[Math] Verifying properties of events $A$, $B$ given $P(A)$, $P(B)$, $P(A\cup B)$: where is the error

conditional probabilityprobabilityprobability theory

Given that:

  • $P(A) = 0.5$
  • $P(B) = 0.7$
  • $P(A \cap B) = 0.3$

I have to choose one option that is true… However they all seem to be false which means I am possibly making a mistake.. The only option that is almost true is B it seems. Any help is appreciated. Here are the options followed by my work so far:

A) $A$ and $B$ are independent

If $A$ and $B$ are independent $P(A \cap B) = (0.5)(0.7) = 0.35$; so this is not true

B) $A$ and $B$ are mutually exclusive

If $A$ and $B$ are mutually exclusive then $P(A \cap B) = 0$; so this is not true

C) $P(A \cup B) = 0.8$

$(0.5 + 0.7) – 0.3 = 0.9$; so this is not true

D) $P(A|B) = 0.6$

$$P(A \mid B) = \frac{P (A \cap B)}{P(B)} = \frac{0.3}{0.7} = 0.42$$ so this is not true

Best Answer

Your explanations are correct.

  • $A,B$ are independent if and only if $P(A \cap B) = P(A)P(B)$. You've shown this isn't true.
  • $A,B$ are mutually exclusive if and only if $P(A \cap B) = 0$. This obviously isn't true.
  • From inclusion-exclusion, $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ which you've calculated and shown is $0.9$, not $0.8$.
  • The conditional probability formula gives $$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$$ and you've calculated this to be (about) $0.42$, not $0.6$.

The only reasonable conclusion is that there is an error in the problem.

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue. Posting as Community Wiki since I don't have much to add.

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