Classify as true or false and explain why
(a) If $A$ and $B$ are two independent events satisfying $0 < P(A)$, $P(B) < 1$, then it is always true that $A$ and $B^c$ are independent events.
(b) $A$ and $(A ∩ B)$ are two mutually exclusive events if and only if $A$ and $B$ are mutually exclusive events.
(c) If $A$, $B$ and $C$ are events satisfying $0 < P(A)$, $P(B)$, $P(C) < 1$, then it is always true that
$P(A ∩ B ∩ C) + P(A ∩ B ∩ C) + P(A ∩ B ∩ C) + P(A ∩ B ∩ C) = 1 $
(d) If $A$ and $B$ are events satisfying $0 < P(A)$, $P(B) < 1$ and $P(A ∩ B) > P(A ∩ B)$ then it is always true that
$P(A\mid B) \ge P(A^c\mid B)$

For part (a), I believe that this is true, because both events have a chance of occurring between $1$% and $99$%. By making it these bounds, this makes it so the event of both of them happening is not $0$ or $1$.

For part (b), I believe this is false. If $A$ and $B$ are mutaully exclusive events, this makes the probability of both of them happening at the same time $0$. This would make $(A ∩ B)$ invalid, since the event can never occur given the circumstances.

For part (d), I was thinking about using algebra to try and formulate a direct proof, but I'm not exactly sure what I should do.

Any help and clarification is appreciated.

  • $\begingroup$ Check what you've written for errors. (a) doesn't make sense. The premise that $A$ and $B$ are independent events is the same as the conclusion that $A$ and $B$ are independent events. $\endgroup$
    – JMoravitz
    Aug 10 '20 at 22:31
  • $\begingroup$ @JMoravitz sorry about the error. Thank you for the heads up. $\endgroup$
    – dumon__
    Aug 10 '20 at 22:33
  • $\begingroup$ You write "for part a,... chance of occurring between 1%-99%" That is not what $0<P(A)<1$ means... Something which occurs with probability $0.000000001\%$ is still something which occurs with greater than zero probability. Do not think that all probabilities are multiples of $1\%$. As for your attempt and conclusion for (a), you seem to have completely missed the point of what "independent events" are. $\endgroup$
    – JMoravitz
    Aug 10 '20 at 22:33

There's a lot to unpack here and misconceptions to correct.

Let's start with some definitions.

When talking about probability, we have a probability space containing subsets of outcomes from our sample space and we call these subsets "Events." We can assign a real number as a value to each of these events, including the empty event. In order for such a probability space to be valid, certain nice properties must hold, such as the probability of the empty event (aka the "impossible event") must be zero, the probability of the sample space as a whole must be $1$, and the probability of the union of two disjoint events (aka mutually exclusive events) must be equal to the sum of their respective probabilities. From these, several more properties can be proven and follow such as how $\Pr(A\cup B) = \Pr(A)+\Pr(B)-\Pr(A\cap B)$ and $\Pr(A) = 1- \Pr(A^c)$ etc... More on those can be found elsewhere.

We can talk about some important relationships between events.

We define $A$ and $B$ are "mutually exclusive" events to mean that $A\cap B = \emptyset$. Recall the axiom mentioned above that this implies that if $A$ and $B$ are mutually exclusive events we will have $\Pr(A\cup B) = \Pr(A)+\Pr(B)$. In general, if $A$ and $B$ are not mutually exclusive events we don't necessarily have $\Pr(A\cup B) = \Pr(A)+\Pr(B)$, it could be less.

We define A and B are "independent" events to mean that $\Pr(A\cap B) = \Pr(A)\times \Pr(B)$. This can be shown to be equivalent also to $\Pr(A\mid B) = \Pr(A)$ and that $\Pr(B\mid A) = \Pr(B)$. Note that while $\Pr(A\cap B) = \Pr(A)\times \Pr(B)$ may be true for independent events, this is not true for any other events.

(a) Suppose that $A$ and $B$ are independent events. (The condition that $0<\Pr(A)<1$ and $0<Pr(B)<1$ is not necessary here, so I will ignore it, but to emphasize what I mentioned in comments, probabilities can be any real values between zero and one, and that includes numbers smaller than $1\%$ and irrational values etc...)

This implies that $\Pr(A\cap B) = \Pr(A)\times \Pr(B)$.

Now... we are asked to verify whether $A$ and $B^c$ are also independent events. Note that $\Pr(A) = \Pr((A\cap B)\cup (A\cap B^c)) = \Pr(A\cap B)+\Pr(A\cap B^c)$ since $(A\cap B)$ and $(A\cap B^c)$ are both mutually exclusive events who union to the event $A$ and so the sum of their probabilities is equal to the probability of their union, again by the axiom mentioned above.

So, by subtracting $\Pr(A\cap B)$ from both sides we have $\Pr(A\cap B^c) = \Pr(A)-\Pr(A\cap B) = \Pr(A) - \Pr(A)\times \Pr(B) = \Pr(A)(1-\Pr(B)) = \Pr(A)\Pr(B^c)$, using all of the aforementioned properties. As such, we do indeed have the probability of the intersection of $A$ and $B^c$ is in fact the product of their respective probabilities and so they are indeed independent.

Note... this has nothing to do with "the event of both of them happening is not 0 or 1." You can have independent events whose probability of both happening is zero (in which case they were both impossible events) or whose probability of both happening is one (in which case they were both sure events). What was important here was that the probability of both happening being equal to the product of their respective probabilities.

(b) Suppose $A$ and $A\cap B$ are mutually exclusive events. That means by definition that $A\cap (A\cap B) = \emptyset$. Well, from elementary set theory, we know how intersections work (if you don't go back and review those chapters) and that intersection is associative. So, rearranging the parentheses by associativity we have $(A\cap A)\cap B = \emptyset$, and then by absorption (that $A\cap A = A$) we have that $A\cap B = \emptyset$. This is however precisely the definition of $A$ and $B$ being mutually exclusive so $A$ and $A\cap B$ being mutually exclusive does indeed imply that $A$ and $B$ will be mutually exclusive as well.

In the reverse direction, we can effectively reverse all of those arguments to get $\emptyset = A\cap B = (A\cap A)\cap B = A\cap (A\cap B)$ in order to show that $A$ and $B$ being mutually exclusive does imply that $A$ and $A\cap B$ are also mutually exclusive.

"This would make $A\cap B$ invalid..." No, that doesn't make it "invalid"... it just makes $A\cap B$ an impossible event. It is still a perfectly valid event however, just one who occurs with zero probability.

(c) clearly has a typo, you wrote $A\cap B\cap C$ four times. As written it is obviously false. Just pick any example where $\Pr(A\cap B\cap C)\neq \frac{1}{4}$, for example when $A=B=C$ and $\Pr(A)=\frac{1}{2}$.

(d) This also clearly has a typo. $\Pr(A\cap B)>\Pr(A\cap B)$ is never true. A number cannot be strictly greater than itself. Presumably this should be $\Pr(A\cap B) > \Pr(A^c\cap B)$ implying that $\Pr(A\mid B) > \Pr(A^c\mid B)$

Well... this should be clear when you remember the definition of conditional probability and that $\Pr(A\mid B) = \dfrac{\Pr(A\cap B)}{\Pr(B)}$

You have then $\Pr(A\mid B) = \dfrac{\Pr(A\cap B)}{\Pr(B)} > \dfrac{\Pr(A^c\cap B)}{\Pr(B)} = \Pr(A^c\mid B)$

  • $\begingroup$ When you code $\Pr(A)$ as \Pr(A) rather than Pr(A) the effect is not only to prevent italicization but also to get proper spacing, thus: $$ \begin{align} 5\Pr(A) \\ (3+5)\Pr(A) \end{align} $$ Contrast this with $$ \begin{align} 5 Pr(A) \\ 5\text{Pr}(A) \end{align} $$ Here the second line is coded as 5\text{Pr}(A) and you don't see proper spacing. $\endgroup$ Aug 10 '20 at 23:30

I explain to you why (a) is true based on a proof argument , I think you can (and should) do the same for the others.

Because $A \cap B^C$ and $A \cap B$ are mutually exclusive events the first step is justified :

$$P((A \cap B^C) \cup (A \cap B)) = P(A \cap B^C)+P(A \cap B) = P(A \cap B^C)+P(A)P(B)$$

the last step is correct because $A$ and $B$ are indipendent.

But you know also that : $$P(A \cap B^C)+P(A)P(B) = P(A) $$

since $(A \cap B^C) \cup (A \cap B) = A$.

Then, rearranging you obtain :

$$P(A \cap B^C)= P(A)(1-P(B))= P(A)P(B^C)$$

which is the definition of indipendence.


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