Q.A girl is preparing for National Level Entrance exam and State Level Entrance exam for professional courses. The chances of her cracking National Level exam is 0.42 and that of State Level exam is 0.54. The probability that she clears both the exams is 0.11. Find the probability that

$0\le P(A\cap B)\le P(A)\le P(A\cup B)\le P(A)+P(B)$.

Since every probability is at least $0$, $P(A\cap B)\ge0$. The intersection is a subset of $A$, so $P(A\cap B)\le P(A)$. $A$ is a subset of $A\cup B$, so $P(A)\le P(A\cup B)$. Finally, by the addition theorem $P(A\cup B)=P(A)+P(B)-P(A\cap B)\le P(A)+P(B)$ because $P(A\cap B)\ge0$. Chaining these:

$0\ \le\ P(A\cap B)\ \le\ P(A)\ \le\ P(A\cup B)\ \le\ P(A)+P(B).$

"Neither A nor B" is the complement of $A\cup B$.

First find $P(A\cup B)$ using the addition rule:

$P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.25+0.50-0.14=0.61$

"Neither $A$ nor $B$ occurs" is the event $(A\cup B)^c$, so:

$P((A\cup B)^c) = 1-P(A\cup B) = 1-0.61 = 0.39$

$A\cap B=\phi\Rightarrow P(A\cap B)=0$; $A\cup B=U\Rightarrow P(A\cup B)=1$.

Here $A$ and $B$ are mutually exclusive and exhaustive events of the sample space $U$.

• Since $A\cap B=\phi$ (impossible event), $P(A\cap B)=P(\phi)=0$.

• Since $A\cup B=U$ (sure event), $P(A\cup B)=P(U)=1$.

Probability of at least one of $A,B,C$ $=P(A\cup B\cup C)$, given by the three-event addition theorem below (equivalently $1-P(A'\cap B'\cap C')$).

Formula (addition theorem for three events).

$P(A\cup B\cup C) = P(A)+P(B)+P(C) - P(A\cap B) - P(B\cap C) - P(A\cap C) + P(A\cap B\cap C).$

Equivalent complement form. The probability of at least one occurring is one minus the probability that none occurs:

$P(A\cup B\cup C) = 1 - P(A'\cap B'\cap C').$