If a person tosses a normal 6- sided die, what is the
probability that he will not roll a 2?

 The required probability is 0.83.

Step-by-step explanation:  Given that a person tosses a normal 6- sided die.

We are to find the probability that he will not roll a 2.

Let  S denote the sample space for the experiment and E denote the event of not rolling a 2.

Then, S = {1, 2, 3, 4, 5, 6}  and  E = {1, 3, 4, 5, 6}.

That is, n(S) = 6  and  n(E) = 5.

Therefore, the probability of event E is given by

Thus, the required probability is 0.83.

step-by-step explanation:

first you will have to convert 3/4 into a decimal

which is .75

then you will have to divide 1.5 by .75

which is 2

no you have to divide