Answer with explanation:
A box contains 12 balls numbered 1 through 12.
Also the ball are drawn without replacement.
a)
The probability that the first ball had a smaller number on it.
i.e. the number on the first ball could be: {1,2,3}
Hence, the probability that the first ball had a smaller number on it is:
![(3)/(12)=(1)/(4)=0.25](https://img.qammunity.org/2020/formulas/mathematics/college/ues722x85hx7778f9y1yfssjbq9lacmllv.png)
b)
The probability that the first ball has a even number on it is:
There are total 6 even numbers {2,4,6,8,10,12}
but 4 can't be considered as it comes on the second draw, so we are left with just 5 balls with even number.
Hence, the probability is:
![(5)/(12)=0.4166](https://img.qammunity.org/2020/formulas/mathematics/college/rmqm0pmf5qadwfrnzaclntperecpob43h7.png)