To find the probability of an event, we use the following:

\(\displaystyle \text{probability of an event} = \frac{\text{number of ways the event can happen}}{\text{number of total possible outcomes}}\)

So, when looking at the event of drawing a 2 from a deck of cards, we get

\(\displaystyle \text{number of ways the event can happen} = 4\)

It is 4 because we can draw a 2 from a deck of cards 4 different ways:

• 2 of hearts
• 2 of diamonds
• 2 of clubs
• 2 of spades

Now, 

\(\displaystyle \text{number of total possible outcomes} = 52\)

because there are 52 cards in the entire deck.  Now,

\(\displaystyle \text{probability of drawing a 2} = \frac{4}{52}\)

We can simplify, we get

\(\displaystyle \frac{4}{52} = \frac{2}{26} = \frac{1}{13}\)

Therefore, the probability of drawing a 2 from a deck of cards is \(\displaystyle \frac{1}{13}\)

To find the probability of an event, we use the following:

\(\displaystyle \text{probability of an event} = \frac{\text{number of ways the event can happen}}{\text{number of total possible outcomes}}\)

So, when looking at the event of drawing a 9 of hearts from a deck of cards, we get

\(\displaystyle \text{number of ways the event can happen} = 1\)

It is 1 because we can draw a 9 of hearts from a deck of cards in only 1 way:

• 9 of hearts

Now, 

\(\displaystyle \text{number of total possible outcomes} = 52\)

because there are 52 cards in the entire deck.  Now,

\(\displaystyle \text{probability of drawing a 9 of hearts} = \frac{1}{52}\)

Therefore, the probability of drawing a 2 from a deck of cards is \(\displaystyle \frac{1}{52}\)

Given that \(\displaystyle 52\%\) of the class is girls, and there are \(\displaystyle 50\) students, we must first determine how many girls are in the class. 

\(\displaystyle 52\%=\frac{52}{100}\)

\(\displaystyle \frac{52}{100}=\frac{x}{50}\)

Because \(\displaystyle 100\) is double the value of \(\displaystyle 50\), \(\displaystyle 52\) is double the value of \(\displaystyle x\). \(\displaystyle x\) should be half of \(\displaystyle 52\).

\(\displaystyle \frac{52}{100}=\frac{2\times x}{2\times50}=\frac{2x}{100}\)

\(\displaystyle 52=2x\)

\(\displaystyle x=\frac{52}{2}=26\)

If \(\displaystyle 12\) of the \(\displaystyle 26\) girls wear skirts, then we can subtract to find the number of girls not wearing skirts.

\(\displaystyle 26-12=14\)

The pizza that Billy's mom baked was composed of these types of slices:

Half the slices have pepperoni only. Two of the slices have both pepperoni and onoins. One slice has onions only. One slice has only cheese. 

Therefore, 

\(\displaystyle 4\) slices had pepperoni because \(\displaystyle 8\div2=4\)

\(\displaystyle 2\) slices had pepperoni and onions 

\(\displaystyle 1\) slice had onion only

\(\displaystyle 1\) slice had only cheese. 

Billy could eat either the onion only slice, or the cheese only slice. This means that out of \(\displaystyle 8\) pieces of pizza, he could eat \(\displaystyle 2\). 

\(\displaystyle \frac{2}{8}=\frac{1}{4}\)

Given that \(\displaystyle 25\%\) is the percent equivalent of \(\displaystyle \frac{1}{4}\), that is the correct answer. 

The following rolls result in the difference of the dice being \(\displaystyle 1\):

\(\displaystyle (1,2), (2,3), (3,4), (4,5), (5,6)\) and the reverse of each of these.

These are \(\displaystyle 10\) rolls out of a possible \(\displaystyle 36\), so the probability of this event is \(\displaystyle \frac{10}{36} = \frac{5}{18}\)

The rolls that yield a product between \(\displaystyle 12\) and \(\displaystyle 20\) inclusive are:

\(\displaystyle \begin{matrix} \textrm{Roll} & \textrm{Prod} \\ (2,6) &12 \\ (3,4) &12 \\ (3,5) &15 \\ (3,6) & 18\\ (4,3) & 12\\ (4,4) &16 \\ (4,5) &20 \\ (5,3) &15 \\(5,4) & 20 \\ (6,2) & 12 \\(6,3) & 18 \end{matrix}\)

Therefore there are \(\displaystyle 11\) rolls that fit our criteria out of a total of \(\displaystyle 36\) possible rolls, so the probability of this outcome is \(\displaystyle \frac{11}{36}\).

The probability of an event based on observation (empirical probability) can be calculated by dividing the number of times the event occurs by the number of trials total. Since there were \(\displaystyle 172\) trials and \(\displaystyle 72\) heads, there were \(\displaystyle 100\) tails.

\(\displaystyle 172\textup { total}-72\textup { head}=100\textup { tails}\)

The probability of tails is therefore given by the number of tails divided by the total number of trials. Both terms are divisible by \(\displaystyle 4\), allowing us to simplify the fraction.

\(\displaystyle P =\frac{100}{172}=\frac{100\div 4}{172\div 4}= \frac{25}{43}\)

There are  \(\displaystyle 6 \times 6 = 36\) possible rolls of two fair six-sided dice, each of which will come up with equal probability. The set of rolls is shown below, with the ways to roll a sum of 8 or 9 indicated.

There are 9 ways out of 36 to roll a sum of 8 or 9, making the probability of this outcome \(\displaystyle \frac{9}{36} = \frac{1}{4}\).

To find the probability of an event, we will use the following formula:

\(\displaystyle \text{probability of an event} = \frac{\text{number of ways the event can happen}}{\text{total number of all possible outcomes}}\)

So, if we look at the event of drawing a 10 from a deck of cards, we can determine the following:

\(\displaystyle \text{number of ways the event can happen} = 4\)

because there are 4 ways we can draw a 10:

• 10 of hearts
• 10 of diamonds
• 10 of clubs
• 10 of spades

\(\displaystyle \text{total number of all possible outcomes} = 52\)

because there are 52 total cards to choose from.  

Knowing all of this, we can substitute into the formula.  We get

\(\displaystyle \text{probability of drawing a 10} = \frac{4}{52}\)

Now, we simplify.

\(\displaystyle \text{probability of drawing a 10} = \frac{2}{26}\)

\(\displaystyle \text{probability of drawing a 10} = \frac{1}{13}\)

Therefore, the probability of drawing a 10 from a deck of cards is \(\displaystyle \frac{1}{13}\).

To find the probability of an event, we will use the following formula:

\(\displaystyle \text{probability of an event} = \frac{\text{number of ways the event can happen}}{\text{total number of all possible outcomes}}\)

So, if we look at the event of rolling an even number on a dice, we can determine the following:

\(\displaystyle \text{number of ways the event can happen} = 3\)

because there are 3 ways we can roll an even number on a dice:

• 2
• 4
• 6

\(\displaystyle \text{total number of all possible outcomes} = 6\)

because there are 6 possible numbers we can roll:

• 1
• 2
• 3
• 4
• 5
• 6 

Knowing all of this, we can substitute into the formula.  We get

\(\displaystyle \text{probability of rolling an even number} = \frac{3}{6}\)

Now, we simplify.

\(\displaystyle \text{probability of rolling an even number} = \frac{1}{2}\)

Therefore, the probability of rolling an even number on a dice is \(\displaystyle \frac{1}{2}\).