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Problem:

The probability that it will rain today is 0.5.The probability that it will rain tomorrow is 0.6.The probability that it will rain either today or tomorrow is 0.7.What is the probability that it will rain today and tomorrow?

  1. Why we just can't multiply the probability of today's and tomorrows' raining, as they are two events and must be followed one after other to get the final event on raining on two days. So the answer is 0.3.

  2. But the solution say we got to use the equality Pr(A and B) = Pr(A) + Pr(B) - Pr(A or B). This give us the answer 0.4.

It look to me like Independence is to do something in here, but I might be wrong.

Please help me with, why set notation works but not the product rule ?

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    $\begingroup$ Yes, the issue is that independence fails. Which is something familiar about rain. In many places, the probability it is raining tomorrow given it is raining today is greater than the probability it is raining tomorrow given it is dry today. $\endgroup$ – André Nicolas Sep 29 '15 at 7:29
  • $\begingroup$ Given the scenario, we can identify if two events are independent either using Pr(A|B) = Pr(A), or Pr(B) = 0, or we can use Pr(A∩B) ≠ Pr(A).Pr(B). But I'm not able to identify if the given case is independent or not. How to do that, in the first place? $\endgroup$ – Deepak Sep 29 '15 at 7:38
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    $\begingroup$ There are physical situations where independence is a reasonable assumption (successive throws of two dice) and there are situations where it is not. In the rain case, we are given enough information to find $\Pr(A\cap B)$ and we can verify it is not the product, so we know we do not have independence. $\endgroup$ – André Nicolas Sep 29 '15 at 7:45
  • $\begingroup$ So it means we have to use intuition while solving the problem in Probability :-( $\endgroup$ – Deepak Sep 29 '15 at 7:51
  • $\begingroup$ A carefully worded theoretical problem will make the independence, if there is independence, explicit. And for "real world" applications, we need to make a mathematical model, but that is true of all applications of mathematics. Even if the fit is not perfect, the model may give valuable information. $\endgroup$ – André Nicolas Sep 29 '15 at 10:13
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The probability of a union of events is always the sum of probabilities of the events minus the probability of their intersection.   This is how probability measures are required to work (among other things), so: $\Pr(A\cup B)=\Pr(A)+\Pr(B)−\Pr(A\cap B)$.   This is equivalently: $$\Pr(A\cap B)=\Pr(A)+\Pr(B)−\Pr(A\cup B)$$

This is always the case.   It is only when the events are independent that it is also true that: $$\Pr(A\cap B)=\Pr(A)\cdot\Pr(B)$$

So we can only use the probability rule when we have certainty that the events are independent.   However, we can always use the addition rule when given three of the four probabilities.

In this case, we don't have any way to guarantee that the events of rainfall on subsequent days are independent.   Rather it would seem reasonable that there may be some dependence.   Since we are given three probabilities (of two events and their union), it is best to use the addition rule to find the fourth (their intersection); because that will always work.

And indeed on doing so, we find that these events are in fact not independent because, in this case, $\Pr(A)\cdot\Pr(B)\neq \Pr(A)+\Pr(B)-\Pr(A\cup B) = \Pr(A\cap B)$ .

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We don't have the equality $\Pr(A\cap B)=\Pr(A)\Pr(B)$ in general. However, if $A$ and $B$ are independent events, then $$ \Pr(A\cap B)=\Pr(A)\Pr(B). $$ So if we don't have independence, we cannot use the equality above.

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  • $\begingroup$ Thanks V.C. Can you help me with the problem in mention in the comment above, please. :-) $\endgroup$ – Deepak Sep 29 '15 at 7:39
  • $\begingroup$ @DeepakRajHR You have that $\Pr(A\cap B)=0.4$ and $\Pr(A)\Pr(B)=0.3$. So the events $A$ and $B$ are not independent since $\Pr(A\cap B)\ne \Pr(A)\Pr(B)$. $\endgroup$ – Cm7F7Bb Sep 29 '15 at 7:42
  • $\begingroup$ We don't know Pr(A∩B). To compute that we have to identify if the two events are independent or not. To identify if two event are independent we have to compute Pr(A∩B). Look to me as a cyclic problem. What is the point that am I missing? :-( $\endgroup$ – Deepak Sep 29 '15 at 7:48
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    $\begingroup$ @DeepakRajHR The measure of a union of events is always the sum of measures of the events minus the measure of their intersection. $\Pr(A\cup B) = \Pr(A)+\Pr(B)-\Pr(A\cap B)$. This is equivalently: $$\Pr(A\cap B) = \Pr(A)+\Pr(B)-\Pr(A\cup B)$$ This is always the case. Only when the events are independent is it also true that: $\Pr(A\cap B)=\Pr(A)\cdot\Pr(B)$ $\endgroup$ – Graham Kemp Sep 29 '15 at 8:13
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    $\begingroup$ @DeepakRajHR Your questions are fine. If we want to show that two events $A$ and $B$ are independent, we need to show that $\Pr(A\cap B)=\Pr(A)\Pr(B)$. So we must calculate $\Pr(A\cap B)$ and $\Pr(A)\Pr(B)$ and see if these terms are equal. It is easy to calculate $\Pr(A)\Pr(B)$. In order to calculate $\Pr(A\cap B)$, we use another equality $\Pr(A\cap B)=\Pr(A)+\Pr(B)-\Pr(A\cup B)$ that holds for any events $A$ and $B$. Then we compare the results. $\Pr(A\cap B)=0.4$ and $\Pr(A)\Pr(B)=0.3$. Hence, the events $A$ and $B$ are not independent. $\endgroup$ – Cm7F7Bb Sep 29 '15 at 8:27

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