$A$ bag contains slips numbered from $1$ to $100$. If Fatima chooses a slip at random from the bag,it will either be an odd number or an even number. Since this situation has only two possible outcomes,the probability of each is $\frac{1}{2}$. Justify.

(A) We know that,between $1$ to $100$,half the numbers are even and half are odd. Specifically,$50$ numbers $(2, 4, 6, 8, \dots, 96, 98, 100)$ are even and $50$ numbers $(1, 3, 5, 7, \dots, 97, 99)$ are odd.
Since the total number of slips is $100$,the number of favorable outcomes for an even number is $50$,and for an odd number is $50$.
Probability of getting an even number $= \frac{\text{Number of even slips}}{\text{Total number of slips}} = \frac{50}{100} = \frac{1}{2}$.
Probability of getting an odd number $= \frac{\text{Number of odd slips}}{\text{Total number of slips}} = \frac{50}{100} = \frac{1}{2}$.
Since both events have an equal number of favorable outcomes,they are equally likely,and the probability of each is indeed $\frac{1}{2}$.