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| 13 Theory slides |
| 14 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
Below, some basic definitions of probability are examined.
An outcome is a possible result of a probability experiment. For example, when rolling a six-sided die, getting a 3 is one possible outcome.
An event is a combination of one or more specific outcomes. For example, when playing cards, an event might be drawing a spade or a heart. For this event, one possible outcome is drawing the A♠ or drawing the 7♡.
However, these are not the only outcomes of this event. All the possible outcomes that satisfy the event are listed below.
The sample space of an experiment is the set of all possible outcomes. For example, when flipping a coin, there are two possible outcomes: heads, H, or tails, T. Therefore, the sample space is {H,T}.
Here, the sample space is shown in a tree diagram. Each row represents the possible outcomes of a toss. When the coin is flipped another time, the tree diagram gets another row with the possible outcomes.
In this case, the sample space has 4 possible outcomes.
For each of the following experiments, list the possible outcomes in the sample space and count the total number of outcomes.
Outcomes:
Consequently, the sample space is the set {1,2,3,4,5,6}.
To list all the possible outcomes, create all possible combinations by determining one outcome for the first die and then varying the outcome of the second die. Then, change the outcome of the first die and repeat the process.
Since each die has 6 possible outcomes, the total number of possible outcomes for this experiment is 6⋅6=36.
Paulina bought two white and three black marbles, all of different sizes, and put them in a bag. When she got home, her little brother Diego and sister Emily asked her to give them two marbles. Paulina agreed but told them to draw one marble each without looking inside the bag. Diego drew the first marble, then Emily.
Outcomes: {W1,W2} and {W2,W1}
Since both marbles are randomly drawn, each outcome of the event will consist of two labels. For the first marble, there are 5 possible outcomes. For the second marble, there are 4 possible outcomes because one marble has already been removed from the bag.
Sometimes more than one event can be involved in an experiment. In such cases, it is important to know how to determine the union or intersection of the events.
A or Bor
A∪B.
A and Bor
A∩B.
Also, there may be situations where it is easier to determine which outcomes do not satisfy an event rather than determining which outcomes do. For such cases, the following concept will be useful.
Outcomes: A∩B={3,5,7}
Outcomes: B∪C′={1,2,3,4,5,7,8,9}
Event | Complement |
---|---|
A: picking a prime number | A′: picking a non-prime number |
B: picking an odd number | B′: picking an even number |
C: picking a multiple of 3 | C′: picking a number that is not a multiple of 3 |
A - event of picking a prime number
B - event of picking an odd number
Therefore, A∩B is the event of picking a prime and odd number.
Number | Is it prime? | Is it odd? |
---|---|---|
1 | No | Yes |
2 | Yes | No |
3 | Yes | Yes |
4 | No | No |
5 | Yes | Yes |
6 | No | No |
7 | Yes | Yes |
8 | No | No |
9 | No | Yes |
Consequently, A∩B={3,5,7}.
B - event of picking an odd number
C′ - event of picking a number that is not a multiple of 3
Therefore, B∪C′ is the event of picking either an odd number or a number that is not multiple of 3.
Number | Is it odd? | Is not a multiple of 3 |
---|---|---|
1 | Yes | Yes |
2 | No | Yes |
3 | Yes | No |
4 | No | Yes |
5 | Yes | Yes |
6 | No | No |
7 | Yes | Yes |
8 | No | Yes |
9 | Yes | No |
Consequently, B∪C′={1,2,3,4,5,7,8,9}. Notice that 6 is the only element of U that is not in B∪C′ since it satisfies neither B nor C′.
Next, draw three sets representing the events A, B, and C and write down the outcomes of each event inside the corresponding set.
Comparing the outcomes of the three events, some conclusions can be drawn.
With this information, the Venn diagram can be drawn.
Since an event is a combination of possible outcomes of an experiment, in some cases the event happens rarely, while in others it happens frequently. This frequency depends on the experiment and the event itself.
The probability of an event occurring can be determined both theoretically and experimentally. Theoretical probability expresses the expected probability when all outcomes in a sample space are equally likely. In comparison, experimental probability is expressed by analyzing data collected from repeated trials of an experiment.
When all outcomes in a sample space are equally likely to occur, the theoretical probability of an event is the ratio of the number of favorable outcomes to the number of possible outcomes.
P(event)=Number of possible outcomesNumber of favorable outcomes
When an experiment is performed, the results may be a little different from what was expected. In other words, slightly different results may be obtained from what the theoretical probability predicted.
Experimental probability is the probability of an event occurring based on data collected from repeated trials of a probability experiment. For each trial, the outcome is noted. When all trials are performed, the experimental probability of an event is calculated by dividing the number of times the event occurs, or its frequency, by the number of trials.
Ramsha and Mark conducted an experiment consisting of rolling two dice and adding their results. The following diagram shows the numbers obtained in each roll.
Consider the event of getting a result greater than or equal to 8.
The 5 in the second row and third column represents the event of rolling a 2 on the first die and a 3 on the second die. The other outcomes can be calculated in the same fashion. Next, highlight the outcomes satisfying the given event, that the sum of the dice is greater than or equal to 8.
Favorable outcomes=15, Total outcomes=36
ba=b/3a/3
Calculate quotient
Round to 2 decimal place(s)
Number of Successes=2, Number of Trials=7
Calculate quotient
Round to 2 decimal place(s)
Number of Successes=6, Number of Trials=9
ba=b/3a/3
Calculate quotient
Round to 2 decimal place(s)
The probability of drawing a club from a standard deck of cards is 0.25. Knowing this, what is the probability of drawing a spade, heart, or diamond if a card is drawn randomly?
To figure it out, instead of counting the favorable outcomes, the complement rule can be used.
The sum of the probability of an event and the probability of its complement is equal to 1.
P(A)+P(A′)=1
P(A)+P(A′)=1
Applying this formula, the probability of drawing a spade, heart, or diamond can be computed.
Dylan cut out 100 squares of paper and wrote a number from 1 to 100 on each square. He then put the papers in a bag and asked his dad to choose a paper at random.
What is easier to count, the numbers that are not multiples of 5 or the numbers that are multiples of 5? Note that these sets of numbers are the complements of each other.
Event | Complement |
---|---|
Picking a number that is not a multiple of 5. | Picking a number that is a multiple of 5. |
Favorable outcomes=50, Total outcomes=70
ba=b/10a/10
Calculate quotient
Round to 2 decimal place(s)
Let's draw all the possible combinations when rolling two dice.
The outcome of rolling two sixes occupies 1 of the 36 possible combinations. We know that the probability of getting two sixes is 25 % greater than getting all other outcomes combined. Therefore, if we label the probability of getting any of these 35 outcomes as x, we can write the probability of getting two sixes as 1.25x. P(not 6,6)=x and P(6,6)=1.25x Notice that these are complementary events, meaning that they add up to 1. With this information, we can solve for x.
The probability of not getting two sixes is 49. If we subtract this from 1, we obtain the probability of obtaining two sixes. P(6,6)=1-4/9=5/9 According to the Multiplication Rule of Probability, we must multiply the probability of two events happening. Therefore, if we call the probability of rolling one six p, then we can write the following equation. p^2=5/9 Let's solve for p.
The probability of rolling one six is about 75 %.
In a board game called Finance
, players can buy bonds. There are twelve bonds numbered 1 through 12. A bond pays out if a player rolls the bond's number, either with a single die or as the sum of a pair of dice. Below is an example where bond 3, bond 4, and bond 7 all win a prize.
To maximize the probability of winning, we want to cover as many favorable outcomes in the sample space as possible. There are two types of outcomes that result in a prize. First Type: &At least one of the dice &shows the bond number. [0.5em] Second Type: &The sum of the dice & is the bond number.
We know that a bond pays out if a single die shows the bond number. For example, if we choose bond 4, all of the following outcomes wins a prize.
We have 11 favorable outcomes. By the same token, we have the same number of winning outcomes for 1, 2, 3, 5, and 6.
Let's populate the sample space with the sum of the dice and mark the most frequent outcome.
As we can see, the most frequent outcome when considering the sum of the dice is 7. However, if we hold bond 7, we would forego every favorable outcome we get from holding any of the bonds that can appear on a single die. Therefore, any bond numbered 7 through 12 is not as favorable as the bonds numbered 1 through 6.
We have already determined that we must choose a bond numbered 1 through 6 to maximize our chances of a payout. The best of these bonds is the one with the greatest number of favorable outcomes when considering the sum of the dice. This is 6, as we can see in the revised sample space.
Therefore, to maximize our chances of receiving a prize, we should choose bond 6.
To determine the probability of winning a prize with bond 6, we will mark all favorable outcomes in the sample space.
As we can see, bond 6 has 16 favorable outcomes out of the 36 possible outcomes. With this information, we can calculate the probability of winning a prize with this bond. P(win)=16/36=4/9