OPTIMA Statistics

To better understand this question, let's look at two important sets. The first set is called the sample space and typically named \(S\text{.}\) It and contains all of the possible outcomes of this experiment. In this case:

The second set is the set containing all of the outcomes from S which meet our criteria of having exactly one heads. Any subset of the sample space is called an event. For this problem, the event of interest is:

If we assume that each of the outcomes is equally likely (see a Subsection 2.3.5), then

1. Because \(P(A)\) is larger (closer to 1) than \(P(B)\text{,}\) \(A\) is more likely to occur than \(B\text{.}\)

2. Because \(A\) and \(B\) are mutually exclusive (have no intersection), the number of ways that \(A\) or \(B\) (\(A\) union \(B\)) can occur is the sum \(n(A) + n(B)\text{.}\) Therefore,

   \begin{equation*} P(A\cup B) = \frac{n(A\cup B)}{n(S)} = \frac{n(A)}{n(S)} + \frac{n(B)}{n(S)} = P(A) + P(B)\text{.} \end{equation*}

   Thus, in our case, the probability that \(A\) or \(B\) will occur is \(0.6 + 0.2 = 0.8\)

3. Finally, based on the rules above, it is impossible to have an event \(C\) with \(P(C) = 7\text{.}\) Remember that probabilities are always between 0 and 1.

First, we must find the sample space. There are two ways we could look at this.

1. We could pretend that we can tell each marble apart, regardless of its color. In that case, the sample space is:

   \begin{equation*} S = \lbrace R_1, R_2, R_3, R_4, B_1, B_2, B_3, G_1, G_2, W_1 \rbrace\text{.} \end{equation*}

2. Since we can't actually tell the marbles of the same color apart, it is more accurate to say that the sample space is:

   \begin{equation*} S = \lbrace R, B, G, W\rbrace\text{.} \end{equation*}

The best answer for the sample space is \(S = \lbrace R,B,G, W\rbrace\) as shown in (2). However, the sample space from (1) can be a good way to answer the second part of the problem, what probability each outcome should have.

• \(P(R) = \frac{n(R)}{n(S)} = \frac{4}{10} = \frac{2}{5} = 0.40\)

• \(P(B) = \frac{n(B)}{n(S)} = \frac{3}{10} = 0.30\)

• \(P(G) = \frac{n(G)}{n(S)} = \frac{2}{10} = \frac{1}{5} = 0.20\)

• \(P(W) = \frac{n(W)}{n(S)} = \frac{1}{10} = 0.10\text{.}\)

To solve this problem, we need to assign a variable to the smallest probability. In this case, the probabilities of outcomes \(O_2\) and \(O_3\) are equal and we will call them both \(x\text{.}\) That means that the probability of outcome \(O_1\) is \(2x\text{,}\) and the probability of outcome \(O_4\) is \(3(2x) = 6x\text{.}\) Putting this all together,

Therefore, \(x = \frac{1}{10} = 0.10\text{.}\) From the above discussion, we know that \(P(O_1) = 0.20\text{,}\) \(P(O_2) = 0.10\text{,}\) \(P(O_3) = 0.10\text{,}\) and \(P(O_4) = 0.60\text{.}\)

We start with a valid probability measure.

This is a valid assignment of probabilities because each probability is between 0 and 1, and the sum of the probabilities is 1.

Now let's look at an invalid probability measure.

This is not a valid probability measure because the sum of these probabilities is not 1.

Another invalid probability measure would be

While the sum of these is 1, note that \(O_3\) has a negative probability, which is not allowed.

Using the general addition rule,

Using the general addition rule,

The set \(E = \lbrace 2, 4, 6 \rbrace\) consists of all even numbers. The complement of \(E\text{,}\) \(\overline{E} = \lbrace 1, 3, 5 \rbrace\text{.}\) The probability of each of these sets is \(\frac{3}{6} = \frac{1}{2}\text{.}\)

The set \(F = \lbrace 5, 6 \rbrace\) has probability \(P(F) = \frac{2}{6} = \frac{1}{3}\text{.}\) The complement of \(F\) is \(\overline{F} = \lbrace 1,2,3,4 \rbrace\text{.}\) The probability of this set is \(P(\overline{F}) = \frac{4}{6} = \frac{2}{3}\text{.}\)

Based on the complement rule,

Then, using the general addition rule,

Our first Venn Diagram shows the information given in the problem statement. There are several things to note:

• The probability of the entire box is one, as mentioned above.

• The probabilities of \(R\) and \(W\) (rain and wind) are each split between two regions, as indicated by the arrows.

• The probability of \(W \cap \overline{R}\) is written into the appropriate region.

Now we are ready to do some computations to figure out the missing pieces in the Venn diagrams, starting with the intersection of \(R\) and \(W\text{.}\)

\begin{align*} P(R\cap W) \amp = P(W) - P(R\cap \overline(W)\\ \amp = 0.64 - 0.39\\ \amp = 0.25 \end{align*}

Next, we move on to find the union of \(R\) and \(W\) since we are after the complement of this union (not windy or rainy).

\begin{align*} P(R\cup W) \amp = P(R) + P(W) - P(R\cap W)\\ \amp = 0.35 + 0.64 - 0.25\\ \amp = 0.74. \end{align*}

Finally, using the complement rule, the probability we are looking for is found below.

Note that in the final Venn diagram the sum of all numbers in the box is 1.

The final Venn Diagram is shown below. Can you determine how these numbers were found?

Since we are told the cat is chosen at random, we know that each cat is equally likely to be chosen. Thus, we can use the theoretical probability rule above to solve these problems. For each of these events, note that the sample space is all of the cats owned by this woman. So \(n(S) = 12\) in each problem.

1. Since we are told that \(n(M) = 5\text{,}\) we compute \(P(M) = \frac{5}{12}\text{.}\)

2. Because \(n(T) = 7\text{,}\) we get \(P(T) = \frac{7}{12}\text{.}\)

3. Finally, \(n(M \cup C) = 5 + 3 = 8\text{,}\) so \(P(M\cup C) = \frac{8}{12} = \frac{2}{3}\text{.}\)

Notice that nothing is said about the order in which cards are drawn for the first three events. So we will use combinations to count both the sample space and the number of outcomes in these events. The common sample space is found by couting the number of ways to select 4 of the 10 cards if order doesn't matter: \(C(10,4) = \frac{10 \times 9\times 8\times 7}{4\times 3\times 2\times 1} = 210\text{.}\)

1. If none of the cards are aces, then all four are selected from the remaining 8 non-ace cards. This can be done in \(C(8,4) = 70\) ways. So, \(P(\text{no aces}) = \frac{70}{210} = \frac{1}{3}\text{.}\)

2. We first split the cards into black and red. There are two black aces and two black queens making 4 total black cards. There are two red kings, two red queens, and two red jacks making 6 red cards. Next we must pick 2 of the 4 black cards and 2 of the 6 red cards. This is a multi-step process, so using the multiplication rule:

   \begin{align*} n(\text{2 black and 2 red}) \amp = n(\text{2 black}) \times n(\text{2 red})\\ \amp = C(4,2) \times C(6,2)\\ \amp = 6 \times 15\\ \amp = 90. \end{align*}

   Thus, \(P(\text{ 2 black and 2 red }) = \frac{90}{210} = \frac{3}{7}\text{.}\)

3. “At least one queen” means we could have one queen, two queens, three queens, or all four queens. If we try to count this directly, we must count these four separate cases and then use the addition rule to add them together. Instead, let's take a short-cut. The opposite of “at least one queen” is “no queens”. If we can find the probability of no queens, then we can use the complement rule to find the probability of at least one queen. If we don't allow the queens to be chosen, there are only 6 cards available. We can pick our entire hand from these 6 cards in \(C(6,4) = 15\) ways. So, \(P(\text{no queens}) = \frac{15}{210} = \frac{1}{14}\text{.}\) Therefore,

   \begin{equation*} P(\text{at least one queen}) = 1 - P(\text{no queens}) = 1-\frac{1}{14} = \frac{13}{14}. \end{equation*}

4. Our last event is different from the other three because the order in which we pick our cards matters. We should therefore use a permutation or the multiplication rule instead of a combination. Keeping in mind that there is only one of each suite's queen in the deck, we compute

   \begin{align*} P(\text{Queen of Hearts, Diamons, Clubs, then Spades}) \amp = \frac{1\times 1\times 1\times 1}{P(10,4)}\\ \amp = \frac{1}{10\times 9\times 8\times 7}\\ \amp = \frac{1}{5040}. \end{align*}

We start by counting the sample space, which will be the same for all three events. We are drawing three marbles from an urn containing 12 total marbles. We don't care about the order in which we draw the marbles, just the colors that are drawn. Therefore, \(n(S) = C(12,3) = 220\text{.}\)

1. Since there are only two green marbles, it is impossible for all three of the marbles we draw to be green. Therefore, \(P(\text{all three are green}) = 0\text{.}\)

2. There are four white marbles. The number of ways we can draw three of the four is \(C(4,3) = 4\text{.}\) Therefore, \(P(\text{all three are white}) = \frac{4}{220} = \frac{1}{55}\text{.}\)

3. All three marbles can be the same color by being all green, all white, or all red. We said that \(P(\text{all green}) = 0\) and \(P(\text{all white}) = \frac{1}{55}\text{.}\) To determine the probability they are all red, we count \(n(\text{all three red}) = C(6,3) = 20\text{.}\) Thus, \(P(\text{all red}) = \frac{20}{220} = \frac{1}{11}\text{.}\)

   Now since these are three different ways that we can get all three marbles to be the same color, and since each of these events is mutually exclusive (we can't get all three red and all three white), we use the general addition rule to get:

   \begin{align*} P(\text{all same color}) \amp = P(\text{all green}) + P(\text{all white}) + P(\text{all red} )\\ \amp = 0 + \frac{4}{220} + \frac{20}{220} = \frac{24}{220}\\ \amp = \frac{6}{55}. \end{align*}