>Nadya,
>
>The answer to your question is
>
>p = (2^n - 1)/(2^n)
>
>because any one even number (i assume you're using integers) will make the product even.
>
>More simply, it's
>
>p = 1 - (0.5)^n
>
>Because the probability of picking all odd numbers is (0.5)^n, and the probability of getting an even product is 1 - (probability of picking all odd numbers)
>
>What John hasn't taken into account in his formula is that it only works if each of the outcomes is equally likely (like rolling dice or whatever [except in Vegas :)]). If the outcomes have a different likelyhood, then you need to weight each outcome in proportion.
>
>Hence, on some days, the chance of precipitation is > 0.5 and on some days it is less. There are sophisticated models used to estimate the likelyhood of weather (none of which work very well in my experience <g>) and so the forecasters think they can do better than just p=0.5 as John would have.
>
>Cheers,
>
>Andrew

Well, they can have all the sophisticated models they want. It will either rain or not. The conditions may be such that they "believe" it's going to rain like when the clouds are laden with water, hanging low and the low pressure area moves in, but sometimes it doesn't rain in spite of all the "conditions." Therefore, in the real world - it will either rain or not - 1/2.