OPTIMA Statistics

When you choose door three, you are hoping that \(P(C) = 1\text{.}\) That is, you are hoping that even \(C\) definitely happens. When the game show host reveals the goat behind door number two, he tells you that event \(B\) definitely did not happen. This changes the probability of event \(C\text{,}\) and as we saw in the video, you should switch to door number one.

1. Does knowing if the first flip was a heads or tails change the probability that the third flip will be a heads? Coins do not have memories. Anything we know about the first flip has no effect on what happens on the second flip, so \(A\) and \(B\) are independent.

2. This experiment is different. Knowing that the first flip came up a heads does have an effect on whether all three flips come up heads. After all, if \(A\) did not happen, then \(B\) can not possibly happen either. Therefore, \(A\) and \(B\) are dependent.

3. Since we are replacing the marbles after we draw them, the probability of getting a red on the first draw and on the second draw are the same: \(\frac{2}{3}\text{.}\) What happens on the first draw does not effect the probability of getting a red on the second draw. Because of this, \(A\) and \(B\) are independent.

4. In our final question, we are again drawing marbles, but this time what happens on the first draw does make a difference. Since we do not replace the marbles we draw, if \(A\) does not happen (we draw red), there are two marbles left: 1 red and 1 white. So \(P(B) = \frac{1}{2}\text{.}\) However, if \(A\) does happen (we draw a white), then there are two marbles left, both of which are red. So \(P(B) = 0\text{.}\) Since \(P(B)\) changes depending on whether \(A\) happens or not, \(A\) and \(B\) are dependent.

Since the experiment involves flipping a coin three times, a sample outcome would be \(HHT\text{,}\) representing the first flip coming up heads, the second heads, and the last tails. The set of all outcomes is therefore:

The first event of interest, \(A\text{,}\) consists of all of the outcomes above resulting in a heads on the first flip. This is:

We can therefore conclude that

But what about B? The event B is:

If we know that \(B\) has happened, then we automatically know that \(A\) has happened as well. This is because if all three flips are heads, the first one must have been a heads (\(B\) is a subset of \(A\)). Therefore,

Note that \(P(A|B)\) and \(P(A)\) are different! This is equivalent to saying that \(A\) and \(B\) are dependent.

Our first step is to find the marginal distributions for this contingency table. With those added, the table becomes:

Using this table, we compute the following probabilities.

1. \(P(\text{H. Gone}) = \frac{86}{138} \approx 0.6232\) since 86 of the 138 peoples' headaches were gone after 15 minutes.

2. To find \(P(\text{H. Gone}\,|\,\text{Tylenol})\text{,}\) we use the conditional probability formula.

   \begin{equation*} P(\text{H. Gone}\,|\,\text{Tylenol}) = \frac{P(\text{H. Gone}\,\cap\,\text{Tylenol})}{P(\text{Tylenol})} = \frac{\frac{33}{138}}{\frac{73}{138}} = \frac{33}{73} \approx 0.4521\text{.} \end{equation*}

3. Since the probability of the headache being gone changed when we were told the person took Tylenol, these two events are not independent—they are dependent.

Note: When we have a contingency table, there is a short-cut to finding conditional probability. Since we are told in (b) that the person took Tylenol, we restrict ourselves to the Tylenol column. Then, the probability that their headache is gone is simply \(\frac{33}{73} \approx 0.4521\text{,}\) as above.

1. Since we are told that the first marble was red, when drawing the second marble we have 9 marbles remaining, 6 of which are red. So, the probability the second marble is red is \(\frac{6}{9} = \frac{2}{3}\text{.}\)

2. The probability that the first marble is red will be \(\frac{7}{10}\) since seven of the 10 marbles are red. In (a), we saw that the probability the second marble is red given that the first was red is \(\frac{2}{3}\text{.}\) Using the general multiplication rule, the probability that the first and second marbles are red is

   \begin{equation*} P(\text{1st }R) \times P(\text{2nd }R\,|\,\text{1st }R) = \frac{7}{10} \times \frac{2}{3} = \frac{14}{30} = \frac{7}{15}\text{.} \end{equation*}

We first use the general addition rule to find \(P(A\cap B)\text{.}\)

Next, we check using our test of independence.

Since the above equation is true, \(A\) and \(B\) are in fact independent events.

Because flipping a coin multiple times gives independent events, (remember the coin has no memory, so one flip can not affect another), we can use the multiplication rule.

1. The probability all four flips are tails is computed as follows.

   \begin{align*} P(\text{1st T}\cap\text{2nd T}\cap\text{3rd T}\cap\text{4thT}) \amp = P(\text{1st T})P(\text{2nd T})P(\text{3rd T})P(\text{4th T})\\ \amp = \frac{2}{3}\times\frac{2}{3}\times\frac{2}{3}\times\frac{2}{3}\\ \amp = \frac{16}{81} \approx 0.1975. \end{align*}

2. This event is simular, but one of the four flips must be a heads instead of a tails. So, we split this into the four different cases that give us one tails and add the results of the multiplication rule together, giving:

   \begin{align*} \amp \left(\frac{1}{3}\times\frac{2}{3}\times\frac{2}{3}\times\frac{2}{3}\right) + \left(\frac{2}{3}\times\frac{1}{3}\times\frac{2}{3}\times\frac{2}{3}\right) + \\ \amp \left(\frac{2}{3}\times\frac{2}{3}\times\frac{1}{3}\times\frac{2}{3}\right) + \left(\frac{2}{3}\times\frac{2}{3}\times\frac{2}{3}\times\frac{1}{3}\right) = \frac{32}{81} \approx 0.3951. \end{align*}

In the tree diagram below, the labels represent the color of marble drawn. The fractions on the branches represent the probability of drawing that color marble at that stage in the experiment.

Now using this tree diagram, we find the desired probabilities.

1. Note that \(P(\text{2nd }W\,|\,\text{1st }R)\) can be read right off the tree. Since we are given that the first marble was red, we go across that middle \(R\) branch. Now, the probability the second is white is the probability on the \(W\) branch, so \(\frac{1}{5}\text{.}\)

2. There are two ways the 2nd marble can be white. If the 1st is red or if the 1st is blue. These are distinct (mutually exclusive) paths through the tree. The probability of each is found by multiplying down the branches. Together, their probability is the sum of these two products, giving us:

   \begin{align*} P(\text{2nd }W) \amp = P(\text{1st }R\cap \text{2nd }W) + P(\text{1st }B\cap \text{2nd }W)\\ \amp = \left(\frac{2}{6}\times \frac{1}{5}\right) + \left(\frac{3}{6}\times \frac{1}{5}\right)\\ \amp = \frac{2}{30} + \frac{3}{30}\\ \amp = \frac{1}{6}\text{.} \end{align*}

We wish to find \(P(\text{1st }W\,|\,\text{2nd }R)\text{.}\) Unfortunately, our tree diagram is “backwards” for this computation. The probabilities on the branches are the probability of a particular color on the 2nd marble given that we got a certain color on the first. So, we must resort to the conditional probability formula.

The numerator of this fraction can be found by multiplying along a single branch in the tree—\(\frac{1}{6} \times \frac{2}{5} = \frac{2}{30}\text{.}\) To get the denominator, we need to add together the probabilities for all possible ways the last marble can be red. This is:

Therefore, the desired conditional probability is: