Download Basic Terms and Concepts: Random Experiment An experiment

Basic Terms and Concepts:
Random Experiment
An experiment where the results depend on chance
e.g. Drawing the numbers for LOTTO
A Trial
One performance of the experiment
e.g. One ball is drawn
Outcome
The result of the experiment
e.g. The ball is a ‘4’
Sample Space
The list of all of the possible outcomes
e.g. {1, 2, 3, 4, …, 40}
Event
Is part of the sample space
e.g. Selecting the numbers 5, 8, 11, 15, 22, 30 in LOTTO
Long Run Relative Frequency
This can be obtained when a large number of trials are performed
e.g. A coin is tossed 100 times and 48 heads occur
Probability = number of favourable outcomes = 48
total number of trials
100
Equally Likely Outcomes
When outcomes are equally likely.
number of elements in event
Probability =
number of elements in the sample space
The probability of an event is a number between 0 and 1
A Probability Distribution Table lists all of the outcomes and their
associated probabilities. In a probability distribution table the sum of the
probabilities always equals 1.
A random variable is a variable whose value is determined by the
outcome of a random experiment.
X is used for the random variable, x is used for any values that it takes.
e.g. Ten seeds are planted. A random variable might count the number of
seeds that germinate.
X = random variable, x = 0,1,2,3,4,5,6,7,8,9,10 represents the values
that it can take.
P(X = x) could have P(X = 2) = “the probability that 2 germinate”
e.g. Tossing two dice could yield random variables
i) A = “the number of twos occurring” a = {0,1,2}
ii) B = “the sum of the result” b = {2,3,4,5,6,7,8,9,10,11,12}
Properties of a Sample Space
1.
2.
3.
4.
P(Ø) = 0, the probability of nothing is 0
P(S) = 1, the probability of all happening is 1
0 ≤ P(A) ≤ 1, the prob. for any event A is always between 0 and 1
P(A’) = 1 – P(A), prob. A not occurring is 1 – prob A occurs
Or P(A’) + P(A) = 1
A’ is the complement of A
e.g. A coin is tossed twice – what is the prob. that at least one head
occurs?
S = (HH, HT, TH, TT)
E = (HT, TH, HH)
P(E) = ¾
Or P(at least one head) = 1 – P(no head) = 1 – ¼ = ¾
e.g. A loaded die is such that an even number is twice as likely to occur
as an odd number. Find the probability that a number less than a 4 appears
in a single toss.
S = (1, 2, 3, 4, 5, 6)
E = (1, 2, 3)
Let w = P(odd no.) then 2w = P(even no.)
So w + 2w + w + 2w + w + 2w = 1
P(E) = 1/9 + 2/9 + 1/9 = 4/9
9w = 1
w = 1/9
Tree Diagrams
- also can be used when sampling without replacement. This is when the
second event is dependant upon what happened in the first event.
e.g. A uni student brought 10 unlabelled cans of food. He knows that 5 of
the 10 are sweetcorn, while the other 5 are baked beans. What is the
probability that two of the first three cans selected contain sweetcorn?
S
3/8
S
4/9
5/10
5/10
S
S'
S
5/8
4/8
S'
5/9
5/9
S'
4/8
4/8
S'
S
4/9
S'
S
4/8
5/8
P(2 of 3 cans are sweetcorn) =
5/10 × 4/9 × 5/8 + 5/10 × 5/9 × 4/8 +
5/10 × 5/9 × 4/8
= 5/12
S'
S
3/8
S'
Venn Diagrams
- diagrams that allow us to explain the relationship between events.
A
B
1.
A  B - A intersection B (in A and B)
2.
A  B - A union B (in A or B)
3.
A' - A complement (not in A)
S
A'
A
e.g.
A
B
0.3
0.15
0.2
1.
2.
3.
4.
P(A) = 0.3 + 0.15 = 0.45
P(B) = 0.2 + 0.15 = 0.35
P(A  B) = 0.15
P(A  B) = 0.3 + 0.2 + 0.15 =
0.65
5. P(A ' ) = 1 – 0.45 = 0.55
0.35
Probability Rule
P(A  B) = P(A) + P(B) – P(A  B)
i.e. the probability of being in A or B is the prob. of being in A + the prob.
of being in B – the prob. of being in both. (when you count the prob. in A
then B you count the intersection twice – hence subtract one lot of the
intersection)
e.g. The prob. that a student passes Math is 2/3 and the prob. she passes
English is 4/9. If the prob. of passing at least one course is 4/5, what is the
prob. that she passes both courses?
P( M  E) = P( M) + P( E) – P( M  E)
4 2 4
= + – P( M  E)
5 3 9
14
P( M  E) =
45
Special Events
1.
Complementary Event
Two events which exhaust the sample space.
A and A’ are complementary events and
P(A) + P(A’) = 1
e.g. A coin is tossed twice. What is the probability that at least one head
occurs? The complementary events are: that at least one head
occurs or no heads occur.
P(at least one head occurs) = 3/4
P(no heads occur) = 1/4
P(at least one head occurs) + P(no heads occur) = 1
or P(at least one head occurs) = 1 – P(no heads occur)
e.g. A coin is tossed 6 times in succession (26 = 64 outcomes)
What is the prob. that at least one head occurs?
Let E = at least one head occurs, E’ = no heads occur
P(E) = 1 – P(E’) = 1 – 1/64
= 63/64
2.
Mutually Exclusive Events
Two events are mutually exclusive if they cannot both happen at the same
time. There is no intersection on the venn diagram. The prob. that both
occur is zero.
The prob. that A or B occurs: P(A  B) = P(A) + P(B) as P(A  B) = 0
e.g. A = Sally won the tennis final last Saturday
B = Sally was runner up in the tennis final last Saturday.
e.g. C = the sum of two dice is 11
D = one of the dice shows a 3
e.g. What is the prob. of getting a total of 7 or 11 when two dice are
thrown?
They can’t both happen so we add the prob. of getting a 7 to the
prob. of getting 11.
Let X = the sum of the two dice.
P(X = 7) = 6/36 P(X = 11) = 2/36 P(X = 7 or 11) = 6/36 + 2/36 = 2/9
3.
Independent Events
Two events which have no bearing on each other are said to be
independent i.e. the outcome of one event is not going to influence the
outcome of the other.
If A and B are independent events: P(A  B) = P(A)  P(B)
e.g. A = a family has children of both sexes.
B = a family has at most one girl.
A and B are independent if a family has 3 children, not if it has 2
S2 = (gg, gb, bg, bb)
S3 = (ggg, ggb, gbg, bgg, gbb, bgb, bbg, bbb)
Family of 2: P(A) = 1/2 P(B) = 3/4 P(A  B) = 1/2 P(A) x P(B) = 3/8
Family of 3: P(A) = 3/4 P(B) = 1/2 P(A  B) = 3/8 P(A) x P(B) = 3/8
e.g. An athlete competes in two races. The prob. that he wins the 200m is
0.6, P(A) the prob. that he wins the 400m is 0.7, P(B) independently
of the result of the 200m. What is the prob. he wins:
a)
b)
c)
neither race P(A’) × P(B’) = 0.4 × 0.3 = 0.12
both races P(A) × P(B) = 0.6 × 0.7 = 0.42
just one race? P(A)×P(B’) + P(A’)×P(B) = 0.6 × 0.3 + 0.4 × 0.7 = 0.46
Let A = wins 200m, B = wins 400m
Conditional Probability
When probabilities are influenced by previous events
The conditional probability of B, given that A has already occurred is written
as P(B/A). The symbol / means ‘given’.
Using the probability tree to obtain a formula for conditional probability:
P(B/A)
P(A)
B
P(A  B) or P(B  A)
A
not B
B
not A
not B
P(A)  P(B/ A) = P(B  A)
Therefore the conditional probability formula is often written as:
P(A  B)
P(A / B) =
P(B)
Note: If events A and B are independent, the P(B/A) should equal P(B)!
e.g. The prob. that a married man watches a certain tv show is 0.4 and
the prob. that a married woman watches the show is 0.5. The prob.
that the man watches the show given his wife does is 0.7. Find the
prob. that:
P(A) = 0.4
P(B) = 0.5
P(A/B) = 0.7
Let A = Man watches show, B = wife watches.
a)
a married couple both watched the show
P( A  B) = P( A/ B)  P( B) = 0.7  0.5 = 0.35
b)
a wife watches the show if her husband does
P( B/ A) =
c)
P( A  B) 0.35
=
= 0.875
P( A)
0.4
at least one of the couple watches the show
P( A  B) = P( A) + P( B) – P( A  B) = 0.4 + 0.5 – 0.35 = 0.55
Relative Risk
Risk is used in language as the probability that something undesirable
might happen (eg. We talk about the risk of getting lung cancer).
We can use the same probability calculations for desirable events but we
use the word probability rather than the word risk (eg. The probability of
winning Lotto).
Risk is expressed as the probability of an event occurring for individuals in
a group (eg. The risk of getting lung cancer over a lifetime is 20% for
heavy smokers).
Relative risk is expressed as the risk of an event occurring for individuals
in one group compared to the risk for another related group (the risk of
getting lung cancer for smokers compared to non-smokers is six times
larger).
Relative risk is the comparison of two related risks. It is just as valid to say
relative probability is the comparison of two related probabilities. Note
relative risk can be bigger than 1 but can never be negative.
To solve a problem involving relative risk first understand the context:
 Identify the event the risk is of
 Identify the group the risk is for
 Identify the related group to which the first group is being compared
(the baseline group)
Calculate:
 the risk of the event for the group of interest (Rg)
 the risk of the event for the related or baseline group (Rb)
 the relative risk for the group of interest compared to the baseline
group is the fraction
Using the lung cancer example above this means the probability for nonsmokers of getting cancer must be 1/30 over a lifetime. (20%/(1/30) = 6).
Use the mathematics of percentage change to measure an increase or
decrease in risk.
Warning. Relative risk can be very misleading when the probabilities
involved are tiny (1/1000 divided by 1/100000 = 100). Here the
individual risks are small but one is 100 times the other. Always think
of the number of individuals out of a group who would be affected,
not just the relative risk.