TIFR CSE 2019 | Part A | Question: 4

Question

What is the probability that a point $P=(\alpha,\beta)$ picked uniformly at random from the disk $x^2 +y^2 \leq 1$ satisfies $\alpha + \beta \leq 1$?

• A. $\frac{1}{\pi}$
• B. $\frac{3}{4} + \frac{1}{4} \cdot \frac{1}{\pi}$
• C. $\frac{3}{4}+ \frac{1}{4} \cdot \frac{2}{\pi}$
• D. $1$
• E. $\frac{2}{\pi}$

Comments

This is wrong- Every point P=$(\alpha,\beta)$ on and inside the circle has sum $\alpha+\beta<=1$.

So the event that the point $P$ picked uniformly at random from the disk $x^2+y^2<=1$ satisfies this condition is sure event and probability is $1$.

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First statement is not correct

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@Arjun Sir, this is not correct answer I think. Shobhit Joshi's answer should be correct.

Consider point $(0.65,0.65) 0.65+0.65>1$ not satisfying the first statement I have mentioned.

Answer 1

The $Area$ inside $ADBC$ satisfies the $x^{2}+y^{2}\leq1$ 

The line $AB$ has $eq^{n}\,x+y=1$, the arrow represents the points which statisfy $x+y\leq1$

The $coloured\,region$ shows the area from where any point can be picked from the entire disk $ADBC$

$Area\, of\, the\, disk=\pi*1^{2}=\pi$

$Area\,of\,the\,coloured\,region=Area(ADCBO)+Area(AOB)$

$Area(ADCBO)=\frac{3}{4}*Area\,of\,disk=\frac{3}{4}*\pi$

$Area(AOB)=\frac{1}{2}*AO*OB=\frac{1}{2}*1*1=\frac{1}{2}$

$Area\,of\,the\,coloured\,region=\frac{3}{4}*\pi+\frac{1}{2}$

$Required\,Proabability=\frac{coloured\,region}{total\,area\,of\,disk}=\frac{\frac{3}{4}*\pi+\frac{1}{2}}{\pi}=\frac{3}{4}+\frac{1}{2}.\frac{1}{\pi}$

So, $(c)$ should be correct. Correct me If I'm wrong