Let us find the possibility of no rain on all of the three days.
It would be \(\frac{2}{3} * \frac{2}{3} * \frac{2}{3} = \frac{8}{27}\)

\(1- \frac{8}{27} = \frac{19}{27}\)

This gives possibility of rain at least on one of those 3 days.
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Explanation

In essence, the question is asking, “What is the probability that one or more days are rainy days?” since any single rainy day would mean the city experiences rain. In this case, employ the 1 – x shortcut, where the probability of rain on one or more days is equal to 1 minus the probability of no rain on any day.

Since the probability of rain is \(\frac{1}{3}\) on any given day, the probability of no rain on any given day is \(1 - \frac{1}{3}=2/3\) .

Therefore, the probability of no rain on three consecutive days is \(\frac{2}{3}\frac{2}{3}\frac{2}{3}=\frac{8}{27}\).

Finally, subtract from 1 to find the probability that it rains on one or more days: P(1 or more days) = 1 – P(no rain) = \(1 - \frac{8}{27}=\frac{19}{27}\).
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Why cant we simply multiply 1/3*1/3*1/3 which is probability of rain for any given day?

Hey shubhankarsapa

Read the stem very carefully: In any given 3-day period, what is the probability that the city experiences rain?

So when you multiply for rain for each day, you will get the probability that it will rain all 3 days. But what if it rains the third day and not the first two? It will also be considered under the stem. So we need to find the probability of rain on any day not all days.

Hence we calculate by 1 - no rain = raining at least once.


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