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Question

(B) The total number of possible outcomes when choosing two numbers $p$ and $q$ from the set $\{1, 2, 3, 4\}$ with replacement is $4 	imes 4 = 16$.

Vedclass · Accepted Answer

$\frac{5}{16}$

Vedclass · Answer

(B) The total number of possible outcomes when choosing two numbers $p$ and $q$ from the set $\{1, 2, 3, 4\}$ with replacement is $4 	imes 4 = 16$.
We need to find the number of pairs $(p, q)$ such that $p^2 > 4q$.
Let us test each value of $p$:
If $p = 1$,$p^2 = 1$. $1 > 4q$ is not possible for any $q \in \{1, 2, 3, 4\}$.
If $p = 2$,$p^2 = 4$. $4 > 4q$ is not possible for any $q \in \{1, 2, 3, 4\}$.
If $p = 3$,$p^2 = 9$. $9 > 4q$ is true for $q = 1$ $(9 > 4)$ and $q = 2$ $(9 > 8)$. So,$(3, 1)$ and $(3, 2)$ are favorable outcomes.
If $p = 4$,$p^2 = 16$. $16 > 4q$ is true for $q = 1$ $(16 > 4)$,$q = 2$ $(16 > 8)$,and $q = 3$ $(16 > 12)$. So,$(4, 1), (4, 2)$,and $(4, 3)$ are favorable outcomes.
The total number of favorable outcomes is $2 + 3 = 5$.
Thus,the probability is $\frac{5}{16}$.

If two numbers $p$ and $q$ are chosen randomly from the set $\{1, 2, 3, 4\}$,one by one,with replacement,then the probability of getting $p^2 > 4q$ is

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