Question

The faces of a tetrahedron die are numbered 1 to 4. if each of the 4 sides is equally likely to land down, what is the probability that the 2 does not end up face down? 1

Answer 1

The probability distribution for the face landing down =  4/15

What is the probability?

The area of mathematics known as probability deals with numerical descriptions of how likely it is for an event to happen or for a claim to be true. A number between 0 and 1 is the probability of an event, where, broadly speaking, 0 denotes the event's impossibility and 1 denotes its certainty.

According to the given information:

If we assign 1 probability x

then 2 has probability  1/2 x.

Similarly,

then 3 has probability = 1/2((1/2)*x) = (1/4)*x

so

then 4 has probability = 1/2((1/4)*x) = (1/8)*x

These probabilities need to sum to 1.

Hence:

                $x+\frac{1}{2} x+\frac{1}{4} x+\frac{1}{8} x=1$

or equivalently

                   (15/8)*x = 1

                       x = 8/15

the probability distribution for the face landing down  = (1/2)*(8/15)

                                                                                         = 8/30

                                                                                         =  4/15

To know more about probability visit:

brainly.com/question/1581057                  

#SPJ4

I understand that the question you are looking for is:

A tetrahedral die has four faces, numbered 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 as likely as 3, and so on) what is the probability distribution for the face landing down?