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Probability question

2024 · 30 Jan · Shift 1 · Q43
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  5. /2024 · 30 Jan · Shift 1 · Q43

Probability question

2024 · 30 Jan · Shift 1 · Q43

JEE MainMathematicsProbabilityMCQ+4 / −1
Two integers xxx and yyy are chosen with replacement from the set {0,1,2,3,…,10}\{0,1,2,3, \ldots, 10\}{0,1,2,3,…,10}. Then the probability that ∣x−y∣>5|x-y|\gt 5∣x−y∣>5, is :
  1. A
    31121\frac{31}{121}12131​
  2. B
    60121\frac{60}{121}12160​
  3. C
    62121\frac{62}{121}12162​
  4. D
    30121\frac{30}{121}12130​
View written solutionFree

Correct answer: D

  1. Total number of outcomes

Since xxx and yyy are chosen with replacement from the set

{0,1,2,3,…,10}\{0,1,2,3,\dots,10\}{0,1,2,3,…,10}

there are 111111 choices for xxx and 111111 choices for yyy.

So total ordered pairs:

11×11=12111\times 11=12111×11=121
  1. Required condition

We need:

∣x−y∣>5|x-y|>5∣x−y∣>5

This means:

∣x−y∣=6,7,8,9,10|x-y|=6,7,8,9,10∣x−y∣=6,7,8,9,10
  1. Count favorable pairs

For a fixed difference d>0d>0d>0, the number of ordered pairs satisfying ∣x−y∣=d|x-y|=d∣x−y∣=d is:

2(11−d)2(11-d)2(11−d)

because:

  • x−y=dx-y=dx−y=d gives 11−d11-d11−d pairs,
  • y−x=dy-x=dy−x=d gives another 11−d11-d11−d pairs.

Now compute for d=6,7,8,9,10d=6,7,8,9,10d=6,7,8,9,10:

  • For d=6d=6d=6:
2(11−6)=2⋅5=102(11-6)=2\cdot 5=102(11−6)=2⋅5=10
  • For d=7d=7d=7:
2(11−7)=2⋅4=82(11-7)=2\cdot 4=82(11−7)=2⋅4=8
  • For d=8d=8d=8:
2(11−8)=2⋅3=62(11-8)=2\cdot 3=62(11−8)=2⋅3=6
  • For d=9d=9d=9:
2(11−9)=2⋅2=42(11-9)=2\cdot 2=42(11−9)=2⋅2=4
  • For d=10d=10d=10:
2(11−10)=2⋅1=22(11-10)=2\cdot 1=22(11−10)=2⋅1=2

Total favorable outcomes:

10+8+6+4+2=3010+8+6+4+2=3010+8+6+4+2=30
  1. Probability

Therefore,

P(∣x−y∣>5)=30121P(|x-y|>5)=\frac{30}{121}P(∣x−y∣>5)=12130​
  1. Option matching
30121\frac{30}{121}12130​

corresponds to Option D.

  1. Comparison with stored answer

Stored correct answer: D

This matches our derived answer.

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