Types of Events

There are 13 hearts in the deck, so [latex]P(\text{heart})=\frac{13}{52}=\frac{1}{4}[/latex].  Note that [latex]\frac{1}{4}=0.25=25\%[/latex].

The probability of not drawing a heart is the complement: [latex]P(\text{not heart})=1-P(\text{heart})=1-\frac{1}{4}=\frac{3}{4}=75\%.[/latex]

Notice that since there are 2 possible coin tosses (Heads or Tails) and there are 6 possible die rolls, then there are [latex]2\cdot6=12[/latex] total outcomes, namely:

{H1,H2,H3,H4,H5,H6,T1,T2,T3,T4,T5,T6}.

Out of these, only 1 is the desired outcome, so the probability is [latex]\frac{1}{12}[/latex].

1. The probability that a head comes up on the second toss is 1/2 regardless of whether or not a head came up on the first toss, so these events are independent.
2. These events are not independent because it is more likely that it will rain in Galveston on days it rains in Houston than on days it does not.
3. The probability of the second card being red depends on whether the first card is red or not, so these events are not independent.

First, note that these two events (pulling out a white pair of socks and pulling out a white tee shirt) are independent.  So, we need to find the probability of each event.

• The probability of choosing a white pair of socks is [latex]\frac{6}{10}[/latex].
• The probability of choosing a white tee shirt is [latex]\frac{3}{7}[/latex].
• Thus, the probability of both being white is [latex]\frac{6}{10}\cdot\frac{3}{7}=\frac{18}{70}=\frac{9}{35}[/latex]

Here, there are still 12 possible outcomes: {H1,H2,H2,H4,H5,H6,T1,T2,T3,T4,T5,T6}

By simply counting, we can see that 7 of the outcomes have a head on the coin OR a 6 on the die or both – we use or inclusively here (these 7 outcomes are H1, H2, H2, H4, H5, H6, T6), so the probability is [latex]\frac{7}{12}[/latex].

Could we have calculated this from the individual probabilities?

As we would expect, [latex]\frac{1}{2}[/latex] of these outcomes have a head, and [latex]\frac{1}{6}[/latex] of these outcomes have a 6 on the die. If we add these, [latex]\frac{1}{2}+\frac{1}{6}=\frac{6}{12}+\frac{2}{12}=\frac{8}{12}[/latex], which is not the correct probability. Looking at the outcomes we can see why: the outcome H6 would have been counted twice, since it contains both a head and a 6; the probability of both a head and rolling a 6 is [latex]\frac{1}{12}[/latex].

If we subtract out this double count, we have the correct probability: [latex]\frac{8}{12}-\frac{1}{12}=\frac{7}{12}[/latex].

Half the cards are red, so [latex]P(\text{red})=\frac{26}{52}[/latex]

There are four Kings, so [latex]P(\text{King})=\frac{4}{52}[/latex]

There are two cards that are both red and Kings, so [latex]P(\text{Red AND King})=\frac{2}{52}[/latex]

We can then calculate

[latex]P(\text{Red OR King})=P(\text{Red})+P(\text{King})-P(\text{Red and King})=\frac{26}{52}+\frac{4}{52}-\frac{2}{52}=\frac{28}{52}[/latex]

There are four Queens in the deck, so [latex]P(\text{Queen})=\frac{4}{52}[/latex].

Similarly, there are four Kings in the deck, so [latex]P(\text{King})=\frac{4}{52}[/latex].

Notice that there are no cards in the deck that are both Queen AND King, therefore, [latex]P(\text{Queen AND King})=0[/latex].  These events are, therefore, called mutually exclusive. 

Now, using our probability rule, the probability of drawing a Queen or a King is:

[latex]P(\text{King or Queen})=P(\text{King})+P(\text{Queen})-P(\text{King and Queen})=\frac{4}{52}+\frac{4}{52}-0=\frac{8}{52}[/latex]

In this problem, we are wanting to find the probability that “at least one” of the two items grabbed (a pair of socks and a tee shirt) is white.  Another way to phrase this question would be to find the probability that the pair of socks is white OR the tee shirt is white.

Using our probability formula, the probability of grabbing a pair of white socks or a white tee shirt (or both) occurring is given by the formula:

[latex]P(\text{White socks or White tee})=P(\text{White socks})+P(\text{White tee})–P(\text{White socks and White tee})[/latex].

We can calculate each of the required probabilities on the right side of this formula:

• [latex]P(\text{White socks})=\frac{6}{10}=\frac35[/latex]
• [latex]P(\text{White tee})=\frac{3}{7}[/latex]
• [latex]P(\text{White socks and White tee})=P(\text{White socks})\cdot P(\text{White tee})=\frac35\cdot\frac37=\frac{9}{35}[/latex]  (since these are independent events)

Substituting into our formula, we get:

[latex]P(\text{White socks or White tee})=\frac35+\frac{3}{7}-\frac{9}{35}=\frac{27}{35}[/latex]

1. Notice that having a red car and getting a speeding ticket are not independent events, so the probability of both of them occurring is not simply the product of probabilities of each one occurring.Fortunately, we have enough data to calculate the probability of this event directly.  We can see that 15 people of the 665 surveyed had both a red car and got a speeding ticket, so  [latex]P(\text{Red car and Speeding ticket})=\frac{15}{665}\approx0.0226[/latex].
2. To find [latex]P(\text{Red car or Speeding ticket})[/latex], we use the formula:[latex]\begin{align}P(\text{Red car or Speeding ticket})&=P(\text{Red car})+P(\text{Speeding ticket})–P(\text{Red car and Speeding ticket})\\&=P(\text{Red car})+P(\text{Speeding ticket})–P(\text{Red car and Speeding ticket})\\&=\frac{150}{665}+\frac{60}{665}-\frac{15}{665}=\frac{195}{665}\approx0.2932\end{align}[/latex].Note that we also could have answered this question directly by simply adding up the numbers: 15 people with red cars and speeding tickets + 135 with red cars but no ticket + 45 with a ticket but no red car = 195 people. So [latex]P(\text{Red car or Speeding ticket})=\frac{195}{665}[/latex] as shown above.