Question

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

Answer 1

Answer: 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

`P(E)=(n(E))/(n(S))=(5)/(6)=0.83.`

Thus, the required probability is 0.83.