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:
[tex]\dfrac{3}{12}=\dfrac{1}{4}=0.25[/tex]
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:
[tex]\dfrac{5}{12}=0.4166[/tex]
