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 : ${12,15,18, \ldots .54}$

$P(A)=\frac{15}{45}=\frac{1}{3}$