Let M be the maximum value of the product of two positive integers when their sum is 66. Let the sample space S={x∈Z:x(66−x)≥95M} and the event A={x∈S:x is a multiple of 3}. Then P(A) is equal to:
Explanation
Solving
Finding M: For x+y=66, product is max when x=y=33. ⟹M=33×33=1089
Sample Space S: x(66−x)≥95(1089)=605
66x−x2≥605⟹x2−66x+605≤0
(x−11)(x−55)≤0⟹x∈[11,55]
Total elements in S: n(S)=55−11+1=45
Event A (Multiples of 3): x∈{12,15,…,54}
Number of elements in A: n(A)=354−12+1=15
Probability: P(A)=4515=31
Correct Option: (4)
Explanation
Solving
Finding M: For x+y=66, product is max when x=y=33. ⟹M=33×33=1089
Sample Space S: x(66−x)≥95(1089)=605
66x−x2≥605⟹x2−66x+605≤0
(x−11)(x−55)≤0⟹x∈[11,55]
Total elements in S: n(S)=55−11+1=45
Event A (Multiples of 3): x∈{12,15,…,54}
Number of elements in A: n(A)=354−12+1=15
Probability: P(A)=4515=31
Correct Option: (4)
