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

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

Prerequisites

Set Theory and Venn Diagram
Basic Probability Theory
AM - GM Inequality

Solution

Draw the Venn Diagram

venn diagram and am gm inequality problem

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,

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

Prerequisites

Set Theory and Venn Diagram
Basic Probability Theory
AM - GM Inequality

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,

Venny Venny AMy GMy | ISI MStat 2016 PSB Problem 3

Problem - Venn diagram and AM GM inequality

Prerequisites

Solution

Draw the Venn Diagram

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

The Final Inequality

Related

Problem - Venn diagram and AM GM inequality

Prerequisites

Solution

Draw the Venn Diagram

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

The Final Inequality

Related

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