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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

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  • $\begingroup$ It means the same thing? Its just a rearrangement of the formula? $\endgroup$
    – ys wong
    Aug 20, 2015 at 10:33
  • $\begingroup$ If the probability of the event "raining on Sat. and Sun." is $=0$ then the probability of the event "raining on weekends" is 50%. $\endgroup$
    – Megadeth
    Aug 20, 2015 at 10:33
  • $\begingroup$ "the probability that it will rain on the weekends" is unclear because the plural doesn't seem to make sense. Did you mean "on the weekend" (i.e. on at least one of the weekend days) or "on the weekend days" (i.e. on both of them)? $\endgroup$
    – joriki
    Aug 20, 2015 at 10:41
  • $\begingroup$ I meant on both days. $\endgroup$
    – ys wong
    Aug 20, 2015 at 10:42
  • $\begingroup$ You may just write all the possible outcomes and thus check your answer: let former correspond to Saturday and latter correspond to Sunday: P(rain+rain) = 0.25*0.25, P(no rain + no rain) = 0.75*0.75, P(rain+no rain) = 0.25*0.75, P(no rain+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 $\endgroup$
    – Slowpoke
    Aug 20, 2015 at 10:49

3 Answers 3

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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.

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You write that "the sum of $P(A)$ and $P(B)$ alone would not add up to $P(A\cup B)$" – but note that the remaining term is subtracted, so it's not just that they don't add up alone, but a probability gets subtracted that's at least $\max(P(A),P(B))$, so that $P(A\cap B)\le\min(P(A),P(B))$.

A more intuitive way to think about this might be that if there's only a $25\%$ chance that it will rain on Saturday, then surely the chance that it will rain on Saturday and on Sunday can't be greater than that – the probability can't increase if you add an extra condition.

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You can simply calculate $1$ minus the probability of the complementary event:

$$1-(1-0.25)^2=0.4375$$

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