# Independent Events- $P(\text{neither/exactly/at least one})$

I'm working on some problems from Pitman's Probability.

Let events $A$ and $B$ be independent, with $P(A)=0.1$ and $P(B)=0.3$. Find:

$(a)$ Neither of the events occur

$(b)$ At least one of the events occur

$(c)$ Exactly one of the events occur

Attempt:

$(a)$ If neither of the events occur, this is the complement of both events occurring. Hence $1-P(A\cap B)=1-P(A)P(B)=1-\frac{1}{30}=\frac{29}{30}$

$(b)$ At least one of the events can be represented as the union of the two events. Hence $P(A\cup B)=P(A)+P(B)-P(A\cap B)=P(A)+P(B)-P(A)P(B)=\frac{4}{10}-\frac{1}{30}=\frac{11}{30}.$

$(c)$ Not sure what this would be. Maybe $P(A^c\;\cap\;B)+P(A\cap B)$?

a) The probability that $A$ does not occur is $0.9$. The probability $B$ does not occur is $0.7$. So the probability neither occurs, by independence, is $(0.9)(0.7)$.

b) This is the complement of the event of a).

c) We want the probability that $A$ occurs but $B$ does not, or that $A$ does not occur but $B$ does. So in symbols we want $\Pr(A\cap B^c)+\Pr(A^c\cap B)$. We leave the rest of the calculation to you.

Remark: Your calculation in b) was almost right. There was an arithmetical slip, since $\Pr(A)\Pr(B)=(0.1)(0.3)=0.03\ne \frac{1}{30}$.

a.$P(A'\cap B') = P((A\cup B)') = 1 - P(A\cup B) = 1 - P(A) - P(B) + P(A\cap B) = 1 - P(A) - P(B) + P(A)P(B) = 1 - 0.1 - 0.3 + 0.1\cdot 0.3 = 0.6 + 0.03 = 0.63$.

c. $P((A\cap B')\cup (B\cap A')) = P(A\cap B') + P(B\cap A') = P(A)P(B') + P(B)P(A') = 0.1\cdot 0.7 + 0.3\cdot 0.9 = 0.07 + 0.27 = 0.34$