Probability Question 3
Question 3 - 25 January - Shift 1
Let $M$ be the maximum value of the product of two positive integers when their sum is 66 . Let the sample space $S={x \in Z: x(66-x) \geq \frac{5}{9} M}$ and the event $A={x \in S: \text{ is a multiple of} ; 3}$. Then $P(A)$ is equal to
(1) $\frac{15}{44}$
(2) $\frac{1}{3}$
(3) $\frac{1}{5}$
(4) $\frac{7}{22}$
Show Answer
Answer: (2)
Solution:
Formula: Probability of occurrence of an event
$M=33 \times 33$
$x(66-x) \geq \frac{5}{9} \times 33 \times 33$
$11 \leq x \leq 55$
A : ${12,15,18, \ldots .54}$
$P(A)=\frac{15}{45}=\frac{1}{3}$
