It is known that in any calendar year, there can be at most one hurricane. The p

Question

It is known that in any calendar year, there can be at most one hurricane. The probability of a hurricane is 0.05 in a calendar year. The numbers of hurricanes in different calendar years are mutually independent. Using these assumptions, calculate (numerical simplifications are not necessary) (i) the exact probability that in a 20 year period there are fewer than 3 hurricanes. (ii) an approximation to part (i). It is known that in any calendar year, there can be at most one hurricane. The probability of a hurricane is 0.05 in a calendar year. The numbers of hurricanes in different calendar years are mutually independent. Using these assumptions, calculate (numerical simplifications are not necessary) (i) the exact probability that in a 20 year period there are fewer than 3 hurricanes. (ii) an approximation to part (i). It is known that in any calendar year, there can be at most one hurricane. The probability of a hurricane is 0.05 in a calendar year. The numbers of hurricanes in different calendar years are mutually independent. Using these assumptions, calculate (numerical simplifications are not necessary) (i) the exact probability that in a 20 year period there are fewer than 3 hurricanes. (ii) an approximation to part (i).

Explanation / Answer

Let X be the random variable that number of hurricans.

Here X ~ Binomial (n= 20, p = 0.05)

SO the pmf of X is,

P(X=x) = (20 C x) * 0.05^x * (1-0.05)^(20-x)   

Here we have to find P(X < 3).

P(X<3) = P(X=0) + P(X=1) + P(X=2)

P(X=0) = (20 C 0) * 0.05^0 * (1-0.05)^(20-0) = 0.3585

P(X=1) = (20 C 1) * 0.05^1 * (1-0.05)^(20-1) = 0.3774

P(X=2) = (20 C 2) * 0.05^2 * (1-0.05)^(20-2) = 0.1887

P(X <3) = 0.3585 + 0.3774 + 0.1887 = 0.9245

The exact probability that in a 20 year period there are fewer than 3 hurricanes is 0.9245

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