12 votes

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?

- $0.3$
- $0.25$
- $0.35$
- $0.4$

20 votes

Best answer

2

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).

Answer would be 0.5*0.6=0.3.

So answer is option A

Answer would be 0.5*0.6=0.3.

So answer is option A

0

@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 ... ?

9

@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 :)

4

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.

1

@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 ??

0

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) $