Lesson 4: Probability Distributions of Discrete Random Variables
B. Main Lesson: Probability Distribution
Introduce the concept of the Probability Mass Function of Discrete Random Variables:
a table, graph, or formula that lists all the possible values of the random variable and the corresponding probability for each value. Take note that the probabilities may be empirical probabilities, theoretical probabilities, or subjective probabilities.
In the examples done earlier, the table and the histogram are two ways to represent the probability mass function, also called the probability distribution. You can explain further that it is called probability distribution because it is as though we are distributing probability weights (or masses) to all the possible observations or values of the random variable. This will then lead to the properties of probability
distributions which will be discussed later. In the previous example, you distributed all the “weights” from the learners to the different values of the random variable (number of siblings).
As a second example, you can consider the probability distribution of the number of heads occurring when tossing a coin three times, and then counting the number of heads (the activity done in Lesson 2 of Chapter 2). Suppose there is an equal chance that the coin lands on a head or a tail (However, this assumption cannot be done always, since we are not exactly sure if we do have a “fair coin.” In fact, this is what differentiates statistics from probability, where in the latter, we make
assumptions about the probability of garnering a head, while in statistics, we conduct data collection to estimate this unknown probability). Then, there will be eight outcomes, each one assumed to have 1/8 chance of appearing. Suppose X is the number of heads, then the Probability Distribution is as follows:
Tabular Representation of the Probability Distribution of X Outcomes Number of heads Probability
TTT 0 1/8
TTH, THT, HTT 1 3/8
THH, HTH, HHT 2 3/8
HHH 3 1/8
In general, when flipping a coin n times where the coin has probability p of getting a head in 1 toss, then the probability mass function for generating exactly X heads is
This is called the binomial pmf. The formula can be understood as follows: We want exactly k heads and n − k tails. For a particular sequence of k heads, the multiplication rule says, that this has a chance pk and similarly for a particular sequence of n-k tails, this has a chance of (1-p)n-k . However, the k heads can occur anywhere among the n trials, and there are different ways of distributing k heads in a sequence of n trials.
This second case is an example where the probabilities are derived theoretically.
So, whether probabilities are assigned empirically or theoretically, the probability distribution should have the same general properties.
I. Properties of Probability Distributions of Discrete Random Variables
• Probabilities should be confined between zero (0) and 1 (inclusive of both ends).
• The sum of all the probabilities should be 1 (i.e., 100%).
If we represent graphically the probability distribution of a discrete random variable X, with the area of the rectangles corresponding to the probability of each value of
X, we note that each area must be a non-negative valued (and at most equal to one). In addition, the area of all the rectangles should total 1 (or 100 percent).
Inform learners that for the two worked examples (on number of siblings and number of heads obtained in three tosses of a fair coin), all these properties of probability distributions were satisfied. It might be helpful if you do not remove the two examples from the board. Show in each example that probabilities cannot be below zero (0), and they cannot be above one (1). In addition, the probabilities sum up to one (1). For the graphical representation, show the area for each rectangle is the same as the probability, and then show that they all add up to 100%.
You can stop at this point and just give an assessment (either in the form of a
seatwork or a short quiz). If the time is not enough, you may give the assessment as a homework.
II. Determining Probabilities based on the Probability Distribution
Since the probability distribution contains the values of random variables and the corresponding probabilities of each value, then it can be used to determine the probabilities that a random variables will take on certain values.
For example, given the illustrative data on the number of siblings that the learners have, or better yet the actual distribution of the data in class, you can ask the learners:
• What is the probability that a randomly-selected learner is an only child?
• What is the chance that a randomly-selected learner has at most two siblings?
• What is the probability that a randomly-selected learner has three or more siblings?
For the illustrative data shown above, if you want to determine the probability that the randomly-selected learner is an only child, then you just get the probability that W=0, i.e.,
P(W=0) = 4%.
For the other questions, the chance that a randomly-selected learner has at most two siblings is:
P( W ≤ 2 ) = P( W = 0 or W = 1 or W = 2) = P(W=0) + P(W=1) + P(W=2) = 4 % + 20% + 56% = 80%
while the probability that a randomly-selected learner has three or more siblings is P( W ≥ 3 ) = P( W = 3 or W = 4 or W = 5 or W=7 )
= P(W=3) + P(W=4) + P(W=5) + P(W=7) = 10% + 6% + 2% + 2 % = 20%
Alternatively, for the latter probability, you will notice that having three or more siblings is the complement of having at most two siblings (whose probability was calculated already to be at 80%). As was stated in Lesson 1, the chance of the complement of an event is 100 percent minus the chance of an event.
P( W ≥ 3 ) = 1 – P( W ≤ 2 ) = 100% - 80% = 20%
You can also use the graphical representations of the probability distribution in order to determine the probability of the events of interest. Since the area under the graph is the same as the probability, then adding the areas of the rectangles will give the appropriate probability that you are looking for.
You should point out to learners that, in general, for a discrete random variable X, the probability that X lies in some discrete set A, may be obtained by summing the probability for the distinct values in the set A, that is.
For instance, in the last case,
P( W ≥ 3 ) == P(W=3) + P(W=4) + P(W=5) + P(W=7) = 10% + 6% + 2% + 2 % = 20%
C. Enrichment
You can go beyond the examples and do exercises like these:
Given the following table is a probability distribution for a random variable X, which corresponds to the number of pens that children from a class have in their bags.
k Probability that
1 X=k 0.15
2 0.2
3 0.35
4
Identify the value of a.
Answer: since the total probability is 1, then .
Therefore,
You can also add questions like: What is the probability that a randomly-selected student has at least three pens in his/her bag. Answer:
You can give similar examples that are based on the actual data that students did from the first lesson in Chapter 1 (the centralized data collected on the first day of stat). For instance, let Y = rating of how a randomly-selected students feels today (on a scale of 1-10).
ASSESSMENT
1. A probability distribution is an equation that:
a. associates a particular probability of occurrence with each outcome in the sample space
b. measures outcomes and assigns values of X to simple events c. assigns a value to the variability in the sample space
d. assigns a value to the center of the sample space Answer: A
2. Given the results of a survey of high school students, given the following probability distribution for Y, the number of pets they have at home.
Y Frequency Y Frequency
0 5 3 8
1 4 4 1
2 6 5 1
(a) Construct a histogram for the probability distribution
0102030Percent
0 2 4 6
number of pets they have at home
(b) Determine the probability of selecting a student having
• 3 pets.
• At most 2 pets
• At least 3 pets
• At least 1 pet Answer:
The probability of selecting a student with
• 3 pets is
P(Y=3) = 8/25 = 32%
• At most 2 pets is
P(Y≤2) = P(Y=0 or Y =1 or Y =2)=P(Y=0) + P(Y=1) + P(Y=2) = (5/25) + (4/25) + (6/25) =15/25= 60%
• At least 3 pets is
P(Y≥3) = 1- P(Y≤2 )=100% -60%
• At least 1 pet is
P(Y≥1) = 1- P(Y=0 )=1- (5/25) = 100% -20% =80%
3. Your mom decides to buy a single ticket for the lotto. Suppose that it has the following possible payoffs with their associated probabilities.
Payoff Probability P 100 0.0500 P1250 0.0100 P5,000 0.0050 P25,000 0.0010 P250,000 0.0005 P500,000 0.0001
(a) the probability that your mom will win any money is ________. (Answer: 0.0666) (b) the probability that your mom will win at least P5000 is ________. (Answer:
0.0066)
4. The following table contains the probability distribution for X = the number of retransmissions necessary to successfully transmit a 5 GB data package through a double satellite media.
X 0 1 2 3
P(X) 0.4
0 0.30 0.2
5 0.05
(a) the probability of no retransmissions is ________. (Answer: 0.40)
(b) the probability of at least one retransmission is ________. (Answer: 0.60)
5. Erik is going to flip a coin twice. Each coin flipped is independent, but the coin is biased: the probability that the coin will flip a head is 25 percent each time. If X is a random variable that represents the number of head obtained when the coin is tossed, create a histogram representing the probability distribution for all possible values of X.
Answer: Generate a graph to represent the following probability distribution P(X=0) = (0.75)2 = 0.5625
P(X=1) =2 (0.75) (0.25)= 0.375 P(X=2) = (0.25)2 = 0.0625
6. There are 8 players on an amateur basketball team. They are practicing their free throws by having each player shoot two free throws. The table below shows the result of each player's free throws. "X" represents a missed free throw, and
"O" represents an accomplished free throw.
Player Noel Candido Robert Michael Lino Carlos Angelo Ramon Free
throws XX OO XX XO OX XO XO XX
Make a histogram representing the proportion for each possible number of free throws made by a player.
Answer: Since we have the following probability distribution P(X=0) =3/8
P(X=1) =4/8 P(X=2) = 1/8
the histogram is shown below:
01020304050Percent
0 1 2
number of free throws
7. A couple intends to have children until they get at least one boy and one girl, but they agree that they will not have more than three children, even if all are girls or all are boys. (Assume boys and girls are equally likely).
(a) Determine the probability model for the number of children they will have (b) Calculate the probability of having two children
Answer:
Enumerating possible scenarios and probabilities we will get:
First child Second Child Third Child Fourth Child Probability
Boy Boy Boy Boy (1/2) x (1/2) x (1/2) x (1/2) = 1/16 Girl (1/2) x (1/2) x (1/2) x (1/2) = 1/16 Girl (1/2) x (1/2) x (1/2) = 1/8
Girl (1/2) x (1/2) = 1/4
Girl Boy (1/2) x (1/2) = 1/4
Girl Boy (1/2) x (1/2) x (1/2) = 1/8
Girl Boy (1/2) x (1/2) x (1/2) x (1/2) = 1/16 Girl Girl (1/2) x (1/2) x (1/2) x (1/2) = 1/16 (a) Thus the probability distribution is:
P (X = 2) =P(BG or GB)= ¼ + ¼ = ½ P (X= 3) =P(BBG or GGB) =1/8 + 1/8 = ¼
P (X= 4) =P(BBBB or BBBG or GGGB or GGGG) =1/16 + 1/16 + 1/16 + 1/16
= ¼
(b) The probability of having two children is P (X = 2) = ½
8. A six-sided dice is biased yielding the following probability distribution for X, the number of spots on the uppermost face when the die is rolled.
k 1 2 3 4 5 6
P(X=k)
(a) What values of θ are feasible, in order for P to be a probability distribution?
(b) Find P(X≤2) in terms of θ.
(c) Find P(2≤X≤5).
(d) What is the probability of rolling an odd number?
Answer:
(a) ≥ 0 ; ≥0 and + + + + + =1 so that -1≤ θ ≤1 except that θ ≠ 0 otherwise, we have a fair dice
(b)
(c)
= =
(d) The probability of rolling an odd number is