For a particular peculiar pair of dice, the probabilities of rolling , , , , , and , on each die are in the ratio . What is the probability of rolling a total of on the two dice?

Let be the probability of rolling a . The probabilities of rolling a , , , , and are , , , , and , respectively.

The sum of the probabilities of rolling each number must equal 1, so

So the probabilities of rolling a , , , , , and are respectively , and .

The possible combinations of two rolls that total are:

The probability of rolling a total of on the two dice is equal to the sum of the probabilities of rolling each combination.

Answer:

The wheel shown is spun twice, and the randomly determined numbers opposite the pointer are recorded. The first number is divided by and the second number is divided by The first remainder designates a column, and the second remainder designates a row on the checkerboard shown. What is the probability that the pair of numbers designates a shaded square?

Solution 1

When dividing each number on the wheel by the remainders are and Each column on the checkerboard is equally likely to be chosen.

When dividing each number on the wheel by the remainders are and

The probability that a shaded square in the st or rd row of the st or rd column is

The probability that a shaded square in the nd or th row of the nd column is

Add those two together

Solution 2

Alternatively, we may analyze this problem a little further.

First, we isolate the case where the rows are numbered 1 or 2. Notice that as listed before, the probability for picking a shaded square here is because the column/row probabilities are the same, with the same number of shaded and non-shaded squares

Next we isolate the rows numbered 3 or 4. Note that the probability of picking the rows is same, because of our list up above. The columns, of course, still have the same probability. Because the number of shaded and non-shaded squares are equal, we have Combining these we have a general probability of

Answer:

Three red beads, two white beads, and one blue bead are placed in line in random order. What is the probability that no two neighboring beads are the same color?

There are two ways to arrange the red beads.

1. R _ R _ R _
2. R _ _ R _ R

In the first, there are three ways to place a bead in the first free space, two for the second free space, and one for the third, so there are arrangements. In the second, a white bead must be placed in the third free space, so there are two possibilities for the third space, two for the second, and one for the first. That makes arrangements. There are arrangements in total. The two cases above can be reversed, so we double to arrangements. Also, in each case, there are three ways to place the first red bead, two for the second, and one for the third, so we multiply by to get arrangements. There are total arrangements so the answer is .

Answer:

Integers and , not necessarily distinct, are chosen independently and at random from 0 to 2007, inclusive. What is the probability that is even?

The only times when is even is when and are of the same parity. The chance of being odd is , so it has a probability of being even. Therefore, the probability that will be even is .

Answer: