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. 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, which of the choices is the correct assignment of probabilities to the six faces

1            2                  3             4            5       6

2 ÷ 15   1 ÷ 6           1 ÷ 6        1 ÷ 6     1 ÷ 6    1 ÷ 5

Step-by-step explanation:

As it is mentioned in the question that the probability of faces like 2,3,4,5 are not affected so in this case

P(2) = P(3) = P(4) = P(5) = 1 ÷ 6

And,  P(6) = 0.2

As we know that

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

This results P(1) = 2 ÷ 15

So,

1            2                  3             4            5       6

2 ÷ 15   1 ÷ 6           1 ÷ 6        1 ÷ 6     1 ÷ 6    1 ÷ 5

Hence, the correct option is 2nd

is only a line? i think so

a

step-by-step explanation: