Question

There is a $60$percent chance that event $A$will occur. If $A$does not occur, then there is a $10$percent chance that $B$will occur. What is the probability that at least one of the events $AorB$will occur?

Short answer

The probability that at least one of the events $A$or $B$will occur$0.64=64\mathrm{\%}$

Step-by-step solution

Given

$P\left(A\right)=60\mathrm{\%}=0.6$

$P\left(B\mid A^c\right)=10\mathrm{\%}=0.1$

Complement Rule

Be using the Complement rule to help you out: $P\left(A^c\right)=P\left(\text{not}A\right)=1-P\left(A\right)$ $P\left(A^c\right)=1-P\left(A\right)$

While $A$arises, $A\cup B$still must follow (and $A$is encompassed under $A\cup B$). A given venue's frequency is similar to $1$.

$P(A\cup B\mid A)=1$

Because $A$doesn't quite arise, $A\cup B$can really only eventuate once $B$happens.

$P\left(A\cup B\mid A^c\right)=P\left(B\mid A^c\right)=0.1$

Law of Total Probability

Be using the law of total probability to its benefit: $P\left(A\right)=\sum _{i=1}^m P\left(B_i\right)P\left(A\mid B_i\right)$

$P(A\cup B)=P(A\cup B\mid A)P\left(A\right)+P\left(A\cup B\mid A^c\right)P\left(A^c\right)$

role="math" localid="1649656510027" $=1\times 0.6+0.1\times 0.4$

$$\begin{gathered} =0.6+0.04 \\ =0.64 \\ =64\% \end{gathered}$$