The term without replacement indicates that there will be one less marble per trial executed.

Find the total number of marbles by adding all the existing marbles in the bag.

First trial with 20 marbles:

Second trial with 19 marbles:

Third trial with 18 marbles:

Fourth trial with 17 marbles:  

Multiply all probabilities.

The answer is:  

The lowest sum Eric can roll with a six-sided die in three rolls is , because he can roll a  three times in a row. Similarly, the highest sum Eric can roll with a six-sided die in three rolls is , because he can roll a  three times in a row. Hence, in order to calculate the probability that Eric rolls a prime number as a sum in three rolls of the dice, we need only consider the prime numbers between  and . These prime numbers are:

.

Because each roll of the dice is independent (i.e., the probability that Eric rolls a certain number on his dice is not affected by previous rolls of the dice), we can calculate the probability that he rolls a prime sum in three rolls of the dice by calculating each probability that he rolls a particular prime number between  and  and adding them together.

Let  denote the probability that the sum of the three numbers Eric rolls on his dice is a prime number. Then

We can calculate these probabilities separately by considering each result on a case-by-case basis. 

The number of possible sequences of three rolls of a six-sided die is , and Eric can roll a sum of  in only one way (namely, by rolling a  three throws in a row). Hence,

.

Similarly, Eric can roll a sum of  in the following ways:

where  denotes a sequence of three rolls of a six-sided dice,  is the number rolled on the first throw,  is the number rolled on the second throw, and  is the number rolled on the third throw. Hence,

.

Calculating the remaining probabilities in an identical manner yields

,   ,    ,   .

Hence, the probability that Eric rolls a sum equal to a prime number in three rolls of the dice is

                              .

This problem involves overlapping events, which are not mutually exclusive.

Write the formula for union of probabilities.

There are 13 spades in the 52 cards.

The probability of drawing a spade is:  

There are four Jacks, four Queens, and four Kings.  The probability of drawing a face card is:  

We will need to note that this involves overlapping of three cards, involving the Jack, Queen, and King of spades. This will need to be subtracted.

Apply the formula.

Reduce this fraction.

The answer is:  

Note that these are all mutually exclusive events.  The probability of using the OR rule is:

Let  be the probability of drawing a 10, and  be the probability of drawing a jack.  

There are 54 cards including the jokers, and there are four 10s and four jacks.

Therefore,

Add the two probabilities.

The answer is:  

After Billy picks out the red marble from the first trial, without replacement, there will be four marbles remaining in the bag.

Billy picks out the red marble for his second trial, which means that there are three marbles left:

He replaces the red marble from the second trial, which means there is now four marbles in the bag.

The probability of Billy drawing the black marble is:  

This means that the probability of Billy NOT drawing the black marble is:

The answer is:  

Multiply these two probabilities together:

Therefore the probability of flipping a tail on a coin and rolling a 3 is  .

To calculate the probability of two independent events occuring, you multiply the probabilities of each event occuring independently.

In this case, the probability of rolling any number is , so the probability of rolling this particular combination is:

 .

A die has 6 faces with a number on each face from one to six.

Calculate the total number of combinations of the 2 dice.

Write out the combination sets of the condition that satisfy the scenario.

There are only three sets that will achieve the sum of 2 or 3, but no greater than 3.  Divide the total number of possibilities by the total number of combinations.

The probability of rolling a sum that is no greater than 3 is:  

Our perect squares are 

.

There are ten of them. Since there are  slips in the bag, our fraction becomes .

If we cut a zero from both top and bottom, which is the same as dividing by ten, our final answer is 

.

The probability of getting a question wrong is .

Since we want to get it wrong twice, we need to multiply as these are events that must occur.

So we do 

.