Answer to Question #419 in Statistics and Probability for Tony
3 rainy days can be selected out of 12 rainy days in (C12)^3 ways. 5 clear days can be selected out of 18 clear days in (C18)^5 ways. The total number of ways to select 3 rainy and 5 clear days is: m = (C12)^3 *(C18)^5.
Thus, the probability of having exactly 3 rainy days in the selection is: P = m/N = ((C12)^3 *(C18)^5)/(C30)^8 ≈ 0.322.
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