0
$\begingroup$

Events $A$ and $B$ are mutually exclusive. Suppose event $A$ occurs with probability $0.74$ and event $B$ occurs with probability $0.22$.

  1. Compute the probability that $A$ occurs but $B$ does not occur.
  2. Compute the probability that either $A$ occurs without $B$ occurring or $A$ and $B$ both occur.

Answers:

  1. $\Pr(A \cap \neg B) = \Pr(A) + \Pr(\neg B)- \Pr(A \cap B) = 0.74 + 0.78 -(0.74 \cdot 0.78)=0.9428$
  2. $\Pr([B \cap A] \cup [A \cap B])= \Pr(\neg B \cap A) + \Pr(A \cap B) = 0.78 \cdot 0.74 + 0.74 \cdot 0.22=0.74$

Is it correct?

$\endgroup$
8
  • $\begingroup$ What is your definition of mutually exclusive events? $\endgroup$ Apr 6, 2018 at 9:21
  • $\begingroup$ That means very different from each other? $\endgroup$
    – johnc
    Apr 6, 2018 at 9:22
  • $\begingroup$ That's not a mathematical definition. One definition is $P(A \cap B) = 0$. $\endgroup$ Apr 6, 2018 at 9:23
  • $\begingroup$ ohhhh!!!!!! so for no.1 it should be 0.74+0.78? $\endgroup$
    – johnc
    Apr 6, 2018 at 9:25
  • $\begingroup$ Can a probability be $>1$? $\endgroup$ Apr 6, 2018 at 9:25

1 Answer 1

1
$\begingroup$

Supposing, as is usual (e.g. here) that mutually exclusive means that $P(A \cap B) = 0$ we have that $P(A \land \lnot B) = P(A \setminus B) = P(A) - P(A \cap B) = P(A) = 0.74$.

The final one is then just $P(A)$ too. $A$ cannot occur at the same time as $B$.

$\endgroup$
3
  • $\begingroup$ Thank you!!!!!!!!!!!!!!!!!!!!! $\endgroup$
    – johnc
    Apr 6, 2018 at 9:32
  • 1
    $\begingroup$ Mutually exclusive implies that $P(A\cap B)=0$ which is enough here. The actual meaning of it is that $A\cap B=\varnothing$. $\endgroup$
    – drhab
    Apr 6, 2018 at 9:58
  • $\begingroup$ @drhab Quite. And so both parts are just roundabout ways of asking for $P(A)$. A bit confusing for the OP maybe. $\endgroup$ Apr 6, 2018 at 10:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .