Transcript S1 Probability

Page 79
Exercise 5A
Homework - using GCSE notes
for review prior to starting this
unit
Intersections and Unions
100 in a sixth form
34 take Maths
42 take Physics
24 take Chemistry
18 take Biology
How would you represent this data?
Why does this come to more than 100?
If I tell you 24 Mathematicians also
take Physics what is the probability I
pick at random a student from this
sixth form who only takes Maths?
What is the probability a student takes
Maths or Physics?
A card is selected at random from a normal set
of 52 playing cards. Let Q be the event that the
card is a Queen and D be the event the card is a
Diamond. What do you think the following
questions are asking?
P(Q D)
P(QUD)
P(Q'UD)
P(Q D')
Try solving them.
How might you calculate
P(A U B)?
It's important we know the names for these notations too
A U B means A's "union" with B
A or B happens or both
A
B means A's "intersection" with B
A and B happens
A' means A's complement
Everything that isn't A happens
P (A U B) = P(A) + P(B) - P(A
B)
Can be rearranged to...
P(A
B) = P(A) + P(B) - P (A U B)
When solving these style of questions
a Venn Diagram is helpful
Don't forget that drawing these on Venn diagrams
makes it much simpler
enter the values in the correct bits and shade in
what is required.
If it's an "Intersection" you use the values that
have been shaded by BOTH
If it's a "Union" you use ALL the values that have
been shaded
Page 83
Exercise 5B
Q1 - 6
Finish for homework
Now check your answers and correct or seek help
where necessary
Page 87
Exercise 5C
We'll do Q4 together
Now try Q1, 3, 6 and 7
Putting the information into a Venn
diagram WILL help
Complete for homework and check it.
Seek help with those questions you
don't understand BEFORE you hand it
in.
Conditional Probability
A bag contains 6 red and 4 blue beads. A bead is
picked out and retained and then a second bead is
picked out. Find the probability that:
a) both beads are red
b) the beads are of different colours
c) the second bead is red given that the first one is
blue?
Remember our sixth form from earlier
100 in a sixth form
34 take Maths
42 take Physics
But only 24 of the mathematicians also study
Physics
If I went into the Maths classroom what is the
probability I pull out a physicist?
How does this relate to the figures we
have?
Another piece of notation we will use shows
conditional probability
P(A / B) means the probability
A happens given that B has already happened
You have already met this when you learned about
tree diagrams.
We can use our new notation with tree diagrams
and generate formulae which will be useful
Don't forget formulae can be rearranged.
Of course this can be rearranged as shown below.
Sometimes this is a more useful format so learn both
It's useful to note that the event after the slash / is the
event you divide by and also the event which happens
first.
It can often be useful to sketch a tree diagram to put
the information on. However you may also need to
consider a Venn diagram to calculate the P(A
B) part
Page 90 Exercise 5D
Q1, 2 and 3
warm you up to the phrase "given that"
We'll try Q7 together
Now try Q4 and 6 so you practice using the notation
Tree Diagrams
Page 94 Exercise 5E
Q1,3 and 5
As always finish for homework
The Chevalier du Mere's Problem.
A seventeenth-century French gambler had run out of takers
for his bet that, when a fair cubical dice was thrown four
times, at least one six would be scored. He therefore
changed the game to throwing a pair of dice 24 times. What
is the probability that, out of these 24 throws, at least one is
a double six?
(You may wish to consider his first problem with 4 dice to warm up)
The Birthday Problem
What is the probability that out of 23 randomly chosen people,
at least two share a birthday?
Assume that all 365 days of the year are equally likely and
ignore leap years.
(You may want to find the probablities that two people have different
birthdays, that three people have different birthdays, and so on first)
A question might give you information about
P(B/A) but then ask you P(A/B) which requires
you to consider it from the perspective that B
happened first.
For any events A and B, write P(A and B) and P(B) in terms
of P(A), P(A'), P(B/A) and P(B/A'). Now write P(A/B) in
terms of P(A), P(A'), P(B/A) and P(B/A')
This is known as
Bayes' Theorem:
You may look the formula up on
Google and use it to help in any
questions where the order of events
has been reversed
The Doctor's Dilemma
It is known that among all patients displaying a certain set of
symptoms the probability they have a particular rare disease is
0.001. A test for the disease has been developed. The test shows
a positive result on 98% of the patients who have the disease and
on 3% of patients who do not have the disease. The test is given
to a patient displaying the symptoms and it records a posiitve
result. Find the probability that the patient has the disease.
Comment on your result.
The Prosecutor's Fallacy.
An accused prisoner is on trial. The defence lawyer asserts that
in the absence of further evidence the probability that the
prisoner is guilty is one in a million. The prosecuting lawyer
produces a further piece of evidence and asserts that if the
prisoner was guilty the probability that this evidence could be
obtained is 999 out of 1000 and if he were not guilty it would
only be 1 in 1000; in other words,
P(evidence / guilty) = 0.999 and P(evidence / not guilty) =
0.001. Assuming that the court excepts the legality of the
evidence and that both lawyers figures are correct what is the
probability that the prisoner is guilty?
Comment on your answer.
You should remember earlier we considered a
particular sixth form.
100 in a sixth form
24 take Chemistry
18 take Biology
If I told you the probability a student takes Chemistry or
Biology is 0.42 what else does that tell you?
Because there couldn't possibly be anyone taking Biology
and Chemistry we call them Mutually Exclusive - they can't
both happen
If two events A and B are Mutually Exclusive
then
P(A and B) = 0
This makes our equation from before:
P(AUB) = P(A) + P(B)
There is a similar result affecting the other formula we've
been looking at in this chapter
P(A) x P(B/A) = P(A and B)
logically if P(A) x P(B) = P(A and B)
then P(B/A) = P(B)
which means B is unaffected by A
this is known as being Independent of A
Page 100 Exercise 5F
We'll do Q3 and 6 together
Now try Q1, 2, 4 and 8
Don't forget to check your answers, ask and correct.
Page 104 has a mixed exercise of exam style
questions for revision practice of your own.