The probability that it will rain on Saturday is 25% and the probability that it will rain on Sunday is also 25%. Is it true that the probability that it will rain on the weekends is 50%. Explain why or why not.
What i tried
I know that it is not true.
Let $P(A)$ represent the probability that it will rain on Saturday, while $P(B)$ represent the probability that it will rain on Sunday. Hence $P(A \cap B)$ will represent the probability that it would rain on both days (weekends) and since $$P(A \cap B)=P(A)+P(B)-P(A\cup B)$$
From the formula above we can see that the sum of $P(A)$ and $P(B)$ alone would not add up to $P(A \cap B)$ which means that we simply could not just add the two probabilities together to get $50$%. Since i find this rather counterintutive could anyone provide a simpler and more clearer explanation to this problem.Thanks
Best Answer
I will copy my comment here and continue:
All the possible outcomes: let former correspond to Saturday and latter correspond to Sunday: $P(rain+rain) = 0.25*0.25$, $P(norain + norain) = 0.75*0.75$, $P(rain+norain) = 0.25*0.75$, $P(norain+rain) = 0.75*0.25$;
The sum = 1 ! Thus probability of rain on both days $= (0.25)^2$, on one of the days $= 2*0.25*0.75 = 0.375$
You are interested in the event: $P(rain+rain)+P(no rain + rain)+P(rain+no rain) = 0.25^2+0.375=0.4375$;
Let's check the formula:
$P(A)=P(B)=0.25$, $P(A\cap B)= P(A)P(B) = 0.25^2$,
$P(A\cup B) = P(A)+P(B)-P(A\cap B) = 0.5 - 0.25^2 = 0.4375$;
Why is this true?
$P(A\cup B) = P(A \backslash B) + P(B \backslash A) + P(A\cap B) = P(rain+norain)+P(norain+rain)+P(rain+rain)$;
Inclusion-exclusion formula that you have used counts $P(A \cap B)$ twice, as
$P(A) = P(A \backslash B) + P(A \cap B)$
$P(B) = P(B \backslash A) + P(A \cap B)$
so it need to be substracted.