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 0.21 and the probabilities of the 2, 3, 4, and 5 faces are not affected, what is the assignment of probabilities to the six faces?

Give your answer to 2 decimal places.
Fill in the blanks:
The probability assigned to: Face with 1 spot is: _Answer 1_ .
The probability assigned to: Face with 2 spots is: _Answer 2_ .
The probability assigned to: Face with 3 spots is: _Answer 3_ .
The probability assigned to: Face with 4 spots is: _Answer 4_ .
The probability assigned to: Face with 5 spots is: _Answer 5_ .
The probability assigned to: Face with 6 spots is: _Answer 6_ .

Answer 1

Let p be the probability assigned to faces with 1 to 5 spots (since their probabilities are unaffected) and let x be the probability assigned to the face with 6 spots. Then, we have the equation:

0.21 = x + p

Since the probabilities of all six faces must add up to 1, we also have the equation:

1 = 5p + x

Solving these equations simultaneously, we get:

p = 0.146

x = 0.064

Therefore, the probability assigned to the faces with 1 to 6 spots (in order) are:

0.146, 0.146, 0.146, 0.146, 0.146, and 0.064.

Explanation: