Please can you help check!
A two-day conference is going to be held in a city. The leading newspaper in the city says there is a 10% chance of rain on the first day of the conference and a 40% chance of rain on the second day. If the newspapers are correct then:
i) Find the chance that it rains in the city during the conference is:

Unfortunately the newspaper doesn't give enough information. You can't multiply the probabilities because the events A and B are not known to be independent (and indeed are almost certainly dependent). The best you can say is that P(A∪B)≥P(B)=0.4 and P(A∪B)≤P(A)+P(B)=0.5. for i) final result It would be at least 0.4 or at least 40% if A⊂B (i.e. if it rains on the first day it must also rain on the second day), and ii) Final result is it would be at most 0.5 or at most 50% if A and B are disjoint (i.e. it can't rain on both days). recheck using compliment rule to verify below!

ii-)If the newspaper’s figures are correct, , the chance that it rains in the city during the conference is Compliment Rule P(A and Bat the most)=(1-P(i)=The probability it doesn't rain on Day 1 is 0.9, and the probability it doesn't rain on Day 2 is 0.6. So by independence, the probability both days are dry is (0.9)(0.6). :)

Thus the probability of rain on at least one of the days is 1−(0.9)(0.6). =At most .50 or at most 50% is final Result

Question

Please can you help check!
A two-day conference is going to be held in a city. The leading newspaper in the city says there is a 10% chance of rain on the first day of the conference and a 40% chance of rain on the second day. If the newspapers are correct then:
i) Find the chance that it rains in the city during the conference is:

Unfortunately the newspaper doesn't give enough information. You can't multiply the probabilities because the events A and B are not known to be independent (and indeed are almost certainly dependent). The best you can say is that P(A∪B)≥P(B)=0.4 and P(A∪B)≤P(A)+P(B)=0.5. for i) final result  It would be at least 0.4 or at least 40% if A⊂B (i.e. if it rains on the first day it must also rain on the second day), and ii) Final result is it would be at most 0.5 or at most 50% if A and B are disjoint (i.e. it can't rain on both days). recheck using compliment rule to verify below!

ii-)If the newspaper’s figures are correct, , the chance that it rains in the city during the conference is Compliment Rule P(A and Bat the most)=(1-P(i)=The probability it doesn't rain on Day 1 is 0.9, and the probability it doesn't rain on Day 2 is 0.6. So by independence, the probability both days are dry is (0.9)(0.6). :)

Thus the probability of rain on at least one of the days is 1−(0.9)(0.6). =At most .50 or at most 50% is final Result

anonymous · Answer

http://openstudy.com/users/drawar#/updates/516cb833e4b02ec89c5b36f0

anonymous · Answer

Confusing yeah haha

anonymous · Answer

Please drawar check I got the finalized in this finally figured out!:)

anonymous · Answer

Sorry but your wording makes it very hard for me to get what you mean. Anyway, maybe I'm wrong but the chance that it rains in the city during a (2-day) conference should be P(A and B) right? I mean the problem says 'during' so it must rain on both 2 days of the conference.

anonymous · Answer

You are not wrong the problem itself is incomplete and its unfortunate problem!
Thanks for showing the link I got 50% right after making error first part retroactively re-figured! Thanks Your help!