Q.A bag contains 4 red, 5 white and 6 blue balls. One ball is drawn at random from the bag. Find the probability that the ball drawn is red or blue.

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

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 $P(A\cap B)\ge0$. Cha …

"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:

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

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

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

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).

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

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