5
$\begingroup$

If $P(A) = 1/3$ and $P(B^{\complement}) = 1/4$, then, can $A$ and $B$ be mutually exclusive?

I already know that for $A$ and $B$ to be mutually exclusive, $A \cap B = \varnothing$ and $P( A \cup B ) = P (A) + P(B)$.

I just can't proceed further than this to prove if $A$ and $B$ are mutually exclusive though as I feel like this is not enough information to determine that.

please help

Thank You

$\endgroup$
  • $\begingroup$ If $B'$ a notation for the complement of $B$ here? $\endgroup$ – drhab Sep 19 '17 at 8:17
  • $\begingroup$ Yes it is a complement of B $\endgroup$ – ISuckAtMathPleaseHELPME Sep 19 '17 at 8:20
  • $\begingroup$ @Vihv Would you want to accept my answer? $\endgroup$ – Toby Mak Sep 19 '17 at 9:12
13
$\begingroup$

Since $P(B^{\complement})$ (which is $B$ complement) is equal to $\frac{1}{4}$, $P(B) = \frac{3}{4}$.

Now $P(A)+P(B)-P(A \cap B)$ must sum to less than $1$. However, $P(A)+P(B)$ is $\frac{1}{3} + \frac{3}{4} = \frac{13}{12}$. The minimum probability of $P(A \cap B)$ is therefore $\frac{1}{12}$. What can you conclude, given your first condition? (which is for $A,B$ to be mutually exclusive, $A \cap B = \varnothing$.)

$\endgroup$
4
$\begingroup$

Hint:

If they are mutually exclusive then $P(A)+P(B)=P(A\cup B)\leq1$.

Check whether that necessary condition is satisfied and draw conclusions.

$\endgroup$
0
$\begingroup$

$P(B^{\complement}) = 1/4$ gives that $P(B) = 1 - 1/4 = 3/4$

$P(A) = 1/3$

This gives

$P(A) + P(B) = 13/12$

You stated that $P(A \cup B) = P(A) + P(B)$ is a necessary condition for A and B being mutually exclusive.

If this were the case, the event $(A \cup B)$ would have probability 13/12 which is larger than 1.

But no event can have a probability larger than 1. Therefore, it cannot be true that $P(A \cup B) = P(A) + P(B)$.

Since this was a necessary condition for them being mutually exclusive, that cannot be true either.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.