Which of the following distribution assumptions will produce the lowest probability of exceeding an extreme...

An 'extreme value' will be a value that will lie in the tails. We need to determine the distribution that will have the least weight in the tails so that the probability of exceeding this tail value is minimum across the given choices.

The t-distribution, a distribution with kurtosis > 3 and a normal mixture distribution are all distributions with tails fatter than that for a normal distribution. A normal distribution will have the 'thinnest' tails among the choices and therefore the lowest probability of exceeding a given tail event value.

A note about the t-distribution: Leptokurtic distributions (those that have kurtosis>3, ie kurtosis greater than that for a normal distribution) generally appear to have higher peaks on their PDF graphs. The t-distribution is flatter, and actually appears lower than a normal distribution, which may make one think that it has a lower kurtosis and therefore should have thinner tails than a normal distribution. But that is not so, and the "visual" inspection test fails for inferring the kurtosis from just looking a the shape of the distribution. The kurtosis of a t-distribution is given by the formula {3 + 6/(d - 4)}, where d is the degrees of freedom and d > 4. Therefore the kurtosis of a t-distribution is always greater than 3 as "6/(d-4)" will always be a positive number being added to 3. Therefore there is no conflict between a t-distribution having fatter tails than a normal distribution as it has a higher kurtosis, even though it appears 'lower' on a graph when superimposed with a normal distribution.