Suppose you ask a friend to randomly choose an integer between 1 and 10, inclusive. what is the probability that the number will be more than 5 or odd? (enter your probability as a fraction.)

There are 10 possible outcomes when randomly choosing an integer between 1 and 10. To choose an integer more than 5, there are 4 possible outcomes. So the probability of choosing a number more than 5 is 4/10 = 0.4.

Meanwhiles, there are 5 odd numbers between 1 and 10. So the probability of choosing an odd number would be 5/10 = 0.5.

Let F = {choosing a number more than 5},
O = {choosing an odd number}
Then P(F) = 0.4 and P(O) = 0.5.
We know that
P(F or O) = P(F) + P(O) – P(FO)
In this case, P(FO) = P(more than 5 and that's an odd number) = P(the number chosen is 7 or 9) = 2/10 = 0.2
Hence,
P(F or O) = 0.4 + 0.5 – 0.2 = 0.7
Thus the answer is 0.7.

Dis the solution here