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.
For any two events \(A\) and \(B\), show that
$$
(\mathrm{P}(A \cap B))^{2}+\left(\mathrm{P}\left(A \cap B^{c}\right)\right)^{2}+\left(\mathrm{P}\left(A^{c} \cap B\right)\right)^{2}+\left(\mathrm{P}\left(A^{c} \cap B^{c}\right)\right)^{2} \geq \frac{1}{4}
$$
P(region Red) = \(Y\)
P(region Blue) = \(Z\)
P(region Grey) = \(W\)
P(region Brown) = \(X\)
Observe that \( W + X + Y + Z = 1\). \( W, X, Y, Z \geq 0\).
\({P}(A \cap B) = Z\).
\({P}\left(A \cap B^{c}\right) = Y\).
\({P}\left(A^{c} \cap B\right) = W\).
\( {P}\left(A^{c} \cap B^{c}\right) = X\).
\( W, X, Y, Z \geq 0\).
\( W + X + Y + Z = 1\).
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.
For any two events \(A\) and \(B\), show that
$$
(\mathrm{P}(A \cap B))^{2}+\left(\mathrm{P}\left(A \cap B^{c}\right)\right)^{2}+\left(\mathrm{P}\left(A^{c} \cap B\right)\right)^{2}+\left(\mathrm{P}\left(A^{c} \cap B^{c}\right)\right)^{2} \geq \frac{1}{4}
$$
P(region Red) = \(Y\)
P(region Blue) = \(Z\)
P(region Grey) = \(W\)
P(region Brown) = \(X\)
Observe that \( W + X + Y + Z = 1\). \( W, X, Y, Z \geq 0\).
\({P}(A \cap B) = Z\).
\({P}\left(A \cap B^{c}\right) = Y\).
\({P}\left(A^{c} \cap B\right) = W\).
\( {P}\left(A^{c} \cap B^{c}\right) = X\).
\( W, X, Y, Z \geq 0\).
\( W + X + Y + Z = 1\).
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,
$$
(\mathrm{P}(A \cap B))^{2}+\left(\mathrm{P}\left(A \cap B^{c}\right)\right)^{2}+\left(\mathrm{P}\left(A^{c} \cap B\right)\right)^{2}+\left(\mathrm{P}\left(A^{c} \cap B^{c}\right)\right)^{2} \geq \frac{1}{4}
$$
