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Probability question

2022 · 27 Jun · Shift 2 · Q34
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  5. /2022 · 27 Jun · Shift 2 · Q34

Probability question

2022 · 27 Jun · Shift 2 · Q34

JEE MainMathematicsProbabilityMCQ+4 / −1
If a point A(x, y) lies in the region bounded by the y-axis, straight lines 2y + x = 6 and 5x −-− 6y = 30, then the probability that y < 1 is :
  1. A
    16{1 \over 6}61​
  2. B
    56{5 \over 6}65​
  3. C
    23{2 \over 3}32​
  4. D
    67{6 \over 7}76​
View written solutionFree

Correct answer: B

  1. Interpret the region

The point A(x,y)A(x,y)A(x,y) is uniformly chosen from the region bounded by:

  • the yyy-axis: x=0x=0x=0
  • the line 2y+x=6⇒x=6−2y2y+x=6 \Rightarrow x=6-2y2y+x=6⇒x=6−2y
  • the line 5x−6y=30⇒x=6+6y55x-6y=30 \Rightarrow x=6+\frac{6y}{5}5x−6y=30⇒x=6+56y​

We first find the vertices of the bounded region.


  1. Find intersection points

(i) Intersection of x=0x=0x=0 with 2y+x=62y+x=62y+x=6

Putting x=0x=0x=0: 2y=6⇒y=32y=6 \Rightarrow y=32y=6⇒y=3 So one point is: (0,3)(0,3)(0,3)

(ii) Intersection of x=0x=0x=0 with 5x−6y=305x-6y=305x−6y=30

Putting x=0x=0x=0: −6y=30⇒y=−5-6y=30 \Rightarrow y=-5−6y=30⇒y=−5 So another point is: (0,−5)(0,-5)(0,−5)

(iii) Intersection of the two lines

Solve x+2y=6x+2y=6x+2y=6 5x−6y=305x-6y=305x−6y=30 From the first equation, x=6−2yx=6-2yx=6−2y Substitute into the second: 5(6−2y)−6y=305(6-2y)-6y=305(6−2y)−6y=30 30−10y−6y=3030-10y-6y=3030−10y−6y=30 −16y=0⇒y=0-16y=0 \Rightarrow y=0−16y=0⇒y=0 Then x=6x=6x=6 So the third point is: (6,0)(6,0)(6,0)

Thus the bounded region is the triangle with vertices: (0,3), (0,−5), (6,0)(0,3),\ (0,-5),\ (6,0)(0,3), (0,−5), (6,0)


  1. Total area of the triangle

Take the side along the yyy-axis from (0,−5)(0,-5)(0,−5) to (0,3)(0,3)(0,3) as base. Its length is: 3−(−5)=83-(-5)=83−(−5)=8 The perpendicular distance of (6,0)(6,0)(6,0) from the line x=0x=0x=0 is 666.

Hence, Total area=12×8×6=24\text{Total area}=\frac{1}{2}\times 8 \times 6=24Total area=21​×8×6=24


  1. Find the part where y<1y<1y<1

The line y=1y=1y=1 cuts the triangle. We need the area of the portion below this line.

Find where y=1y=1y=1 meets the two slanted sides.

On x+2y=6x+2y=6x+2y=6

x+2(1)=6⇒x=4x+2(1)=6 \Rightarrow x=4x+2(1)=6⇒x=4 So intersection point is (4,1)(4,1)(4,1).

On 5x−6y=305x-6y=305x−6y=30

5x−6(1)=30⇒5x=36⇒x=3655x-6(1)=30 \Rightarrow 5x=36 \Rightarrow x=\frac{36}{5}5x−6(1)=30⇒5x=36⇒x=536​ But this point is not on the bounded segment of this line inside the triangle, because the segment in the triangle runs from (0,−5)(0,-5)(0,−5) to (6,0)(6,0)(6,0), where yyy ranges only from −5-5−5 to 000. So y=1y=1y=1 does not meet that side of the triangle.

Thus, the region with y>1y>1y>1 is the small triangle with vertices: (0,3), (0,1), (4,1)(0,3),\ (0,1),\ (4,1)(0,3), (0,1), (4,1)

Its area is: 12×2×4=4\frac{1}{2}\times 2 \times 4=421​×2×4=4

Therefore, area where y<1y<1y<1 is: 24−4=2024-4=2024−4=20


  1. Probability

Since the point is uniformly distributed in the triangular region, P(y<1)=Area where y<1Total area=2024=56P(y<1)=\frac{\text{Area where } y<1}{\text{Total area}}=\frac{20}{24}=\frac{5}{6}P(y<1)=Total areaArea where y<1​=2420​=65​


  1. Option check

56\frac{5}{6}65​ corresponds to Option B.


  1. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So, the derived answer agrees with the stored answer.

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