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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)
Izabella is spinning the following three number wheels.
If the sum of the wheels is an odd number, she wins. What is the probability of this happening? Give an exact answer.Examining the diagram, we see that Wheel B only contains even numbers, while Wheel C only contains odd numbers. WheelB:& 2, 4, 6, 8 WheelC:& 1, 3, 5, 7, 9, 11 The sum of an even and an odd number always equals an odd number. Let's prove this algebraically.
Any even number can be divided by 2. Let 2a be an arbitrary even number. An arbitrary odd number can be written as an even number plus one — that is, 2b+1. Even number:& 2a Odd number:& 2b+1 If we add these numbers, we can show that they add up to an odd number.
We previously defined an even number as divisible by 2. Therefore, since 2(a+b) contains a factor of 2, it must be even. If we add 1 to an even number we get an odd number. 2(a+b)^(even)+1_(odd) Consequently, the sum of a number from Wheel B and a number from Wheel C will always be an odd number.
When added to an odd number, which of the numbers on Wheel A will still produce an odd number? As we already discussed, the sum of an even and an odd number is an odd number. Therefore, when A is 2, we will get an odd number. What about the other numbers on Wheel A, both of which are odd? Let's have a look at the sum of two odd numbers.
As we can see, the sum of two odd numbers is an even number. Therefore, Izabella can only win if she spins a 2 on Wheel A. Of the three possible outcomes of Wheel A, only one will result in a win. As such, the probability of Izabella winning is 13.
The Romans played games with a special die called a talus. This die was made from the talus bone of in a hoofed animal's ankle. One die had four sides marked with 1, 3, 4, and 6 dots. Two Romans are playing a game that requires a pair of talus dice.
Let's draw the sample space for the two talus dice and mark the most frequent outcome.
The most probable outcome is 7.
One die has two odd numbers and two even numbers on it. Any combination where at least one of these dice shows either a 4 or a 6 is a favorable outcome. Let's mark these in the sample space.
As we can see, 12 outcomes result in at least one die showing an even number. Now we can calculate the probability of at least one die showing an even number. 12/16=3/4
A pyramid contains four blue and six green orbs.
There is a total of 10 orbs in the pyramid and 4 of them are blue.
By dividing the number of favorable outcomes by the total number of possible outcomes, we can determine the probability of picking a blue orb. P=4/10=40 %
This time, we want to determine how many blue orbs we must add to make the probability of picking a blue orb 60 %. This means the number of green orbs would represent 40 % of the total number of orbs. Since the number of green orbs is still 6, we have enough information to solve for the number of orbs.
There are 15 possible outcomes, which means 15- 6=9 orbs must be blue. Therefore, we must add 5 blue orbs to the 4 we already have.
Magdalena picks two cards from a standard deck, one at a time.
What is the probability that the two cards do not form a pair? Give an exact answer.There are many ways of getting two cards from a deck that do not form a pair. Therefore, it is easier to calculate the probability of the complement of this event, which is the event of getting a pair. P(not pair)=1-P(pair)^(complement) Notice it does not matter what the first card drawn is. Only the second card matters, as it must form a pair with the first card. After the first card is drawn, out of the 51 remaining cards, only three cards in the deck will make a pair with the first card. P(pair)= 3/51 Now we can determine the probability that the two cards do not form a pair by using the Complement Rule.
The probability of not getting a pair is 1617.
In a lottery, the probability of winning a big bag of money is 0.00008%.
From the exercise, we know that there is a 0.00008 % probability of winning the lottery. We can write this as the following equation. P(win)= 8/10 000 000 If we reduce the ratio to its simplest form, we know the least number of winning tickets the lottery can have. 8/8/10 000 000/8=1/1 250 000 For the probability of winning the lottery to be 0.00008 %, one ticket out of 1 250 000 must be a winner. Therefore, the least number of non-winning tickets is 1 249 999.