<h2>
Answer with explanation:</h2>
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:
![\dfrac{3}{12}=\dfrac{1}{4}=0.25](https://tex.z-dn.net/?f=%5Cdfrac%7B3%7D%7B12%7D%3D%5Cdfrac%7B1%7D%7B4%7D%3D0.25)
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:
![\dfrac{5}{12}=0.4166](https://tex.z-dn.net/?f=%5Cdfrac%7B5%7D%7B12%7D%3D0.4166)