LEARNING COMPETENCIES
A. Introduction / Motivation: What is Probability and How to Assign It?
Begin the session with a discussion on your uncertainty over summaries generated from data, especially when data are “random” samples of a larger population of units (i.e. people, farms, firms, etc).
Examples: (i) approval ratings or proportion of people voting for a candidate (in an opinion poll); (ii) average family income (in the Philippine Statistical
Authority’s triennial Family Income and Expenditure Survey); (iii) average prices of commodities (from sample outlets)
Explain that people can quantify uncertainty through the notion of PROBABILITY (or Chance). Suggest to learners that if they were asked for the probability that they would pass the next quiz, they may give a number between 0% and 100 percent.
Typically, the chances of a future outcome may be based on some past experience of data collected. Very studious learners, for instance, had passed their quizzes 100 percent of the time, while average students had passed their quizzes 85 percent of the time.
When considering probabilities of events, learners should be guided to consider a particular context wherein possible outcomes are well defined and can be specified, at least in principle, beforehand. This context is called random process wherein we do not know which of the possible outcomes will occur, but we do know what is on the list of possible outcomes. Learners can be informed that it can be also helpful to view the probability of an event as its “long-run” empirical frequency or the fraction of times the event may have occurred under repeated “trials” of the random process. In the next lesson, we shall call this the “empirical probability,”
and mention that in practice, we expect these empirical probabilities to stabilize toward some “theoretical probability.” This is called the law of large numbers).
Ask learners to think of random processes and an event where:
a. the outcome is certain. Examples may be getting a head (event) in the next toss of a two-headed coin (random process) or getting a number of at most 6 (event) when a die is thrown once (random process)
b. the outcome is impossible. Examples may be getting a tail (event) in the next toss of a two-headed coin (random process) or getting a number greater than 6 (event) when a die is thrown once (random process)
c. the outcome has an even chance of occurring. Examples may be a couple having a boy (event) as their next child (random process) or getting a red card (event) when randomly selecting a card from a deck of cards (random process) d. the outcome has a strong but not a certain chance of occurring. Example might be getting a sum of at most 11 (event) when a pair of dice is thrown (random process)
Then, ask them the probability associated with these events. (Answers: 100 percent for certain events, 0 percent for impossible events, and 50 percent for outcomes with even chance of occurring. For the example in D, there are 36 possible
outcomes for tossing a pair of fair dice, 35 of them will have at most a sum of 11, so the chance of getting at most 11 is 35/36). The closer the value of the probability to 1, the more likely the event will occur and the closer it is to 0, the less likely it will occur.
Important: Point out to learners the following properties of the probability of an event:
• the probability of an event is a non-negative value. In fact, it ranges from zero (0) (when the event is impossible) to one (when the event is sure). The closer the value to one, the more likely the event will occur
• the probability of the sure event is one (In other words, the chance of a sure event is 100 percent).
• if A and B are mutually exclusive events, meaning it is impossible for
these two events to occur at the same time, then P(A or B) = P(A) + P(B).
This is called the Addition Rule.
A more general result (also called the General Addition Rule) states that:
P(A or B) = P(A) + P(B) - P(A and B)
Geometrically, from a Venn Diagram, the area of the union of A and B is the sum of the areas, but if we added the intersection of A ∩ B twice, so we have need to subtract this area from the sum of the areas of A and B.
Illustrate to learners that these properties can help us more readily compute for the probabilities of events.
P(at most a sum of 12 when tossing a pair of fair dice) =
P(at most a sum of 11 OR a sum of 12) = P (at most a sum of 11) + P(sum of 12)
But P(at most a sum of 12) = 1 and P (sum of 12) = 1/36;
Thus, when looking for the value of P (at most a sum of 11) P (at most a sum of 11) = 1 – 1 /36 = 35/36
In general, if we are interested in Ac, the complement of an event A (i.e. the event that happens when A does not), since
P(A or Ac) = P(A) + P(Ac) and P(A or Ac) = P(Sure event) = 100%
Thus,
P(A) + P(Ac) = 1 or equivalently P(Ac) = 1 – P(A)
In consequence, the chance that an event does not occur is one (1) minus the chance it does occur.
In terms of a Venn Diagram below, given a Sure Event S (represented by a square with area 100%), and an event A (represented by the triangle whose area represents the probability of A), then the chance of an event A not happening is one minus the chance of event A happening (i.e. area of the square minus the area of the triangle).
Extra Notes: Mention also to learners that:
(1) Historically, probability was studied by gamblers who wanted to increase their winnings (or at least decrease their losses).
(2) Probability describes random behavior, but does anything really happen at random? Even Albert Einstein, when confronted by theories of quantum mechanics, was said to have pointed out that “God does not play dice.” Yet, many events, especially in nature “seem” to display random behavior. In many real life situations, we will be able to model these by random processes and thus, apply probability to understand the behavior of these situations.
B. Main Lesson: Computing Probabilities of Events
Mention to learners that the calculation of the probability of an event may sometimes be considered directly from the nature of the phenomenon/random process, with some assumptions of symmetry. Some underlying outcomes may be
“equally likely” by assumption such as fair coins and fair dice. In practice, these assumptions need to be tested and will be the subject of inquiry in future lessons.
These assumptions are simplifications to help us calculate probabilities.
Example 1: Tell learners that a box contains green and blue chips. A chip is then drawn from the box. If it is green, you win P100. If it is blue, you win nothing.
• Learners have a choice between two boxes:
– Box A with 3 blue chips and 2 green chips – Box B with 30 blue chips and 20 green chips
• Which would learners prefer???
Some learners may say B, but tell learners that it actually should not matter, because the chance of winning Php100 is 2/5 =40% in box A, while in box B, the chance of winning is 20/50 =40%. Same probability.
Conditional Probability
Mention to learners that sometimes, we may have extra information that can change the probability of an event. Give the following definition of conditional probability.
The conditional probability of event A given that B has occurred is denoted as P(A|B) and defined as
Example 2: Suppose that we want to randomly select a student from among Grades 9 to 12 in a certain school
The chance of selecting a Grade 11 student, given that the student is male, can be computed as follows:
Define events A and B as:
A = event that student selected is a Grade 11 student B = event that student selected is male, then
Example 3: A king comes from a family of two children. What is the chance that the king has a sister?
Remind learners here that as the king comes from a family of two children, we are given extra information that this family of two children has a boy, the king.
Grade Sex
Total
Male Female
9 84 145 229
10 40 82 122
11 36 52 88
12 25 36 61
Total 185 315 500
What we want to compute here is the probability that the sibling of the king is a girl.
Let B the event of having at least one boy. So B={(b,b),(b,g),(g,b)}, where (x,y) means the sex of each child and the possible values are b for boy and g for girl.
Then A is the event that the king's sibling is a girl, A={(b,g),(g,b)}.
While the original sample space S of all possible outcomes is
S={(b,b),(g,b),(b,g),(g,g)}, each outcome has ¼ chance of occurring.
However, P (A | B) = P (A and B) / P(B) = (2/4) / (3/4) = 2/3
Independent Events
Sometimes, the extra information provided may not really change the probability of an event. In this case, the events are said to be independent. The conditional
probability of A given B may still be equal to the (unconditional) probability of event A.
Two events A and B are said to be independent if
P (A and B) = P (A) P (B)
This is also called the Multiplication Rule. Intuitively, we call events such as tossing a coin (or dice) several times independent since future tosses are not affected by previous outcomes.
If however, the events are not independent then we can still obtain the probability that both events A and B will occur using the definition of conditional probability:
P (A and B) = P (A) P (B | A)
Example 4: Tell learners to suppose that there is a box that contains three tickets marked 1, 2, and 3. We shake the box, draw out one ticket at random; shake the box and draw out a second ticket. What would be the probability of getting a sum of “three” if tickets were drawn with replacement? Without replacement?
The possible sums for the two tickets drawn with replacement are shown in a contingency table and tree diagram below
In consequence the probability of getting a sum of three is:
P (sum of three) = 2/9
While if the tickets were drawn without replacement, we have
P (sum of three) = 2/6 = 1/3
Exercise 5: (The Birthday Problem, originally posed by Richard von Mises in 1939, reprinted in English in 1964) Mention to learners that in a room filled with more than 23 people, there is more than half a chance that at least two of them will have the same birthday, and if there are more people, the chances increase further toward 100% (about 99.9% with 70 people). Try it out with learners in your class.
Tell learners to identify how many of them have a birthday in January. Try to see if you can get a match. Go to February if you don’t find anyone that match. Then March, and so forth.
The chance of 2 people having different birthdays is:
= 0.997260
The chance of N people having different birthdays is:
!
So the chance that at least two of them will have the same birthday is:
p(N) =
We have the probabilities computed below for several values of N.
N 10 20 23 30 50 57
p(N) 11.7% 41.1% 50.7% 70.6% 97.0% 99.0%
KEY POINTS
• Probability is a numerical representation of the likelihood of occurrence of an event. Its value is between zero (0) and one (1). When the value approaches 1, this means the event is very likely to occur, while a value close to zero (0) means it is not likely to occur.
• When A and B are mutually exclusive events, then the probability of A or B is P (A or B ) = P(A) + P (B) (this is called the Addition Rule)
• If A and B are independent events, then the probability of A and B is
P(A and B) = P(A) P(B) (this is called the Multiplication Rule)
ASSESSMENT
1. What would be the probability of
a. picking a black card at random from a standard deck of 52 cards?
b. picking a face card (i.e. a king, queen, or jack)?
2. What is the probability of rolling, on a fair dice:
a. a 3?
b. an even number?
c. zero?
d. a number greater than 4?
e. a number lying between 0 and 7?
f. a multiple of 3 given that an even number was drawn
f. P(multiple of 3 given even number) = P(multiple of 3 and even) / P(even)
= P(‘6’)/P(2 or 4 or 6) = (1/6) / (3/6 ) = 1/3
3. A standard deck of playing cards is well shuffled and from it, you are given two cards. You can have 0, 1, or 2 aces: three possibilities altogether. So the probability that you have two aces is equal to 1/3. What is flawed about this argument?
Answer:
The outcomes are not equally likely. There are (52)(51)/2=1326 ways of selecting the first two cards. These are the equally likely outcomes. Of these ways, there would be (4)(3)/2=6 ways of selecting two aces; (48)(47)/2=1128 ways of
selecting no aces, and 1326-6-1128=192 ways of selecting one ace. So, the chance of getting two aces is 6/1326 and not 1/3.
4. You shuffle a deck of playing card, and then start turning the cards one at a time.
The first one is black. The second one is also a black card. So is the third, and this happens up to the 10th card. You start thinking, “the next one will likely be red!” Are you correct in this reasoning?
Answer:
Yes, there are 42 cards left, 26 red and only 16 black. However, likely does not mean certainty. There is 16/42 chance that it is still going to be black.
5. The family of Tony delivers newpapers, one to each house in their village.
Philippine Star 250 Manila Times 140
Philippine Daily Inquirer 300 Manila Standard Today 100 Manila Bulletin 150 Daily Tribune 60
What is the probability that a house picked at random has:
a. the Manila Times?
b. the Manila Standard Today or the Philippine Daily Inquirer?
c. a newspaper other than Daily Tribune?
Answer:
a. P(Manila Times) = 140/1000 =7/50;
b. P(Manila Standard Today or PDI)= (100 + 300)/1000 = 2/5 ;
c. P(other than Daily Tribune) = 1 – P(Daily Tribune) = 1 – (60/1000)= 940/1000
= 47/50
6. A class is going to play three games. In each game, some cards are put into a bag.
Each card has a square or a circle on it. One card will be taken out, then put back. If it is a circle, the boys will get a point. If it is a square, the girls will get a point.
a. Which game are the girls least likely to win? Why?
b. Which game are the boys most likely to win? Why?
c. Which game are the girls certain to win?
d. Which game is impossible for the boys to win?
e. Which game is it equally likely that the boys or girls win?
f. Are any of the games unfair? Why?
Answer:
a. game 3. For girls, chance of winning games 1, 2, and 3 are respectively, 4/8=50%, 8/8=100%, 4/12 = 33.3%.
b.game 3. For boys, chance of winning games 1, 2, and 3 are respectively, 4/8=50%, 0/8=0%, 8/12 = 66.7%.
c. game 2, chance is 100%
d. game 2, chance is 0%
e. game 1.
f. games 2 and 3.
7. In a computer ‘minefield’ game, ‘mines’ are hidden on grids. When you land randomly on a square with a mine, you are out of the game.
a. The circles indicate where the mines are hidden on three different grids. On which of the three grids is it hardest to survive?
b. Grid 1 above is a 3 by 6 grid with 6 mines. On which of the following grids is it hardest to survive?
X. 99 mines on a 30 by 16 grid Y. 40 mines on a 16 by 16 grid Z. 10 mines on an 8 by 8 grid Explain your reasoning.
Answer:
a. P(hit a mine) in grid 1, 2 and 3, respectively is 6/18=33.3%, 8/25=32%, 7/20=35%. Thus, it is hardest to survive in grid 3
b. P(hit a mine) in X, Y, Z grid, respectively is 99/(30x16) = 0.20625, 40/(16x16)=0.15625, 10/8x8)=0.15625. So, it is hardest to survive in grid X.