Topic 7; Probability
Calculating probabilities using equally likely outcomes
When each outcome is equally likely to occur the probability of an event, X, occurring is given by;
P(X) = number of possible outcomes in the event
Total number of possible outcomes
Example 1
A fair dice is rolled what is the probability of getting
(a) a 6,
(b) an odd number,
(c) a 2 or a 3?
Solution 1
Total number of outcomes is 6.
The dice is fair so each of these outcomes is equally likely.
(a)The number of outcomes in the event getting a 6 is 1.
P(6) = 1
6
(b)The total number of outcomes in the event is 3 (1,3,5)
P(odd number) = 3 = 1
6 2
(c)The total number of outcomes in the event is 2 (2,3)
P(2 or 3) = 2 = 1
6 3
Example 2
This table shows how 100 counters are coloured red or blue and numbered 1 or 2.
red / blue1 / 23 / 19
2 / 32 / 26
The 100 counters are put in a bag and a counter is taken from the bag at random.
(a)Calculate the probability that the counter is red
(b)Calculate the probability that the counter is blue and numbered 1.
Solution 2
The total number of possible outcomes = 100.
(a)Red counters = 23 + 32 = 55
P(red) = 55= 11
100 20
This could be written 0.55 or 55%.
(b)There are 19 counters that are blue and numbered 1
P(blue and 1) = 19
100
This can be written 0.19 or 19%
Estimating probabilities using relative frequency
Relative frequency is when we calculate probabilities based on previous observations or an experiment.
Relative frequency = Number of times an event happens in an experiment
Total number of trials in the experiment
E.g.
Jamie does the following experiment with a bag containing 2 red and 8 blue counters;
“Take a counter from the bag at random. Record the colour then put the counter back in the bag. Repeat this for 100 trials.”
Jamie calculates the relative frequency of getting a red counter every 10 trials and shows his results on a graph.
P (red) = 2 = 0.2 This is shown by the dotted line
10
Jamie’s graph shows that as the number of trials increase the relative frequency gives a better estimate of calculated probability.
Example 1
In an experiment a drawing pin is dropped for 100 trials. The drawing pin lands point up 37 times. What is the relative frequency of the pin landing point up?
Solution 1
Relative frequency = 37 = 0.37
100
Example 2
A counter was taken from a bag of counters and replaced.
The relative frequency of getting a blue counter was found to be 0.4.
There are 20 counters in the bag.
Estimate the number of blue counters.
Solution 2
Number of blue counters = 20 x 0.4
= 8
There are 8 blue counters in the bag.
Mutually exclusive events
Events which cannot happen at the same time are called mutually exclusive events. For example the event “heads” cannot occur at the same time as “tails”.
When A and B are mutually exclusive events
P(A or B) = P(A) + P(B)
P (not A) = 1 – P(A) (because either” A” or “not A” is certain to happen)
Example 1
A bag contains 3 red (R) counters, 2 blue (B) counters and 5 green (G) counters.
A counter is taken from the bag at random.
What is the probability that the counter is:
(a)red
(b)green
(c)red or green?
Solution 1
Find the total number of counters in the bag
5 + 2 + 3 = 10
Total number of outcomes = 10
(a)number of possible outcomes = 3
P(R) = 3
10
(b)number of possible outcomes = 5
P(G) = 5 = 1
10 2
(c)Events R and G cannot happen at the same time.
P(R or G) = P(R) + P(G)
= 3 + 5 = 8 = 4
10 10 10 5
Example 2
A bag contains 10 counters.
3 of the counters are red (R).
A counter is taken from the bag at random.
What is the probability that the counter is;
(a)red,
(b)not red?
Solution 2
Total number of possible outcomes = 10.
(a)Number of possible outcomes = 3
P(R) = 3
10
(b)P(not R) = 1 – P(R)
= 1 – 3 = 7
10 10
Combining two events
Example 1
A fair coin is thrown twice. Identify all the possible outcomes and write down their probabilities.
Solution 1
Method 1 – list the outcomes systematically.
1st throw 2nd throw
Head (H) Head (H)
Head (H) Tail (T)
Tail (T) Head (H)
Tail (T) Tail (T)
Method 2 – use a possibility space diagram.
Method 3 – use a tree diagram
When a fair coin is tossed twice there are four possible outcomes.
Because the coin is fair all outcomes are equally likely.
So:
P(H and H) = P(Hand T) = P(T and H) = P(T and T) = ¼
Example 2
A fair dice is rolled twice.
Use a possibility space diagram to show all possible outcomes.
What is the probability of getting a “double 6”?
What is the probability of getting any double?
What is the probability that exactly one six is obtained?
Solution 2
P(double 6) = 1
36
P(any double) = 6 = 1
36 6
P(exactly one 6) = 10 = 5
36 18
Independent events
When two coins are tossed the outcome of the first toss has no effect on the outcome of the second toss.
This is an example of events which can happen together but do not affect each other.
When A and B are independent events then the probability of A and B occurring is given by:
P(A and B) = P(A) x P(B)
Using tree diagrams to work out probabilities
A tree diagram can be used to find all the possible outcomes when two or more events are combined.
Example 1
Box A contains 1 red ball (R) and 1 blue ball (B).
Box B contains 3 red balls (R) and 2 blue balls (B).
A ball is taken at random from box A.
A ball is then taken at random from box B.
(a)Draw a tree diagram to show all the possible outcomes.
(b)Calculate the probability that two red balls are taken.
Solution 1
(a)
(c)P(RR) = X =
Example 2
The probability that Amanda is late for school is 0.4.
Use a tree diagram to find the probability that on two days running:
(a) she is late twice (b) she is late exactly once.
Solution 2
(a)P (late and late) = 0.16
(b)P(late and not late or not late and late)= P( late and not late ) + P(not late and late)
= 0.24 + 0.24
=0.48
Conditional probability
In some situations the probability of an event can be affected by other events.
For example;
If it rains the probability of a particular driver winning the grand prix may change.
When probabilities depend on other events they are called conditional probabilities.
Example
A bag contains 6 red counters and 4 blue counters.
Three counters are taken from the bag at random and put into a box.
(a)what is the probability that the second counter put in the box is blue;
(i)The first counter put in the box is red?
(ii)The first counter put in the box is blue?
(b)What is the probability that the third counter put in the box is red;
(i)If the first two counters put in the box are red?
(ii)If the first counter put into the box is blue and the second counter is red?
(c)Show all the possible ways that the counters can be put in the box on a tree diagram.
Label all the branches of the tree diagram with their probabilities.
(d)What is the probability there are at least two blue counters in the box?
Solution
(a)
(i)If the first counter put in the box is red, then 4 out of 9 counters left in the bag are blue,
So P(Blue) = 4
9
(ii)If the first counter put in the box is blue, then 3 of the 9 counters left in the bag are blue;
So P(blue) = 3 = 1
9 3
(b)
(i)If the first two counters put in the box are red then 4 out of the 8 counters left in the bag are red.
So P(red) = 4 = 1
8 2
(ii)If the first counter put in the box is blue and the second counter put in the box is red then 5 out of 8 counters left in the bag are red
So P(red) = 5
8
(c)
(d)The outcomes included in the event at least two blue counters in the box are;
R,B,B P (R and B and B) =
B,R,B P(B and R and B) =
B,B,R P(B and B and R) =
B,B,B P(B and B and B) =
So P(at least two blue counters) =
Venn diagrams
Another approach to probability questions is to use a Venn diagram.
Example 1
Draw a Venn diagram to show the following information. In a group of 30 students, 15 play chess, 10 play snooker and 6 play both. Draw a Venn diagram and use this to find;
a)The probability that a student chosen from random from the group plays snooker.
b)The probability that the student plays both chess and snooker.
c)The probability that the student plays neither chess nor snooker.
Solution 1
a)P(snooker) =
b)P(chess and snooker) =
c)P(neither) =
Example 2
In a school sixth form 12% of the students are left handed, 15% of the students wear glasses and 3% are both left handed and wear glasses.
a)Given that a student wears glasses find the probability that they are left handed.
b)What is the probability that a left handed student also wears glasses?
Solution 2
a)P(Left handed | Wear glasses) =
b)P(Wears glasses | Left handed) =