| Probability
We deal with probability every day, from the roll of a dice or a flip of a coin, probability allows us to determine the likely outcomes of an event. The probability that an event will occur is a number between 0 and 1, with 0 being impossible, and 1 being definite.
If a coin is flipped it can either land on heads or tails. There are only two outcomes and assuming the coin is fair (i.e. its not weighted so that it lands on heads more often than tails) then we would expect, over a period of time, for the number of flips resulting in heads to be the same as the number of flips resulting in tails (you can try this - you may discover that the more number of trials you run, the nearer the results converge to 0.5 - this is due to the law of large numbers).
We can therefore find the probability than a random flip of a coin would produce a heads as oppose to tails. There are 2 outcomes, therefore our denominator is 2. In one trial a heads can only be flipped once, ergo our numerator is one.
Thus, written in notation, the probability that a head is flipped is a half: P(H) = 1/2
This can also be written as 0.5 or as 50%.
The opposite outcome, is that a tail is flipped. The probability of this is also a half. Remember the total probability of an event occurring will total one.
Henc, we can says that P(H') = 1/2. ' means the probability of an event not occurring.
Example 1
A fair die is rolled, find the probability that it lands on 2. [P(X=2)]
There are 6 possible outcomes, so our denominator is 6, we are asked to find the probability that when rolled the die will land of the face of a 2. On a die there is only 1 face with a 2 on it meaning our numerator is 1.
Hence P(X=2) = 1/6
Example 2
A fair die is rolled, find the probability that it lands on a number greater than 3. [P(X>3)]
The numbers on a die which are greater than 3, are 4, 5 and 6. Again our denominator is 6, but this time our numerator will be 3, because there are possible chances that our die will land on a number greater than 3:
P(X>3) = 3/6 = 1/2
Try some more questions at the bottom of the page.
If we are asked to find out multiple probabilities then we can use tree diagrams. Tree diagrams are used to show multiple events which are independent, this means that the outcome of one event has no bearing on the outcome of another event. For example, tossing a coin is an independent event, the first toss has no bearing on what I toss the second time.
After drawing a tree diagram, we may be asked to find out the probability of 2 events occurring. For example, what is the probability that I roll a 2 and then a 3 on a die? If we draw the tree diagram for this we come up with:
As you can see we have simplified the tree, by saying that on the first roll, we can either roll a 2, or we don't. We could draw the same tree diagram with 6 branches for the first roll, followed by 6 branches on the second roll for every branch of the first. This would give us a complete picture of what we could roll with a 6-sided die.
Now we were asked for the probability that a 2 is rolled, followed by a 3, this can be written as: P(2AND3) or as P(2∩3) [more on this notation later].
The probability that we roll a 2 [P(2)] is 1/6 and separately the probability that we roll a 3 [P(3)] is also 1/6. To find the probability that we roll a 2 AND a 3, we multiply both the individual probabilities. So we do 1/6*1/6 to get 1/36.
P(A and B) = P(A)*P(B) when the events are independent
Example 1
I flip a coin twice, what is the probability that I land on a head followed by a tail:
Step 1 - Draw the tree diagram:
Step 2 - Write the question: P(H AND T)
Step 3 - Solve: P(H) = 1/2; P(T) = 1/2; P(H AND T) = 1/2*1/2 => 1/4.
Therefore, the answer is one quarter.
If I had to find the probability of Event 1 occurring OR Event 2 occurring then I would add the probabilities. For example if P(Event 1)=4/10 and P(Even 2)=2/10; then the probability that E1 AND E2 would occur is 4/10*2/10 = 8/100 => 2/25. If asked to find the probability that E1 occurs OR E2 occurs we would do 4/10+/2/10 = 6/10 =>3/5.
P(A or B) = P(A)+P(B)
Example 2
I have a bag with 5 red counters, 4 blue counters and 1 green counter. I take out a counter and then place it back in the bag. a) What is the probability that I pick a red counter followed by a blue counter? b) What is the probability that I pick 2 red counters? c)What is the probability than I don't pick any red counters? d) What is the probability that I pick a red and blue counter?
My tree diagram is going to have 3 branches for the first pick, and then 3 additional branches (for every first branch) for the second pick as shown below. The total number of counters in the bag which I can pick from is 10 (5+4+1) so this is our denominator. I can pick 5 red, so P(R) = 5/10, I can pick 4 blues so P(B)=4/10 and I can pick 1 green so P(G)=1/10.
a) The probability that I first of all pick a red counter is 5/10, if we follow the R branch we can see that the probability of then picking a blue counter is 4/10. Hence the probability of picking a red followed by a blue is 5/10*4/10 = 20/100 => 1/5.
b) The probability of picking 2 reds will simply be 5/10*5/10 = 25/100 => 1/4.
c) If I don't pick any red counters then I could pick: (a Green AND a Green) OR (a Green AND a Blue) OR (a Blue AND a Blue) OR (a Blue AND a Green): (1/10*1/10)+(1/10*4/10)+(4/10*4/10) = 21/100.
d) This question looks similar to part a, but it is different. In part a, we were constrained by the order, it hand to be Red followed by Blue. Now we are asked to find the probability that a red and blue counter are chosen in any order. Therefore we would do Red and Blue OR Blue and Red: (5/10*4/10)+(4/10*5/10) = 2/5.
Venn diagrams can be created to show the possibilities of different events occurring, and their overlap. A basic 2 circle Venn diagram would look like:
We would write the sample space (the total number of possibilities of an event occurring) on the outside of the Venn diagram. On the inside (outside the circles) we would write the probabilities that are not included in Events A and B. Therefore if all probabilities are calculated from inside the square box it should equal one. The total number of possible events accumulated from inside the square box (called the sample space) would equal the value for the sample space. (see Example 1).
When using Venn Diagrams we use certain notation (which is also used for Rules):
P(A ∩ B) means the probability that A and B both occur, it is called the intersection of A and B, and can be seen as the overlap between the 2 events
IMAGE 1
P(A ∪ B) means the probability that either A or B, or both occur and is called A union B, it is shown as both circles of A and B
IMAGE 2
P(A') means the probability of not A (hence A doesn't occur) and can be seen as everything but A (note: this includes the overlap of A and B, as it is still A)
IMAGE 3
Using this information it is also possible to come up with other notation, for example P(A' ∩ B) means the probability of not A intersect B, and is hence just B with no overlap
IMAGE 4
Example 1
There is a pack of playing cards containing 52 cards. A is the even that the card is the number 3, and B is the event that the card is a heart. Fill in the Venn diagram to show the events A and B and the sample space:
IMAGE 5
Venn diagrams can be composed to show the possibility of up to 3 events occurring. When filling in a Venn diagram with 3 circles it is best to start with P(A ∩ B ∩ C) and then discount from there, as shown below:
Example 2
There are 125 people in a bar, the findings below show who had crisps, who had beer and how had soft drinks:
15 ordered all 3 items
43 had crisps
40 had beer
44 had soft drinks
20 had beer and a soft drink
26 ordered crisps and soft drinks
25 had beer and crisps
a.) Compose a Venn diagram for this, b.) find the probability that a randomly selected person had i.) all three items ii.) beer but not soft drink or crisps iii.) none of these items
a.) IMAGE 6
Start by filling in 15, for all 3 items. Then we look at who had beer and a soft drink (20), because these people are included in the 15 people who ordered all 3 items we have to discount them. Therefore only 5 people had JUST beer and a soft drink, hence we can fill in this. We continue doing this (don't forget to discount) for JUST beer and crisps and JUST crisps and soft drink. Then we move out to those who JUST ordered beer (discounting 10, 15 and 5 from the 40 beer drinkers, because they are included in the other measures), JUST ordered crisps and those who JUST ordered soft drinks, until we have a complete picture. We know the total number of pub-goers was 125 people, but only 71 ordered beer, crisps and or soft drinks. Therefore 54 people didn't order anything.
bi.) P(all 3 items) = 15/125 [the number who had all 3 divided by the total number]= 0.12
bii.) P(beer but not soft drink or crisps = JUST beer) = 10/125 = 0.08
biii.) P(none of the above) = 54/125 = 0.432
We have already seen some of the rules below, but here they are formalised into formulae and expanded upon.
P(A) + P(A') = 1
The probability that A occurs and the probability that A doesn't occur add up to one.
P(A') = 1 - P(A)
The probability that A doesn't occur is 1 minus the probability that A does occur.
Addition Rule: P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
This means that the probability of A union B (A OR B) is the probability of A add the probability of B take away the probability of A intersection B (which is A AND B).
This rule applies to Venn diagrams, we have already seen a slight variation of this rule for Tree Diagrams:
Addition Rule for Independent Events: P(A ∪ B) = P(A) + P(B)
This formula can be rearranged to find the intersection if necessary:
P(A ∩ B) = P(A) + P(B) - P(A ∪ B)
Multiplication Rule: P(A ∩ B) = P(A|B) * P(B)
Alternatively - P(B ∩ A) = P(B|A) * P(B)
This rule states that the probability of A intersection B (A AND B) is equal to the probability of A given B multiplied by the probability of B.
Conditional Probability: P(B|A) = [P(B ∩ A)]/P(A)
Alternatively - P(A|B) = [P(A ∩ B)]/P(B)
Find out more about conditional probabilities below. Now try some examples:
Example 1
P(A) = 0.4, P(B) = 0.3, P(A ∪ B) = 0.6. Find a) P(A ∩ B) b) P(A') c) P(A' ∩ B') d) Draw these events out on a Venn Diagram
a) P(A ∩ B) = P(A) + P(B) - P(A ∪ B)
P(A ∩ B) = 0.4 + 0.3 - 0.6
P(A ∩ B) = 0.1
b) P(A') = 1 - P(A)
P(A') = 1 - 0.4
P(A') = 0.6
c) P(A' ∩ B') - this means that event A doesn't occur AND event B doesn't occur. We know that P(A ∪ B) = 0.6; this means the probability of event A OR B occurring is 0.6. Therefore the probability that event A AND B DON'T occur, must be 1 - P(A ∪ B). This equals 0.4.
d) 
Example 2
The probability that a child in a school has green eyes is 0.37, and the probability they have black hair is 0.45. The probability that the child has either green eyes or black hair or both is 0.40. A child is randomly selected from the school, what is the probability that the child has a) black hair and green eyes b) black hair but not green eyes c) neither black hair or green eyes?
P(BH) = 0.45; P(GE) = 0.37; P(BH ∪ GE) = 0.8
a) P(BH AND GE) = P(BH ∩ GE)
P(BH ∩ GE) = P(BH) + P(GE) - P(BH ∪ GE)
P(BH ∩ GE) = 0.45 + 0.37 - 0.8
P(BH ∩ GE) = 0.02
b) P(BH AND NOT GE) = P(BH ∩ GE')
If we visualise a Venn diagram, we want the BH circle without the intersection with GE circle. Therefore we want BH minus P(BH ∩ GE):
P(BH ∩ GE') = P(BH) - P(BH ∩ GE)
P(BH ∩ GE') = 0.45 - 0.02
P(BH ∩ GE') = 0.43
We can look at this another way, by rearranging formulae:
P(BH ∩ GE') = P(BH) + P(GE') - P(BH ∪ GE')
P(GE') = 1 - P(GE)
P(GE') = 1 - 0.37 => 0.63
P(BH ∪ GE') - is the probability of black hair and not green eyes, which means we can add P(BH) with the sample space. The sample space would be 1 - P(BH ∪ GE) => 1 - 0.8 => 0.2.
P(BH ∪ GE') = 0.65
P(BH ∩ GE') = 0.45 + 0.63 - 0.65
P(BH ∩ GE') = 0.43
This method is more convoluted. It is probably best to either picture, or draw the Venn diagram and solve from there.
When 2 events have no common outcomes they are said to be mutually exclusive. This means they can never happen at the same time. For example it is impossible to be dead and alive at the same time, hence they are mutually exclusive events. This can be shown on a Venn diagram as two separate circles:
IMAGE 7
As we can see there is no intersection: P(A ∩ B) = 0. Hence the additional rule becomes P(A ∪ B) = P(A) + P(B)
When an event has no bearing on the outcome of another event they are said to be independent. Therefore the probability of A occurring is the same, regardless of whether B has occurred. Thus the conditional rule (shown below) changes to P(A|B) = P(A). The multiplication rules becomes P(A ∩ B) = P(A)*P(B). Which is what we would expect from a tree diagram with independent events.
We have already seen that the formula for conditional probability is:
Conditional Probability: P(B|A) = [P(B ∩ A)]/P(A)
Alternatively - P(A|B) = [P(A ∩ B)]/P(B)
We can use a Venn Diagram to explain why this is the case:
We are finding P(B|A). The event A has already happened therefore we can shade in the A circle. The total probability is a, therefore our denominator will be the value of a. Our numerator is going to be the bit where B and A both occur, because we are looking for B given A. Hence this will be P(A ∩ B) which is i. So P(B|A) can be seen on our Venn diagram as i/a. If we formalise this we get P(B|A) = [P(B ∩ A)]/P(A).
Example 1
P(A) = 0.3; P(B) = 0.5; P(A|B) = 0.4
Find a) P(A ∩ B) b) P(A ∪ B) c) P(B|A)
a) P(A|B) = [P(A ∩ B)]/P(B)
0.4 = [P(A ∩ B)]/0.5
0.4*0.5 = P(A ∩ B)
P(A ∩ B) = 0.2
b) P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
P(A ∪ B) = 0.3 + 0.5 - 0.2
P(A ∪ B) = 0.6
c) P(B|A) = [P(B ∩ A)]/P(A)
P(B ∩ A) = P(A ∩ B) => 0.2
P(B|A) = 0.2/0.3
P(B|A) = 0.666...
Example 2
A card is selected randomly from a deck of 52 cards. Find the probability that the card is an ace, given that the card is a diamond.
We need to find P(ace|diamond)
P(ace|diamond) = [P(diamond AND ace)]/P(diamond)
P(diamond) = 13/52 - in the deck of 52 cards, we would expect a quarter of them to be diamond, therefore 13 out of 52.
P(diamond AND ace) = 1/52
P(ace|diamond) = (1/52)/(13/52)
P(ace|diamond) = 1/13
Questions
1. A coin is flipped 10 times. How many times should it land on heads?
2. The probability that an event occurs is 0.4, the event occurs 20 times, how many times should it occur?
3. A card is taken from a standard deck of playing cards (there are 52 cards in a pack), what is the probability that: a) a heart is picked b) the number 4 is selected c) the 4 of hearts is picked d)a diamond is not selected e) a king is not selected.
4. Two dice are thrown and the product of the numbers (both numbers multiplied together) is written down. Find the probability that the product of the dice is a.) exactly 6 and b.) more than 4; given that one die lands on 2.
5.
1. 5 (half)
2. 8
3.
4.
Page last updated on 03/11/13
| 
