1^(°) Prove that if P(A)=0.9 and P(B)=0.8 , then P(AB)≥0.7 .…

$1^{\circ}$ Prove that if $P(A)=0.9$ and $P(B)=0.8$, then $P(AB)\ge0.7$.
$2^{\circ}$ Prove that $P(A_2\mid A_1)\ge1-\dfrac{P(\overline{A_2})}{P(A_1)}$.

Step 1: $1^{\circ}$ — the addition rule

\[ P(A+B)=P(A)+P(B)-P(AB) \quad\Longrightarrow\quad P(AB)=P(A)+P(B)-P(A+B). \]

Since $P(A+B)\le1$:

\[ \boxed{P(AB)\ge P(A)+P(B)-1=0.9+0.8-1=0.7.} \]

Intuition: $A$ misses only $10\%$ of the probability and $B$ only $20\%$; together they can miss at most $30\%$, so they overlap in at least $70\%$.

Step 2: $1^{\circ}$ — equality is possible

Choose a number $X$ uniformly in $[0,1]$, $A=\{X\gt0.1\}$, $B=\{X\lt0.8\}$. Then $P(A)=0.9$, $P(B)=0.8$ and $P(AB)=P(0.1\lt X\lt0.8)=0.7$ exactly. So the strict inequality "$\gt0.7$" is not always true; the correct statement is $P(AB)\ge0.7$, with equality iff $P(A+B)=1$.

Step 3: $2^{\circ}$

Assume $P(A_1)\gt0$. Split $A_1=A_1A_2+A_1\overline{A_2}$ (incompatible parts):

\[ P(A_1A_2)=P(A_1)-P(A_1\overline{A_2})\ge P(A_1)-P(\overline{A_2}), \]

because $A_1\overline{A_2}\subseteq\overline{A_2}$. Dividing by $P(A_1)$:

\[ \boxed{P(A_2\mid A_1)=\frac{P(A_1A_2)}{P(A_1)}\ge1-\frac{P(\overline{A_2})}{P(A_1)}.} \]

Equality holds iff $P(\overline{A_1}\,\overline{A_2})=0$, i.e. $\overline{A_2}\subseteq A_1$ up to probability zero.

Step 4: Meaning of $2^{\circ}$

If $A_2$ is very likely (small $P(\overline{A_2})$) and $A_1$ is not too unlikely, then $A_2$ remains likely even after we learn that $A_1$ occurred. Example: $P(\overline{A_2})=0.05$, $P(A_1)=0.5$ gives $P(A_2\mid A_1)\ge0.9$.

Part $1^{\circ}$ is a special case: with $A_1=A$, $A_2=B$, $P(AB)=P(A)P(B\mid A)\ge P(A)-P(\overline B)=0.9-0.2=0.7$.