Q.The probability that a student will pass in French is 0.64, will pass in Sociology is 0.45 and will pass in both is 0.40. What is the probability that the student will pass in at least one of the two subjects?

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

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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