Question 1191712
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The given probabilities, and the probabilities that can be deduced from the given probabilities, are the following:<br>
P(rain on day 1) = 0.3 (given)
P(no rain on day 1) = 1-0.3 = 0.7 (deduced)<br>
P(rain on day 2 given rain on day 1) = 0.5 (given)
P(no rain on day 2 given rain on day 1) = 1-0.5 = 0.5 (deduced)<br>
P(rain on day 2 given no rain on day 1) = 0.2 (given)
P(no rain on day 2 given no rain on day 1) = 1-0.2 = 0.8 (deduced)<br>
P(rain on day 3 given rain on days 1 and 2) = 0.9 (given)
P(no rain on day 3 given rain on days 1 and 2) = 1-0.9 = 0.1 (deduced)<br>
P(rain on day 3 given rain only on day 1) = 0.5 (given)
P(no rain on day 3 given rain only on day 1) = 1-0.5 = 0.5 (deduced)<br>
P(rain on day 3 given rain only on day 2) = 0.3 (given)
P(no rain on day 3 given rain only on day 2) = 1-0.3 = 0.7 (deduced)<br>
P(rain on day 3 given no rain on days 1 or 2) = 0.5 (given)
P(no rain on day 3 given no rain on days 1 or 2) = 1-0.5 = 0.5 (deduced)<br>
We could use those probabilities to answer the question directly,  However, it is useful to calculate the probabilities of all possible outcomes to verify that their sum is 1; that will give us confidence we have done the calculations correctly.<br>
Using "Y" (yes) or "N" (no) to indicate whether or not there is rain on each of three days, we have the following probabilities:<br>
P(YYY) = (.3)(.5)(.9) = .135
P(YYN) = (.3)(.5)(.1) = .015
P(YNY) = (.3)(.5)(.5) = .075
P(YNN) = (.3)(.5)(.5) = .075
P(NYY) = (.7)(.2)(.3) = .042
P(NYN) = (.7)(.2)(.7) = .098
P(NNY) = (.7)(.8)(.5) = .280
P(NNN) = (.7)(.8)(.5) = .280<br>
The sum of those probabilities is 1, so it is very likely we have done the calculations correctly.<br>
Then the answer to the problem is the sum of the probabilities that it rains on either 2 or all 3 of the 3 days:<br>
ANSWER: P(either 2 or 3 days of rain) = .135+.015+.075+.042 = .267<br>