Transcript STAB22 July 22 Review Session Probabilityx

STAB22 REVIEW SEMINAR
REVIEW SESSION I: PROBABILITY
FRIDAY, JULY 22ND, 2016
TOPICS YOU GUYS HAVE COVERED
(REGARDING PROBABILITY):
- Probability Rules (Addition rule, Multiplication rule,
Complement rule; disjoint vs. independent events;
Conditional probability
- Random Variables (discrete; continuous) => need to be able
to find the mean and SD of a discrete random variable;
understand linear transformations & rescaling; understand
how you can apply the things you learned about the Normal
distribution to a continuous random variable
- Binomial Random Variables: what conditions are required
for the binomial setting? How do you find mean and SD with
Binomial Random Variables? What happens when you have a
very large n?
Does this all make sense? If not, PLEASE ASK QUESTIONS! 
QUESTIONS FOR
UNDERSTANDING
=> EASIER; MAKES SURE YOU UNDERSTAND WHAT’S GOING ON WITH THE BASICS
Five balls are placed in a box: a red ball, two yellow balls, a
green ball, and a blue ball. The box is shaken, and a ball is
drawn at random (without seeing what it is).
a.) What is the probability that the ball is NOT red?
0.8 (4/5)
b.) What is the probability that the ball is EITHER yellow OR green?
0.6 (3/5)
c.) Is the probability of the entire sample space equal to 1 in this
case? Explain?
Yes, otherwise this is not a valid model; P(S) = 1
Real estate ads suggest that 64% of homes for sale
have garages, 21% have swimming pools, and 17%
have both features. What is the probability that a
home for sale has:
a.) A pool or a garage?
0.68
b.) Neither a pool nor a garage?
0.32
c.) A pool but no garage?
0.04
Seventy percent of marriages occur on a Saturday. Suppose 3
marriages are selected at random.
a.) What is the probability that all 3 marriages occur on a Saturday?
b.) Use your answer to (a) to find the probability that at least one
(i.e. one or more) of the 3 marriages does NOT occur on a
Saturday?
a.) P(Saturday AND Saturday AND Saturday)= (0.70)^3 = 0.343
b.) P(at least one wedding not on a Saturday) = 1 – P(Saturday AND Saturday AND Saturday) =
1 – 0.343 = 0.657
What is the difference between DISJOINT and an
INDEPENDENT events? Can two events be both
disjoint AND independent? Why or why not?
=> how can you use the probability rule formulas to
determine if two events are independent?...
• DISJOINT = two (or more) events cannot occur at the same time. Example: Both Hillary Clinton AND Donald
Trump cannot be President of the US at the same time. The event of each candidate being elected are DISJOINT
-
This means that if you know that one event occurred, the other automatically has P=0
• INDEPENDENT = two events do not influence each other. For example, wearing a red t-shirt has no influence on
the chance that you will buy a bagel for breakfast. We have a few ways to define it with our probability rules:
• P(B)= P(B|A)= P(B|A^c)
• P(A AND B)= P(A) x P(B)
•
AKA: P(B) is the same whether or not A happens!
***for independent events only!
SOOOO….. Disjoint events CANNOT be independent. Independence means one event has no influence on the
other if it happens, but disjoint events cannot both happen at the same time!
A manager has a morning meeting and an afternoon
meeting. She has probability of 0.4 of being later for the
morning meeting, probability 0.5 of being later for the
afternoon meeting, and probability of 0.25 of being late for
both meetings.
a.) If she is late for the morning meeting, what is the probability she will be later for the afternoon meeting?
P (late for aft. | late for morning)= P(LA & LM)/P(LM) = .25/.4= .625
b.) Are the events “late for the morning meeting” and “late for afternoon meeting” independent?
NO. Does P(late for aft.)= P(late for aft. | late for morning)? 0.5 ≠ 0.625
Discrete Random Variables:
A gambler makes a bet. The gambler’s winnings are defined
by the random variable W below:
Value of W
-1
3
Probability
0.8
0.2
a.) Calculate the mean of W
-0.2
b.) Calculate the variance and SD of W
Var. = 2.56; SD = sqrt. Var. = 1.6
c.) Calculate the SD of W – 1
1.6
***Adding or subtracting does NOT affect SD or Var!
d.) Calculate the SD of 2W- 1
3.2
***Multiplication/Division DOES affect SD and Var!
With Continuous Random Variables…
- these are different than discrete random variables
- you can take two continuous, normally distributed random
variables, and you can add/subtract them to get a new distribution!
- then just use your Z table to find different probabilities! 
X1 + X2 ~N(mean X1+ mean X2, 𝑆𝐷 𝑋1
X1 – X2 ~N(mean X1- mean X2, 𝑆𝐷 𝑋1
2
2
+ 𝑆𝐷 𝑋2 2 )
+ 𝑆𝐷 𝑋2 2 )
*****Be careful!!! Note that when you subtract, the means are subtracted from each other, BUT the
formula for the variance of your new distribution remains the same, and has ADDITION!!!
Binomial Random Variables:
A clothing store has determined that 30% of the people who
enter the store will make a purchase. Eight people enter the
store during a one-hour period.
a.) Find the probability that exactly four people will make a purchase.
b.) Find the probability that less than three people will make a purchase.
X~Bin(n=8, p=0.30)
P(Exactly four people make a purchase)= P(X=4) = .1361
P(X < 3) = P(X=0 OR X=1 OR X=2)= P(X=0) + P(X=1) + P(X=2) = .0576 + .1977 + .2965 = .5518
HARDER QUESTION!!!! => but… very important to understand!
The probability that a certain machine will produce a defective item is 0.07.
Suppose a simple random sample of 200 items is taken from the output of this
machine; the number of defective items in the sample has a binomial
distribution.
a.) Calculate the mean and SD of the number of defective items in the sample.
b.) Calculate the approximate probability that there will be 22 or more defective
items in the sample, using the normal distribution.
Solution:
The probability that
a certain machine
will produce a
defective item is
0.07. Suppose a
simple random
sample of 200 items
is taken from the
output of this
machine; the
number of defective
items in the sample
has a binomial
distribution.
What do we know?
P(defective)= 0.07
P(not defective)= 0.93
n = 200
Let X = # defective items.
X~Bin(n=200, p=0.07)
a.) μ= np= 200(.07) = 14
σ = 𝑛𝑝(1 − 𝑝) = 200 .07 . 93) = 3.61
b.) np= 14 ≥ 10; n(1-p)= 186 ≥ 10
therefore, we can model with the Normal approximation!
X~N(14, 3.61)
P(X ≥ 22) = ?
Z=
22−14
3.61
Therefore, there is an
~1.32% chance of
choosing 22 or more
defective items!
= 2.22
P= 1 – 0.9868 = 0.0132
2.22
REAL, ACTUAL PREVIOUS
EXAM QUESTIONS!
THESE ARE REAL QUESTIONS FROM PREVIOUS EXAMS. NOW THAT YOU
HAVE MASTERED THE BASICS, IT’S TIME TO TEST OUT YOUR KNOWLEDGE
ABOUT PROBABILITY
The marks in an exam have a Normal distribution with mean 65
and SD 15. A mark of 80 or above qualifies for an A grade. Adam,
Bob, and Cindy are three students writing the exam. Assume that
their marks are independent. What is the probability that at least
on of them will get an A grade?
(A) 0.16
(B) 0.05
(C) 0.4
(D) 0.06
(E) 0.5
Difficulty: Medium-Hard
ANSWER: C
The probability that a randomly chosen calculus student passes a certain calculus
course is 0.70. Use this info for the next two questions:
If 5 students are sampled at random, what is the probability that exactly 4 of
them pass the course?
(A) 0.640
(B) 0.028
(C) 0.360
(D) impossible to determine from the info in this course
(E) 0.335
Difficulty: Medium
Hint: Binomial distribution!
ANSWER: C
In the previous question, the probability was 0.7 that a randomly
chosen calc student would pass a certain course. Under these
circumstances, suppose a simple random sample of 15 calculus
students is taken. What is the probability that 13 or more pass the
course?
(A) 0.000
(B) 0.079
(C) impossible to tell from the info in this course
(D) 0.127
(E) 0.092
Difficulty: Medium
ANSWER: D
In the last question, the probability was 0.7 that a randomly chosen
calc student would pass a certain calc course. Under these
circumstances, suppose a simple random sample of 150 calc
students is taken. What is the probability that 115 or more of these
students pass the course? (You may assume that the total number
of students who take this course is over 2000.)
(A) 0.21
(B) 0.04
(C) less than 0.01
(D) impossible to determine from the info in this course
(E) 0.50
Difficulty: Hard
ANSWER: B
A fair coin is tossed 400 times. Find the standard deviation of the
number of heads that will be obtained.
(A) 10
(B) 20
(C) 30
(D) 40
(E) 50
Difficulty: Easy
ANSWER: A
The random variable X has a Normal distribution with mean 60 and
standard deviation 10. One of the following probabilities is also
equal to P(40 < X ≤ 48). Which one?
(A) P(72 < X ≤ 80)
(B) P(64 < X ≤ 72)
(C) P(50 < X ≤ 58)
(D) P(80 < X ≤ 88)
(E) P(56 < X ≤ 64)
Difficulty: Medium
Hint: It is not necessary to use a normal table to answer this question!
ANSWER: A
The random variable X had the probability distribution shown below:
Value
1
2
3
7
Probability
0.65
0.20
0.10
0.05
What is the mean of X?
(A) 1.7
(B) 1
(C) 1.5
(D) more than 3
(E) 2.5
Difficulty: Easy
ANSWER: A
The random variable X had the probability distribution shown below:
Value
1
4
Probability
0.2
0.8
The mean of Y is 3.4. What is the standard deviation of Y?
(A) 1.9
(B) 5.0
(C) 2.5
(D) 1.2
(E) 1.4
Difficulty: Easy
ANSWER: D
“ You can accomplish
ANYTHING!!! 
- Olivia Rennie
GOOD LUCK EVERYBODY AND THANKS SO MUCH FOR COMING!!!!
”