Answer to Question #238971 in Statistics and Probability for girl

Question

Two events A and B are such that 𝑃(𝐴) = 0.45, 𝑃(𝐵) = 0.22 𝑎𝑛𝑑 𝑃(𝐴 ∪ 𝐵) = 0.53.

5.1.1 Find 𝑃(𝐴̅∪ 𝐵̅)

5.1.2 Are the events A and B independent? 

Answer 1

We can present the union of events as: $A\cup B=(A\backslash(A\cap B))\cup(A\cap B)\cup(B\backslash(A\cap B))$. Thus, we get: $P(A\cup B)=P(A\backslash(A\cap B))+P((A\cap B))+P(B\backslash(A\cap B))=P(A)+P(B\backslash(A\cap B))$.

From the latter we find: $P(A\cup B)-P(A)=P(B\backslash(A\cap B))$.

From the latter we find that: $0.53-0.45=P(B)-P(A\cap B)$ . We substitute $P(B)=0.22$ and get: $P(A\cap B)=0.22-0.08=0.14$.

Denote the whole probability space by $X$. Then, $\bar{A}=X\backslash A$ and $\bar{B}=X\backslash B$. Thus, we get: $\bar{A}\cup{\bar{B}}=(X\backslash A)\cup(X\backslash B)=X\backslash(A\cap B)$. We get:

a). $P(\bar{A}\cup{\bar{B}})=P(X\backslash(A\cap B))=P(X)-P(A\cap B)=1-0.14=0.86$.

b). For independent events one has: $P(A\cap B)=P(A)P(B)$. $P(A)P(B)=0.099$. It follows from the formulation of the problem. As we can see $P(A)P(B)\neq P(A\cap B)$. Thus, the events are not independent.