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 0.21 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_ .

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_ .

Explanation / Answer

The sum of probability of occurance of 1,2,3,4,5,6 will always be equal to 1 (one).

In the crooked dice the probability of occurance of 2,3,4 and 5 are not affected, therefore,

Probability of occurance of each 2,3,4 and 5 = 1 / 6

Probability of 1 + Probability of 2 + Probability of 3 + Probability of 4 + Probability of 5 + Probability of 6 = 1

Probability of 1 + ( 1 / 6 ) + ( 1 / 6 ) + ( 1 / 6 ) + ( 1 / 6 ) + 0.21 = 1

Probability of 1 + ( 4 / 6 ) + 0.21 = 1

Probability of 1 + 0.67 + 0.21 = 1

Probability of 1 + 0.88 = 1

Probability of 1 = 0.12

Probability of 2 = ( 1 / 6 ) = 0.1667 = 0.167

Probability of 3 = ( 1 / 6 ) = 0.1667 = 0.167

Probability of 4 = ( 1 / 6 ) = 0.1667 = 0.167

Probability of 5 = ( 1 / 6 ) = 0.1667 = 0.167

Probability of 6 = 0.21

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