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Mathematics · Probability

JEE Advanced 2023 — Paper 1 — Question 5

Let X:{(x,y)∈Z×Z:x28+y220<1X:\left\{(x, y) \in Z \times Z: \frac{x^{2}}{8}+\frac{y^{2}}{20}<1\right. and y2<5x}\left.y^{2}<5 x\right\}. Three distinct points P,QP, Q and RR are randomly

chosen from X . Then the probability that P,Q\mathrm{P}, \mathrm{Q} and R form a triangle whose area is a positive integer, is

  1. Option A:

    71220\frac{71}{220}

  2. Option B:

    73220\frac{73}{220}

    Correct
  3. Option C:

    79220\frac{79}{220}

  4. Option D:

    83220\frac{83}{220}

Answer: B

Step-by-step solution

x28+y220<1\frac{x^{2}}{8}+\frac{y^{2}}{20}<1 and y2<5xy^{2}<5 x

\frac{x^{2}}{8}+\frac{y^{2}}{20}=1 …(1) \end{gathered}$$ $$\begin{gathered} y^{2}=5 x …(2) \end{gathered}$$ On solving (1) and (2) $\frac{x^{2}}{8}+\frac{x}{4}=1$ $x^{2}+2 x=8$ $x^{2}+2 x-8=0$ $x=2,-4$ $X=\{(1,1),(1,0),(1,-1),(1,2),(1,-2),(2,1),(2,-1),(2,3),(2,-3),(2,-2),(2,2),(2,0)\}$ $\mathrm{n}(\mathrm{S})={ }^{12} \mathrm{C}_{3}$ A is event of selecting 3 points for which area of $\Delta$ is positive integer. $n(A)=4 \times 7+9 \times 5=73$ $P(A)=\frac{73}{{ }^{12} C_{3}}=\frac{73}{220}$
Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2023
Paper
Paper 1
Subject
Mathematics
Chapter
Probability
Topic
Introduction to Probability
Let X: \ (x, y) in Z × Z: frac x 2 8 +frac y 2 20 <1 . and .y 2 <5 x… | JEE Advanced 2023 PYQ with Solution · DhiX AI