Answer:
The correct option is B) [tex]\frac{7}{24}[/tex].
Step-by-step explanation:
Consider the provided information.
An urn contains 4 balls numbered 0 through 3. One ball is selected at random and removed from the urn and not replaced.
Case I: If first ball is 0 and 2nd ball is 3.
The probability is: [tex]\frac{1}{4} \times\frac{1}{3}[/tex]
Case II: If first ball is 1 and 2nd ball is 3.
The probability is: [tex]\frac{1}{4} \times\frac{1}{3}[/tex]
Case III: If first ball is 2 and 2nd ball is 3.
All balls with nonzero numbers less than that of the selected ball are also removed from the urn.
That means if we get 2 number ball then we will remove 1 number ball.
The probability is: [tex]\frac{1}{4} \times\frac{1}{2}[/tex]
Case IV: If first ball is 3 then there is no 3 number ball.
The probability is: [tex]\frac{1}{4} \times0[/tex]
Thus, the total number of ways are:
[tex]\begin{aligned}\frac{1}{4} \times\frac{1}{3}+\frac{1}{4} \times\frac{1}{3}+\frac{1}{4} \times\frac{1}{2}&=\frac{1}{12}+\frac{1}{12}+\frac{1}{8}\\&=\frac{2+2+3}{24}\\&=\frac{7}{24}\end{aligned}[/tex]
Hence, the correct option is B) [tex]\frac{7}{24}[/tex].
