Discrete Random Variable

Let \(X\) be the number of heads to be observed from tossing a fair coin twice. First, construct the probability distribution of \(X\). Second, find the probability that at least one heads is observed.

Solution:

For this sample space, the possible values of \(X\) are \(0\), \(1\), and \(2\). The potential outcomes have equal chances of occurring and follow as:

\(S = {hh, ht, th, tt}\).

That is, "\(hh\)" refers to the outcome of two heads,

"\(ht\)" refers to the outcome of one head and one tail, and so on.

As the number of heads observed is represented by \(X = 0\):

\(X = 0\) corresponds to \({tt}\), with no heads observed

\(X = 1\) corresponds to \({ht, th}\), with \(1\) heads observed

\(X = 2\) corresponds to \({hh}\), with \(2\) heads observed

By simply counting, we derive the probability of each of these three events, as represented by the discrete variable \(X\). Thus:

Table 1: Probability Distribution of Tossing a Fair Coin Twice

The probability that at least one head is observed is an event that can be described by the mathematical expression: \(X \geq 1\). The probability of this particular event (at least one head) is calculated by the addition of the two mutually exclusive events of \(X =1\) and \(X = 2\). Therefore, \(P (X \geq 1) = P (1) + P (2) = 0.50 + 0.25 = 0.75\). In other words, there is a \(75\%\) chance that at least one heads will result from tossing a coin twice.

For example, let's consider a random sample of 125 nurses selected from a large hospital in which the proportion of nurses who are female is 57%. Suppose X denotes the number of female nurses in the sample. In this experiment, there are 125 (n = 125) identical and independent trials of a common procedure: selecting a nurse at random. There are exactly two possible outcomes for each trial, “success” (the event that we are counting, that the nurse is female) and “failure” (not female). Finally, the probability of success on any one trial is the same number (p = 0.57). As the proportion of nurses is 57% female, a random selection would therefore provide a 57% chance of selecting a female nurse. Thus, X is a binomial random variable with parameters n = 125 and p = 0.57.

Assuming a fair coin is tossed \(10\) times, what is the probability of getting \(6\) tails?

Solution

There are two outcomes that can be obtained in a coin toss experiment: a heads or a tails. Therefore:

• The probability of getting a tails is \(50\%\) (or \(0.5\)) in a given toss.

• Number of trials, \(n = 10\).

• Frequency of outcome, \(x = 6\).

• Probability of success on a single trial, \(p = 0.5\)
• Probability of failure on a single trial, \(q = 0.5\).

Plugging these values into the formula:

\[P(X=x)={\frac{n!}{(n-X)!}\cdot (X!)p^x q^{n-x}}\]

\[P(X=x)=0.205\]

For example, consider a geometric random variable, \(X = 3\), which represents obtaining a number \(3\) as the result of the roll of a fair die.

In this geometric random variable experiment, we would count the number of times the die is rolled before a value of \(3 (X = 3)\) is achieved once.

The trial in which a \(3\) is rolled is labeled as a "success," and any trial in which a \(3\) is not rolled is labeled as a "failure." As this is a geometric random variable experiment, we only need to obtain one success in order to finish it.

Since the observations are independent of each other, the probability that \(X = 3\) (a \(3\) will result from the roll of the die) will be \(1/6\) for each roll. This probability of \(1/6\) is because the die has six sides, which gives values of 1 through \(6\).

A representative from the National Theatre Marketing Division randomly selects people on a random street in Washington D.C. until he finds a person who attended the last movie show.

Let \(p\), the probability that he succeeds in finding such a person, equal \(0.20\). And, let \(X\) denote the number of people he selects until he finds his first success.

a. What is the probability that the marketing representative must select 4 people before he finds one who attended the movie show?

Solution:

\[f(x)=P(X=x)=(1-p)^{x-1}p\]

\[P(X=4)=(1-0.2)^{4-1}0.2=0.1024\]

Therefore, there is approximately a \(10\%\) chance that the marketing representative would have to select \(4\) people before he finds one who attends the last movie show.

Find the mean of the discrete probability distribution below:

Solution:

Following the formula:

\[\mu= E(x)=\sum x \cdot P(x)\]

\[\mu = ((-2)\cdot(0.21))+((1)\cdot(0.34))+((2)\cdot(0.54))+((3.5)\cdot(0.31))\]

Thus, \(\mu = 2.085\)

A discrete random variable X has the following probability distribution:

1. \(α\)

2. \(P (0)\)

3. \(P (X > 0)\)

4. \(P (X ≥ 0)\)

5. The mean \(\mu\) of \(X\)

6. The variance of \(X\)

7. The standard deviation of \(X\)

Solution

1. Since all probabilities must add up to \(1\),\( α = 1 – (0.2 + 0.5 + 0.1) = 0.2\)

2. Referencing the table, \(P (0) = 0.5\)

3. From the table once again, \(P (X > 0) = P (1) + P (4) = 0.2 + 0.1 = 0.3\)

4. From the table, \(P (X ≥ 0) = P (0) + P (1) + P (4) = 0.5 + 0.2 + 0.1 = 0.8\)

5. Using the formula in the definition for mean \(\mu\):

\[\mu=E(x)=\sum x P(x)= ((-1)\cdot 0.2) + ((0)\cdot 0.5)+((1)\cdot 0.2)+((4)\cdot 0.1)=0.4 \]

6. Using the value of \(\mu\) obtained with the formula for variance:

\[\sigma^2 = \sum (x-\mu)^2P(x)=((-1-0.4)^2\cdot 0.2) + ((0-0.4)^2\cdot 0.5+ ((1-0.4)^2\cdot 0.2)+((4-0.4)^2\cdot 0.1)=1.84 \]

7. Using the calculated variance, we can then obtain the standard deviation using its formula as follows:

\[\sigma=\sqrt{\sum(x-\mu)^2P(x)}\]

\[\sigma= \sqrt{1.84}=1.3565\]