Q113 / 180 · COMEDK 2024 (Memory Based) · Mathematics
A number $n$ is choosen at random from $S=\{1,2$, $3, \ldots, 100\}$. Let $A=\{n \in S: n$ is a perfect square $\}, B=\{n \in S: n$ is odd number $\}$ and $C=\{n \in S: n$ is a even prime number $\}$ then, correct order of their probability is
- A$P(A) \lt P(B) \lt P(C)$
- B$P(A) \gt P(B) \gt P(C)$
- Correct answerC$P(B) \gt P(A) \gt P(C)$
- D$P(A) \gt P(C) \lt P(B)$
Solution
Total number of outcomes $=100$
Total number of perfect square $=10$
Total odd number $=50$
Total number of even prime number $=1$
$\begin{aligned}<br/>& P(A)=\frac{10}{100}=\frac{1}{10}=0.1 \\<br/>& P(B)=\frac{50}{100}=\frac{1}{2}=0.5 \\<br/>& P(C)=\frac{1}{100}=0.01 \\<br/>& \therefore \quad P(B) \gt P(A) \gt P(C)<br/>\end{aligned}$
Total number of perfect square $=10$
Total odd number $=50$
Total number of even prime number $=1$
$\begin{aligned}<br/>& P(A)=\frac{10}{100}=\frac{1}{10}=0.1 \\<br/>& P(B)=\frac{50}{100}=\frac{1}{2}=0.5 \\<br/>& P(C)=\frac{1}{100}=0.01 \\<br/>& \therefore \quad P(B) \gt P(A) \gt P(C)<br/>\end{aligned}$
Hence, the correct answer is option: 'C'.