Probability of at least one of two events occuring?

Question

The two events A and B are such that $P(A) = 0.6$ and $P(B) = 0.2$ and $P(A|B) = 0.1$

Find the probability that at least one of the events occur?

This is my working so far.

My answer is 0.68 and the correct answer is 0.78. How do I get this answer?

Answer 1

Your diagram appears to make no use of the fact that $\mathsf P(A\mid B)=0.1$, and instead incorrectly treats the events as being independent.   Instead, use the dependency which you have been given.   In tree form, if you must, that is:

$\qquad\qquad \overset{0.1}\nearrow\raise{1ex}{A\cap B; 0.2\times0.1= 0.02} \\ \quad \overset{0.2}\nearrow\raise{2ex}B~~\underset{0.9}\searrow\lower{1ex}{A^\complement\cap B; 0.2\times0.9=0.18} \\ \cdot \\ \quad\underset{0.8}\searrow\lower{2ex}{B^\complement}~\overset{?}\nearrow\raise{1ex}{A\cap B^\complement;} \\ \qquad\qquad\underset{?}\searrow\lower{1ex}{A^\complement\cap B^\complement;} $

To obtain the missing information recall that the total probability of $A$ is $0.6$.

So for example, $\mathsf P(A\cap B^\complement)~{=\mathsf P(A)-\mathsf P(A\cap B)\\=0.6-0.02\\=.58}$

Answer 2

Use various formulae:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$ and $P(A \cap B) = P(A | B)P(B)$

Note that "at least one of $A$ or $B$ occur" means $A \cup B$.

Answer 3

You already got that $P(A\cap B)=0.02$. Now you just need to use the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ (https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle) then you get 0.78.