A standard deck contains an equal number of hearts, diamonds, clubs, and spades. So the probability of drawing a heart is [latex]\frac{1}{4}[/latex]. There are four 7s in a standard deck, and there are a total of 52 cards. So the probability of drawing a 7 is [latex]\frac{1}{13}[/latex].

The only card in the deck that is both a heart and a 7 is the 7 of hearts, so the probability of drawing both a heart and a 7 is [latex]\frac{1}{52}[/latex].

Substitute [latex]P\left(H\right)=\dfrac{1}{4}, P\left(7\right)=\dfrac{1}{13}, \text{and} P\left(H\cap 7\right)=\dfrac{1}{52}[/latex] into the formula.

[latex]\begin{align}P\left(E\cup F\right)&=P\left(E\right)+P\left(F\right)-P\left(E\cap F\right) \\ &=\frac{1}{4}+\frac{1}{13}-\frac{1}{52} \\ &=\frac{4}{13} \end{align}[/latex]

The probability of drawing a heart or a 7 is [latex]\dfrac{4}{13}[/latex].

The events “drawing a heart” and “drawing a spade” are mutually exclusive because they cannot occur at the same time. The probability of drawing a heart is [latex]\frac{1}{4}[/latex], and the probability of drawing a spade is also [latex]\frac{1}{4}[/latex], so the probability of drawing a heart or a spade is

[latex]\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}[/latex]

The first step is to identify the sample space, which consists of all the possible outcomes. There are two number cubes, and each number cube has six possible outcomes. Using the Multiplication Principle, we find that there are [latex]6\times 6[/latex], or [latex]\text{ 36 }[/latex] total possible outcomes. So, for example, 1-1 represents a 1 rolled on each number cube.

1. We need to count the number of ways to roll a sum of 3 or less. These would include the following outcomes: 1-1, 1-2, and 2-1. So there are only three ways to roll a sum of 3 or less. The probability is
   [latex]\dfrac{3}{36}=\dfrac{1}{12}[/latex]
2. Rather than listing all the possibilities, we can use the Complement Rule. Because we have already found the probability of the complement of this event, we can simply subtract that probability from 1 to find the probability that the sum of the numbers rolled is greater than 3.
   [latex]\begin{align}P\left({E}^{\prime }\right)&=1-P\left(E\right) \\ &=1-\frac{1}{12} \\ &=\frac{11}{12} \end{align}[/latex]

1. We need to count the number of ways to choose only bears and the total number of possible ways to select 5 toys. There are 6 bears, so there are [latex]C\left(6,5\right)[/latex] ways to choose 5 bears. There are 14 toys, so there are [latex]C\left(14,5\right)[/latex] ways to choose any 5 toys.
   [latex]\dfrac{C\left(6\text{,}5\right)}{C\left(14\text{,}5\right)}=\dfrac{6}{2\text{,}002}=\dfrac{3}{1\text{,}001}[/latex]
2. We need to count the number of ways to choose 2 bears and 3 dogs and the total number of possible ways to select 5 toys. There are 6 bears, so there are [latex]C\left(6,2\right)[/latex] ways to choose 2 bears. There are 5 dogs, so there are [latex]C\left(5,3\right)[/latex] ways to choose 3 dogs. Since we are choosing both bears and dogs at the same time, we will use the Multiplication Principle. There are [latex]C\left(6,2\right)\cdot C\left(5,3\right)[/latex] ways to choose 2 bears and 3 dogs. We can use this result to find the probability.
   [latex]\dfrac{C\left(6\text{,}2\right)C\left(5\text{,}3\right)}{C\left(14\text{,}5\right)}=\dfrac{15\cdot 10}{2\text{,}002}=\dfrac{75}{1\text{,}001}[/latex]
3. It is often easiest to solve “at least” problems using the Complement Rule. We will begin by finding the probability that fewer than 2 dogs are chosen. If less than 2 dogs are chosen, then either no dogs could be chosen, or 1 dog could be chosen. When no dogs are chosen, all 5 toys come from the 9 toys that are not dogs. There are [latex]C\left(9,5\right)[/latex] ways to choose toys from the 9 toys that are not dogs. Since there are 14 toys, there are [latex]C\left(14,5\right)[/latex] ways to choose the 5 toys from all of the toys.
   [latex]\dfrac{C\left(9\text{,}5\right)}{C\left(14\text{,}5\right)}=\dfrac{63}{1\text{,}001}[/latex]

   If there is 1 dog chosen, then 4 toys must come from the 9 toys that are not dogs, and 1 must come from the 5 dogs. Since we are choosing both dogs and other toys at the same time, we will use the Multiplication Principle. There are [latex]C\left(5,1\right)\cdot C\left(9,4\right)[/latex] ways to choose 1 dog and 1 other toy.

   [latex]\dfrac{C\left(5\text{,}1\right)C\left(9\text{,}4\right)}{C\left(14\text{,}5\right)}=\dfrac{5\cdot 126}{2\text{,}002}=\dfrac{315}{1\text{,}001}[/latex]

   Because these events would not occur together and are therefore mutually exclusive, we add the probabilities to find the probability that fewer than 2 dogs are chosen.

   [latex]\dfrac{63}{1\text{,}001}+\dfrac{315}{1\text{,}001}=\dfrac{378}{1\text{,}001}[/latex]

   We then subtract that probability from 1 to find the probability that at least 2 dogs are chosen.

   [latex]1-\dfrac{378}{1\text{,}001}=\dfrac{623}{1\text{,}001}[/latex]