in a single throw fair die what is probability that an odd number or perfect square greater then one show up?

odds: 1,3,5

perfect squares: 1, 4

I count 4 possibilities, so .....

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so we have 4 cases,
prob(odd or perfect square) = 4/6 = 2/3

To calculate the probability of an outcome, we need to know the number of favorable outcomes and the total number of possible outcomes. Let's break down the problem to find these values:

Favorable outcomes:
1. Odd numbers on a fair die: 1, 3, 5
2. Perfect squares greater than one: 4

So, the total number of favorable outcomes is 3 + 1 = 4.

Total number of possible outcomes:
A fair die has 6 sides, numbered 1 to 6. Therefore, the total number of possible outcomes is 6.

Now, we can calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes:

Probability = Number of favorable outcomes / Total number of possible outcomes

Probability = 4 / 6

Simplifying, we get 2 / 3.

Therefore, the probability that an odd number or a perfect square greater than one shows up in a single throw of a fair die is 2/3 or approximately 0.667.