0
$\begingroup$

The probability of rain in Greg's area on Tuesday is $0.3$. The probability that Greg's teacher will give him a pop quiz in Tuesday is $0.2$. The events occur independently of each other. What is the probability of neither events occur?

My approach:

Probability of rain or quiz or both = $0.3+0.2= 0.5$

So, probability of neither= $1-0.5$ = $0.5$

Question: Actual probability of this problem is $0.7\cdot0.8$ = $0.56$. But, I don't understand what is the mistake in above approach?

$\endgroup$
  • 2
    $\begingroup$ They are independent not mutually exclusive. For mutually exclusive events you have $Pr(A\text{or}B)=Pr(A)+Pr(B)$. That is not true of independent events. $\endgroup$ – JMoravitz Aug 23 '16 at 6:11
  • 1
    $\begingroup$ Principle of inclusion-exclusion 2set case: $Pr(A\cup B) = Pr(A)+Pr(B)-Pr(A\cap B)$ $\endgroup$ – JMoravitz Aug 23 '16 at 6:18
3
$\begingroup$

Note the following two things:

Principle of inclusion-exclusion (2-set case) $$Pr(A\cup B) = Pr(A)+Pr(B)-Pr(A\cap B)$$

Definition of independent events

The following are equivalent statements

  • $A$ and $B$ are independent events

  • $Pr(A\mid B) = Pr(A)$

  • $Pr(B\mid A) = Pr(B)$

  • $Pr(A\cap B)=Pr(A)\cdot Pr(B)$

For your problem: you know $Pr(A)=0.3, Pr(B)=0.2$ and they are independent, so

$$Pr(A\cup B) = Pr(A)+Pr(B)-Pr(A\cap B) = 0.3+0.2-0.3\cdot 0.2 = 0.5-0.06 = 0.44$$

So the probability of at least one of the events occurring is $0.44$. The probability of no events occurring is then one minus that, i.e. $1-0.44=0.56$.


Alternatively, if $A$ and $B$ are independent, then $A^c$ and $B^c$ are also independent. We have then $Pr(A^c\cap B^c)=Pr(A^c)Pr(B^c)=(1-0.3)(1-0.2)=0.7\cdot 0.8 = 0.56$

$\endgroup$
0
$\begingroup$

Your approach double-counts some possibilities. The $0.3$ covers all possibility of rain - so, it includes both "rain and quiz" and "rain and no quiz". The $0.2$ covers all possibility of a quiz, so it includes both "rain and quiz" and "quiz and no rain". So $0.3 + 0.2$ covers "rain and quiz", "rain and no quiz", "rain and quiz" again, and "quiz and no rain". To cut out one of the countings of "rain and quiz", we need to subtract the probability of that event, which is $0.3 \cdot 0.2 = 0.06$. So the probability of "rain or quiz or both" is $0.2 + 0.3 - 0.06 = 0.44$. The probability of neither is then $1 - 0.44 = 0.56$.

$\endgroup$

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.