Assume that meteoroligists can predict the probability of rain accurately. If they predict the next seven days all have a 20% chance of rain, what is the probability of it raining at least one of those days? Two days? What's the general formula?

  • $\begingroup$ Also, what if the probabilities are different for each day, such as { .1, 0, .25, .7, 0, .1, 0 } ? $\endgroup$ – Eric Jan 20 '15 at 21:31
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    $\begingroup$ The probability of at least one day having rain is simple if you first compute the probability of getting no rain on every day. $\endgroup$ – turkeyhundt Jan 20 '15 at 21:32
  • $\begingroup$ so.. chance of no rain on one day is 1- 20% = 80%. chance of no rain ever day is 80% ^ 7 = .209. So the answer is 1 - .209 = .79 = 79% chance of rain during the week. $\endgroup$ – Eric Jan 20 '15 at 21:34
  • $\begingroup$ Yes, exactly right on the at least one day part. $\endgroup$ – turkeyhundt Jan 20 '15 at 21:36
  • $\begingroup$ Hint: The number of successful outcomes of series of trials each with an identical chance of success has a Binomial Distribution. $\endgroup$ – Graham Kemp Jan 20 '15 at 21:37

In this example, the probability of it raining $n$ days is the number of ways you can pick those $n$ days out of $7$, multiplied by $.20$ for each of those $7$ days, multiplied by $.8$ for each of the days it doesn't rain.

$$P(\text{rains on n days}) = {7\choose n} (.20)^n (.80)^{7-n}$$

  • $\begingroup$ so, n=1. 7choose1 = 7. ---> 7*(.2^1)*(.8^6) = .367 or 37% $\endgroup$ – Eric Jan 20 '15 at 21:45
  • $\begingroup$ Correct. Here is some info to check your work. You can expand the table of values. wolframalpha.com/input/… $\endgroup$ – turkeyhundt Jan 20 '15 at 21:49
  • $\begingroup$ You have assumed independence of rainy days, which is not stated in the question. $\endgroup$ – Henry Jan 21 '15 at 0:14

This is a binomial distribution. For that, we have the formula $p(X=x)=\binom{n}{x}(p^x)(q^{n-x})$ For our scenario, we have $p=0.2$, and $q=1-p=1-0.2=0.98$

If we want to find the probability of rain for two days out of seven, we set

We then get $p(X=2)= \binom{7}{2}(0.2^2)(0.98^5) =21(0.04)(0.9)=0.76$


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