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Probability question

2023 · 25 Jan · Shift 1 · Q29
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  5. /2023 · 25 Jan · Shift 1 · Q29

Probability question

2023 · 25 Jan · Shift 1 · Q29

JEE MainMathematicsProbabilityMCQ+4 / −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∈Z:x(66−x)≥59M}S = \left\{ {x \in \mathbb{Z}:x(66 - x) \ge {5 \over 9}M} \right\}S={x∈Z:x(66−x)≥95​M} and the event A={x∈S:x is a multiple of 3}\mathrm{A = \{ x \in S:x\,is\,a\,multiple\,of\,3\}}A={x∈S:xisamultipleof3}. Then P(A) is equal to :
  1. A
    13\frac{1}{3}31​
  2. B
    15\frac{1}{5}51​
  3. C
    722\frac{7}{22}227​
  4. D
    1544\frac{15}{44}4415​
View written solutionFree

Correct answer: A

  1. Find the maximum product MMM

Let the two positive integers be xxx and 66−x66-x66−x. Their product is x(66−x)=66x−x2.x(66-x)=66x-x^2.x(66−x)=66x−x2. This is a downward opening parabola, so its maximum occurs at x=662=33.x=\frac{66}{2}=33.x=266​=33. Hence, M=33⋅33=1089.M=33\cdot 33=1089.M=33⋅33=1089.

  1. Determine the sample space SSS

We need x(66−x)≥59M=59⋅1089=605.x(66-x)\ge \frac{5}{9}M=\frac{5}{9}\cdot 1089=605.x(66−x)≥95​M=95​⋅1089=605. So, x(66−x)≥605x(66-x)\ge 605x(66−x)≥605 66x−x2≥60566x-x^2\ge 60566x−x2≥605 x2−66x+605≤0.x^2-66x+605\le 0.x2−66x+605≤0. Now factor: x2−66x+605=(x−11)(x−55).x^2-66x+605=(x-11)(x-55).x2−66x+605=(x−11)(x−55). Thus, (x−11)(x−55)≤0,(x-11)(x-55)\le 0,(x−11)(x−55)≤0, which gives 11≤x≤55.11\le x\le 55.11≤x≤55. Since x∈Zx\in \mathbb Zx∈Z, S={11,12,13,…,55}.S=\{11,12,13,\dots,55\}.S={11,12,13,…,55}.

Number of elements in SSS: ∣S∣=55−11+1=45.|S|=55-11+1=45.∣S∣=55−11+1=45.

  1. Find event AAA

AAA consists of those elements of SSS which are multiples of 333. Multiples of 333 from 111111 to 555555 are: 12,15,18,21,24,27,30,33,36,39,42,45,48,51,54.12,15,18,21,24,27,30,33,36,39,42,45,48,51,54.12,15,18,21,24,27,30,33,36,39,42,45,48,51,54. This is an arithmetic progression with first term 121212, last term 545454, common difference 333.

Number of terms: ∣A∣=54−123+1=14+1=15.|A|=\frac{54-12}{3}+1=14+1=15.∣A∣=354−12​+1=14+1=15.

  1. Compute the probability

P(A)=∣A∣∣S∣=1545=13.P(A)=\frac{|A|}{|S|}=\frac{15}{45}=\frac{1}{3}.P(A)=∣S∣∣A∣​=4515​=31​.

  1. Compare with stored answer

Derived answer is 13\boxed{\frac{1}{3}}31​​, which matches option A.

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