Cheenta
Get inspired by the success stories of our students in IIT JAM MS, ISI  MStat, CMI MSc DS.  Learn More

Venny Venny AMy GMy | ISI MStat 2016 PSB Problem 3

This problem is a very basic and cute application of set theory, Venn diagram and AM GM inequality to solve the ISI MStat 2016 PSB Problem 3.

Problem - Venn diagram and AM GM inequality

For any two events $$A$$ and $$B$$, show that
$$(\mathrm{P}(A \cap B))^{2}+\left(\mathrm{P}\left(A \cap B^{c}\right)\right)^{2}+\left(\mathrm{P}\left(A^{c} \cap B\right)\right)^{2}+\left(\mathrm{P}\left(A^{c} \cap B^{c}\right)\right)^{2} \geq \frac{1}{4}$$

Solution

Draw the Venn Diagram

P(region Red) = $$Y$$

P(region Blue) = $$Z$$

P(region Grey) = $$W$$

P(region Brown) = $$X$$

Observe that $$W + X + Y + Z = 1$$. $$W, X, Y, Z \geq 0$$.

Now, Calculate Given Probability of Sets in terms of $$W, X, Y, Z$$.

$${P}(A \cap B) = Z$$.

$${P}\left(A \cap B^{c}\right) = Y$$.

$${P}\left(A^{c} \cap B\right) = W$$.

$${P}\left(A^{c} \cap B^{c}\right) = X$$.

The Final Inequality

$$W, X, Y, Z \geq 0$$.

$$W + X + Y + Z = 1$$.

Observe that $$3(W^2 + X^2 + Y^2 + Z^2) = (W^2+X^2) + (W^2+Y^2) + (W^2+Z^2) + (X^2+Y^2) + (X^2+Z^2) + (Y^2+Z^2)$$.

$$3(W^2 + X^2 + Y^2 + Z^2) \geq 2WX + 2WY + 2WZ + 2XY + 2XZ + 2YZ$$ by AM - GM Inequality.

$$\Rightarrow 4(W^2 + X^2 + Y^2 + Z^2) \geq (W + X + Y + Z)^2 = 1$$.

$$\Rightarrow (W^2 + X^2 + Y^2 + Z^2) \geq \frac{1}{4}$$.

Hence,

$$(\mathrm{P}(A \cap B))^{2}+\left(\mathrm{P}\left(A \cap B^{c}\right)\right)^{2}+\left(\mathrm{P}\left(A^{c} \cap B\right)\right)^{2}+\left(\mathrm{P}\left(A^{c} \cap B^{c}\right)\right)^{2} \geq \frac{1}{4}$$

This problem is a very basic and cute application of set theory, Venn diagram and AM GM inequality to solve the ISI MStat 2016 PSB Problem 3.

Problem - Venn diagram and AM GM inequality

For any two events $$A$$ and $$B$$, show that
$$(\mathrm{P}(A \cap B))^{2}+\left(\mathrm{P}\left(A \cap B^{c}\right)\right)^{2}+\left(\mathrm{P}\left(A^{c} \cap B\right)\right)^{2}+\left(\mathrm{P}\left(A^{c} \cap B^{c}\right)\right)^{2} \geq \frac{1}{4}$$

Solution

Draw the Venn Diagram

P(region Red) = $$Y$$

P(region Blue) = $$Z$$

P(region Grey) = $$W$$

P(region Brown) = $$X$$

Observe that $$W + X + Y + Z = 1$$. $$W, X, Y, Z \geq 0$$.

Now, Calculate Given Probability of Sets in terms of $$W, X, Y, Z$$.

$${P}(A \cap B) = Z$$.

$${P}\left(A \cap B^{c}\right) = Y$$.

$${P}\left(A^{c} \cap B\right) = W$$.

$${P}\left(A^{c} \cap B^{c}\right) = X$$.

The Final Inequality

$$W, X, Y, Z \geq 0$$.

$$W + X + Y + Z = 1$$.

Observe that $$3(W^2 + X^2 + Y^2 + Z^2) = (W^2+X^2) + (W^2+Y^2) + (W^2+Z^2) + (X^2+Y^2) + (X^2+Z^2) + (Y^2+Z^2)$$.

$$3(W^2 + X^2 + Y^2 + Z^2) \geq 2WX + 2WY + 2WZ + 2XY + 2XZ + 2YZ$$ by AM - GM Inequality.

$$\Rightarrow 4(W^2 + X^2 + Y^2 + Z^2) \geq (W + X + Y + Z)^2 = 1$$.

$$\Rightarrow (W^2 + X^2 + Y^2 + Z^2) \geq \frac{1}{4}$$.

Hence,

$$(\mathrm{P}(A \cap B))^{2}+\left(\mathrm{P}\left(A \cap B^{c}\right)\right)^{2}+\left(\mathrm{P}\left(A^{c} \cap B\right)\right)^{2}+\left(\mathrm{P}\left(A^{c} \cap B^{c}\right)\right)^{2} \geq \frac{1}{4}$$

This site uses Akismet to reduce spam. Learn how your comment data is processed.