get,

P(1)=0.133

P(2)=0.167

P(3)=0.167

P(4)=0.167

P(5)=0.167

P(6)=0.2 (given)

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