Math for Liberal Arts

There are 13 hearts in the deck, so \(P\left(\text{heart}\right)=\frac{13}{52}=\frac{1}{4}\text{.}\)

The probability of not drawing a heart is the complement:

\(P\left(\text{not heart}\right)=1-P\left(\text{heart}\right)=1-\frac{1}{4}=\frac{3}{4}\)

We could list all possible outcomes: {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

Notice there are \(2\cdot 6=12\) total outcomes. Out of these, only 1 is the desired outcome, so the probability is \(\frac{1}{12}\text{.}\)

1. The probability that a head comes up on the second toss is \(\frac{1}{2}\) 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.

The probability of choosing a white pair of socks is \(\frac{6}{10} = \frac{3}{5}\text{.}\)

The probability of choosing a white tee shirt is \(\frac{3}{7}\)

The probability of both being white is \(\frac{3}{5} \cdot \frac{3}{7}=\frac{9}{35}\)

Here, there are still 12 possible outcomes: {H1,H2,H3,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, H3, H4, H5, H6, T6), so the probability is \(\frac{7}{12}\text{.}\) How could we have found this from the individual probabilities?

As we would expect, \(\frac{1}{2}\) of these outcomes have a head, and \(\frac{1}{6}\) of these outcomes have a 6 on the die. If we add these, \(\frac{1}{2} + \frac{1}{6} = \frac{6}{12} + \frac{2}{12} = \frac{8}{12}\) , 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 \(\frac{1}{12}\text{.}\) If we subtract out this double count, we have the correct probability: \(\frac{8}{12} - \frac{1}{12} = \frac{7}{12}\)

There are 4 Queens and 4 Kings in the deck, hence 8 outcomes corresponding to a Queen or King out of 52 possible outcomes. Thus the probability of drawing a Queen or a King is:

P(King or Queen) = \(\frac{8}{52} = \frac{2}{13}\)

Note that in this case, there are no cards that are both a Queen and a King, so P(King and Queen) = 0. Using our probability rule, we could have said:

\(P\)(King or Queen)\(=P\)(King)\(+P\)(Queen)\(-P\)(King and Queen)\(= \frac{4}{52}+ \frac{4}{52} - 0 = \frac{8}{52}\)

Half the cards are red, so \(P\left(\text{red}\right)=\frac{26}{52}\)

There are four kings, so \(P\left(\text{King}\right)=\frac{4}{52}\)

There are two red kings, so \(P\left(\text{Red and King}\right)=\frac{2}{52}\)

We can then calculate

\(P\left(\text{Red or King}\right) = P\left(\text{Red}\right) + P\left(\text{King}\right) - P\left(\text{Red and King}\right) = \frac{26}{52}+\frac{4}{52} - \frac{2}{52} = \frac{28}{52} = \frac{7}{13}\)

1. We can see that 15 people of the 665 surveyed had both a red car and got a speeding ticket, so the probability is \(\frac{15}{665} \approx 0.0226\text{.}\)

   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.

2. We could answer this question 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 the probability is \(\frac{195}{665} \approx 0.2932\text{.}\)

   We also could have found this probability by:

   P(had a red car) + P(got a speeding ticket) - P(had a red car and got a speeding ticket)=\(\frac{150}{665}+\frac{60}{665}-\frac{15}{665}=\frac{195}{665}\text{.}\)