If two numbers $p$ and $q$ are chosen randomly from the set $\{1,2,3,4\}$, one by one, with replacement,…
- $\frac{1}{4}$
- $\frac{7}{16}$
- $\frac{1}{2}$
- $\frac{9}{16}$
Solution
Two numbers are selected with replacement from $S = \{1,2,3,4\}$, giving $4 \times 4 = 16$ total ordered pairs $(p,q)$.
The condition $p^2 \ge 4q$ is analyzed:
When $p=1$, $p^2=1$; no $q$ satisfies $1 \ge 4q$.
When $p=2$, $p^2=4$; the condition $4 \ge 4q$ holds only for $q=1$.
When $p=3$, $p^2=9$; $9 \ge 4q$ yields favorable $q \in \{1,2\}$.
When $p=4$, $p^2=16$; all $q \in \{1,2,3,4\}$ satisfy $16 \ge 4q$.
The number of favorable outcomes is $0 + 1 + 2 + 4 = 7$.
The probability is $\frac{7}{16}$, corresponding to option B.
Asked in: MHT CET 2025 (20 April Shift 1)