An Introduction to Probability Theory and Its Applications, Vol. 1, 3rd Edition

A: If we did not know any information about the resulting number, then the probability of all numbers being even is simply \(\frac{1}{16}\) as there are \(2^{4} = 16\) ways for choosing 4 numbers as either odd or even. However we do know that their sum is even.

In addition to the above the following also holds.

- The sum of two even numbers is always even
- If \(x_1 = 2n + 2\) is an even number and \(x_2 = 2m + 2\) is another even number, their sum is \(2(n+m) + 4\) which is even.
- The sum of two odd numbers is also always even
- If \(x_1 = 2n + 1\) is an odd number and \(x_2 = 2m + 1\) is another odd number, their sum is \(2(n+m) + 2\) which is even.
- The sum of an odd and an even number is always odd
- If \(x_1 = 2n + 1\) is an odd number and \(x_2 = 2m + 2\) is an even number their sum is \(2(n+m) + 3\) which is always odd.

Since the order of addition does not matter, the above results yields the table below

$$ P(4E|E) = \frac{P(E|4E)\times P(4E)}{P(E|4E)\times P(4E) + P(E|2E)\times P(2E) + P(E|0E)\times P(0E)}$$

Notice we have eliminated cases where \(P(E|3E)\) & \(P(E|1E)\) get set to 0. So the above works out to

$$ P(4E|E) = \frac{1\times \frac{1}{16}}{1\times \frac{1}{16} + 1\times \frac{6}{16} + 1\times \frac{1}{16}} = \frac{1}{8}$$

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