# GATE1997-1.1

3.1k views

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. $0.3$
2. $0.25$
3. $0.35$
4. $0.4$

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Inclusion-Exclusion Principle (2 Sets)

Answer: $D$

$P$(it will rain today either today or tomorrow) = $P$(it will rain today) $+ P$(it will rain tomorrow) $- P$(it will rain today and tomorrow)

So, $0.7 = 0.5 + 0.6 - P$(it will rain today and tomorrow)

$\Rightarrow P$ (it will rain today and tomorrow) =$0.4$

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Here Events are independent events .If A is event :"Rains Today" and B is event :"Rains Tomorrow" then Raining tomorrow doesn't depend on raining today.So A and B are independent events and for independent events P(A and B ) is given by P(A)* P(B).

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@vivek  ... I agree with you. But then The probability that it will rain either today or tomorrow should be - It rains today but not tomorrow + It does not rain today but rains tomorrow = (0.5) * (0.4) + (0.5) * (0.6) =( 0.5 ) , but in the question it is given 0.7

Can any one please make it clear ... ?

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@Vijay - yes, and that shows that the events are not independent. If it rains today, the probability of it raining tomorrow increases- may be showing that it is Monsoon season :)
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Sir... What if The probability that it will rain either today or tomorrow is (0.7) not given ??

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In that case we cant say the events are dependent or otherwise and usually have to consider real world scenario. I would say consider "events as independent" is its surely the case - as throwing dice 2 times, or if we need to use the independent event formula- no other data given.
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@vivek  ... I agree with you. But then The probability that it will rain either today or tomorrow should be - It rains today but not tomorrow + It does not rain today but rains tomorrow = (0.5) * (0.4) + (0.5) * (0.6) =( 0.5 ) , but in the question it is given 0.7

If we consider Events are independent then

P(A U B) = P(A) + P(B) - P(A)*P(B)

P(it will rain today either today or tomorrow) = P(it will rain today) +P(it will rain tomorrow) −P(it will rain today)*P(it will rain tomorrow) ==> 0.8

how you got 0.5 ..where i am wrong above ??

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@jatin khachane 1

You are correct, it will be 0.8 only.

@vijaycs has calculated it wrongly for $p(A \oplus B) = p(A \cap \overline{B}) + p(B \cap \overline{A} ) - 2p(A \cap B)$

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Sometimes we need to forget about real world scenario and just focus on the details that are given in the question. In real world, tomorrow's rain may or may not depend on today's rain, it doesn't matter. In the question, it does depend and that is all we have to think about.
0.7 = 0.5 + 0.6 - p(today and tomorrow) = 0.40

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