Verified Answer :
R - it is rain (probablity: 0.86)
N - it is not rain (probablity: 1-0.86 = 0.14)
You've got these cases:
R R R R R R R R R N
R R R R R R R R N R
R R R R R R R N R R
. . .
N R R R R R R R R R - it is 10 cases. Probablity each case is
(0.86)^9 * 0.14 - it is approximately 0.036 (9R and 1N - you multiply everything)
It is 10 cases so you have to multiply by 10 and you get 0.36
There is also eleventh case:
R R R R R R R R R R
Probablity is (0.86)^10 approximately 0.22
Add these cases and you'll get your probablity: 0.36 + 0.22 = 0.58
0.5816 = 58.16% probability that it rains there in June at least 9 times in a decade.
For each June 9 in Florida, there are only two possible outcomes. Either it rains, or it does not. The probability of raining on June 9 of an year is independent of rain on June 9 on any other year, which means that the binomial probability distribution is used to solve this question.
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Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
\(P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}\)
In which \(C_{n,x}\) is the number of different combinations of x objects from a set of n elements, given by the following formula.
\(C_{n,x} = \frac{n!}{x!(n-x)!}\)
And p is the probability of X happening.
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- 0.86 probability of rain means that \(p = 0.86\)
- A decade means 10 years, so \(n = 10\)
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What is the probability that it rains there in June at least 9 times in a decade?
This is:
\(P(X \geq 9) = P(X = 9) + P(X = 10)\)
In which
\(P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}\)
\(P(X = 9) = C_{10,9}.(0.86)^{9}.(0.14)^{1} = 0.3603\)
\(P(X = 10) = C_{10,10}.(0.86)^{10}.(0.14)^{0} = 0.2213\)
Thus
\(P(X \geq 9) = P(X = 9) + P(X = 10) = 0.3603 + 0.2213 = 0.5816\)
0.5816 = 58.16% probability that it rains there in June at least 9 times in a decade.
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