Question

3. Suppose that A and B are two events defined over the same sample space, with probabilities P(A) 3/4 and P(B)- 3/8. (a) Show that P(A UB) 2 3/4. (b) Show that 1/8 < P(AB) 3/8 (c) Give inequalities analogous to (a) and (b) for P(A) 2/3 and P(B)1/2.

Answer 1

Given the probabilities of 2 events, $P\left ( A \right )=3/4,P\left ( B \right )=3/8$ .

Use the identities $P\left ( AB \right )=P\left ( B \right )P\left ( A|B \right )$ and $P\left ( A\cup B \right )=P\left ( A \right )&plus;P\left ( B \right ) -P\left ( AB \right )$

a) Now rewriting the second identity,

$P\left ( A\cup B \right )=P\left ( A \right )&plus;P\left ( B \right ) -P\left ( B \right ) P\left ( A|B \right) \\ P\left ( A\cup B \right )=P\left ( A \right )&plus;P\left ( B \right )\left [ 1- P\left ( A|B \right) \right ] \\ P\left ( A\cup B \right ) \geqslant P\left ( A \right ) \textup{ since }0<P\left ( A|B \right)\leqslant 1\\ P\left ( A\cup B \right ) \geqslant 3/4$

The proof is complete.

b) Again using the identities,

$P\left ( A\cup B \right ) \geqslant 3/4\\ P\left ( A \right )&plus;P\left ( B \right ) -P\left ( AB \right ) \geqslant 3/4\\ 3/4&plus;1/8-P\left ( AB \right ) \geqslant 3/4\\ 7/8-P\left ( AB \right ) \leqslant 3/4\\ 1/8\leqslant P\left ( AB \right )$

$P\left ( A\cap B \right )=P\left ( A|B \right )P\left ( B \right )\\ P\left ( A\cap B \right )\leqslant P\left ( B \right ) \textup{ since }0<P\left ( A|B \right )\leqslant 1\\ P\left ( A\cap B \right ) \leqslant 3/8$

Thus ${\color{Blue} 1/8 \leqslant P\left ( A\cap B \right ) \leqslant 3/8}$

c) When $P\left ( A \right )=2/3,P\left ( B \right )=1/2$ . Here $P\left ( A \right )=2/3$ is larger, so the analogous inequalities are

${\color{Blue} P\left ( A\cup B \right ) \geqslant 2/3}$

$P\left ( A\cup B \right )\leqslant 2/3\\ P\left ( A \right )&plus;P\left ( B \right ) -P\left ( AB \right ) \leqslant2/3\\ 2/3&plus;1/2-P\left ( AB \right ) \leqslant 2/3\\ {\color{Blue} 1/2\leqslant P\left ( AB \right )}$

$P\left ( A\cap B \right )=P\left ( A|B \right )P\left ( B \right )\\ P\left ( A\cap B \right )\leqslant P\left ( B \right ) \textup{ since }P\left ( A|B \right )<0\\ P\left ( A\cap B \right ) \leqslant 1/2$

Thus ${\color{Blue} 1/2 \leqslant P\left ( A\cap B \right ) \leqslant 2/3}$