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STAT 240
PROBLEM SOLVING SESSION #1
Example: Random Variables
A student’s state from this course at the end of the semester defines a
random variable, S.
Values:{Pass,Fail}
Probabilities:
There are multiple ways of assigning probabilities. You need some
assumptions. First is that these probabilities are assigned for an average
student. Second, students are equivalent in all respects (iq, motivation,…etc)
1.
2.
3.
They might be already given. Say Pr(S=pass)=0.7
You are totally ignorant. Then Pr(S=pass)=0.5
You have some data. You have heard that 30 students among 40 have passed from this
course previous semester. So you consider nothing has changed. Then Pr(S=pass)=0.75
Example cts(II)
One way or another you have identified your random variable.
S
Pass
Fail
Pr(S)
0.6
0.4
A concept
we will talk
about a lot!
With this assignment let’s see what we can do.
Consider 2 friends. They want to know whether
all can pass from the course. State of each
student at the end of the semester is
independent of each other. Let’s try to answer
this problem.
Example cts(III) – basic intuition behind multiplication rule.
Now consider Student 1. The box below defines the probabilities of the
state of the student at the end of the semester.
The size of this rectangle is 40%. Of this 40%,
with 60% probability student 2 passes. So
green area coresponds to
Student 1’s state
Fail: 40%
p(Student1=fail, Student2=pass)=0.4x0.6=0.24
Similarly lets find,
Student 2’s state for each
state of Student 1
Pass: 60%
p(Student1=fail, Student2=fail)=
p(Student1=pass, Student2=pass)=
p(Student1=pass, Student2=fail)=
Also find,
p(Student2=pass) by summing up green areas.
Example 2
Now that we now multiplication rule lets try to answer the same
problem when there are 3 students.
Solving Example 1 with Binomial Formula
Binomial Formula:
If a random variable has only two values then joint behavior of N many
of these random variables, 𝑆𝑁 has binomial distribution. In this case 𝑆𝑁
is the count of one of the outcomes.
Binomial distribution has 2 parameters: N and p
If 𝑆𝑁 =count of pass
𝑁 𝑛
Pr 𝑆𝑁 = 𝑛 =
𝑝 1 − 𝑝 𝑁−𝑛
𝑛
In our example N=3 is the number of students. We are interested in the
probability of passing. This is the desired outcome. We are counting
this outcome.
Solving Example 1 with Binomial Formula
Setup the example:
Example 2
Using the previous setup solve
1. Find the probability that one student passes from a group of 5
students
Find the probability that at least one student fails
from a group of 5 students.
Find the probability that three or more
student passes from a group of 5 students.
Find the probability distribution 𝑆5
Calculate the probability of observing no 4’s
in 5 dice rolls.
Calculate the probability that the sum of a pair of
dices in 3 of the 5 rolls greater than nine.