(a) For a 2 to be tossed, both dice must show a 1. There are four ways to roll a 1 on each die, so there are  ways to toss a 2.

(b) For a 3 to be tossed, one die must show a 1 and the other must show a 2. There are four ways to toss a 1 on the first die and two ways to toss a 2 on the second, so there are  ways to toss this; by symmetry, there are 8 ways to toss a 2 and a 1, since the dice are alike. This makes a total of 16 ways to toss a 3.

Since the dice are fair, the probabilities are equal.

The prime numbers that can be the sum of the dice are .

The following rolls result in one of these outcomes:

15 rolls out of a possible 36 result in prime sums.

Therefore (a) is greater.

(a) The following rolls result in one die being one higher than the other:

, and the reverse of each of these.

Therefore there are ten possible rolls.

(b) The following rolls result in one die being two higher than the other:

, and the reverse of each of these.

Therefore there are eight possible rolls.

This makes (a) greater.

There are  possible rolls of the two dice. The ones that result in a sum of 6 or less are:

This makes 15 rolls. Since one-fourth of 64 is 16, the probability of one of these rolls is less than .

Since the population increased from 1945 to 1950, then in 1945, the population must have been less than 3,049, and, subsequently, less than the 1940 population of 3,221.

A 3 or less can be rolled the following ways:

Both dice can be 1, for a roll of 2. This can happen  ways, since each die has three 3's.

The first die can be 1 and the second die can be 2, for a roll of 3. This can happen  ways, as each die has three 3's and two 2's.

Similarly, the first die can be 2 and the second die can be 3 - by symmetry, this can happen 6 ways.

This is  outcomes out of 36 rolls that are favorable to the event, leaving  unfavorable outcomes. This makes (a) greater.

Since the population increased over both the first and second half-decade, the population in 1965 had to be between the 1960 and 1970 levels - that is, it was between 3,415 and 3,866.

Therefore (b) is greater.

Four cards, including one spade (the ace of spades) changed decks.

Deck 1 now has forty-eight cards, including twelve spades. Therefore,  of its cards are spades.

Deck 2 now has fifty-six cards, including fourteen spades. Therefore,  of its cards are spades.

Therefore, the probability that a random card from either deck will be a spade is  for both decks.

There are four kings and two red threes, making six favorable outcomes and forty-six unfavorable ones. The odds against a favorable outcome are

or 23 to 3.

Out of fifty-two cards, the winning cards are the thirteen spades and the three other aces - sixteen winning cards and thirty-six losing cards. The odds against drawing a winning card are therefore

or 9 to 4.