In a jungle, the probability of an animal being a mammal is 0.6, a nocturnal is 0.2. What is the probability that an animal found in this jungle is either a mammal, or nocturnal or both. Assume that these are independent traits.
$\endgroup$3 Answers
$\begingroup$Hint: The probability that $A$, $B$, or both $A$ and $B$ occur is $$ P(A\text{ and }B)+P(A\text{ and not }B)+P((\text{not }A)\text{ and }B). $$
$\endgroup$ $\begingroup$A different hint: The probability that $A$, $B$, or both $A$ and $B$ occur is $$\begin{align} P(A \mathbin{\operatorname {or}} B) &= 1 - P(\operatorname{not} (A \mathbin{\operatorname{or}} B)) \\ &= 1-P((\operatorname{not}A) \mathbin{\operatorname{and}}(\operatorname{not}B)). \end{align}$$
The second equality follows from De Morgan's laws.
$\endgroup$ $\begingroup$Start with the general formula for the union of two events: $P(A \cup B)= P(A) + P(B) - P(AB)$. Now use the fact that the sample space is equal to $S = B \cup B^c$. Since $B$ and $B^c$ are disjoint, you can write $P(A) = P(AB) + P(AB^c)$. Likewise, $P(B) = P(AB) + P(A^c B)$. Thus,
$$ \begin{array}{rcl} P(A \cup B) & = & P(A) + P(B) - P(AB) \\ & = & P(AB) + P(AB^c) + P(AB) + P(A^c B) - P(AB) \\ & = & P(AB) + P(AB^c) + P(A^c B) \\ & = & P(B) P(A|B) + P(AB^c) + P(A^c B) \\ & = & P(A) P(B) + P(AB^c) + P(A^c B) \end{array} $$
where we used $P(AB) = P(B) P(A|B) = P(A) P(B)$ since they are given to be independent.
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