Download FD12 5.5 Lesson KEY

FD12
CH.5 Probability
KIM
5.4 Conditional Probability- Day One
Goal: Understand and solve problems that involve dependent events.
Warm-up:
Consider the following events:
A={the first card is a heart}
B={the second card is a heart}
1. One card is drawn from a standard deck of 52 cards and is replaced. A second card
is then drawn.
a) Determine P(A).
13/52
b) Determine P(B).
13/52
2. One card is drawn from a standard deck of 52 cards and is not replace. A second
card is then drawn.
a) Determine P(A).
13/52
b) ?Determine P(B). ???
Connection:
Why is it not possible to determine P(B) when the first card drawn is not replaced?
Not knowing if the first card drawn was a heart or not, the probability of getting the
second heart cannot be determined.
Independent Events
vs.
If the knowledge that one event has occurred
Has no effect on the probability of the other event
occurring
Dependent Events
the probability of event DEPENDS
on whether or not event A occurred
1
FD12
CH.5 Probability
KIM
Ex. 1 Classify the following events as dependent or independent.
a) The experiment is rolling a die and tossing a coin. The first event is rolling 2 on the
die and the second event is tossing tails on the coin.
b) The experiment is choosing two cards without replacement from a standard deck.
The first event is that the first card is a king and the second event is that the second
card is a king.
c) The experiment is choosing two cards with replacement from a standard deck. The
first event is that the first card is a king and the second event is that the second card is
a king.
Investigation: How do we determine the probability of an event whose probability
depends on the outcome of the other event?
Consider the scenario from Ex.2 Part C).
One card is drawn from a standard deck of 52 cards and is not replaced. A second
card is then drawn.
A={the first card is a king}
B={the second card is a king}
*USE TREE DIAGRAM
With probability
2
FD12
CH.5 Probability
KIM
Conditional Probability:
- If event B depends on event A occurring, the probability of an event
occurring given that another event has already occurred.
P(B|A)=
𝑷(𝑨 ∩𝑩)
𝑷(𝑨)
or P(A⋂B) = P(A)·P(B|A)
-TWO EVENTS DO NOT HAPPEN SIMENTANEOUSLY
- A Tree diagram is very useful for modelling problems that involve
dependent events.
Ex. 2 Calculating the probability of two events that are dependent.
A computer manufacturer knows that, in a box of 100 chips, 3 will be defective.
Jocelyn will draw 2 chips, at random, from a box of 100 chips. What is the probability
that both of the chips will be defective?
2/99
3/100
97/99
97/100
Thus,
3/99
96/99
3
FD12
CH.5 Probability
KIM
Your Turn
2 blue and 3 red marbles are in a bag. Drawing 2 marbles at random, what is the
probability that both marbles are blue?
Ex. 3
Solving a conditional probability problem
According to a survey, 91% of Canadians own a cellphone. Of these people, 42%
have a smartphone. Determine the probability that
any Canadian you met would have a smartphone. -> cell phone AND a smartphone
C= owning a cell
P(C)=0.91
S=owning a smartphone
Given that the
person owns a cellphone
 P(S|C) =0.42
P(S|C)=P(S⋂C)
P(C)
0.42= P(S⋂C)
P(S⋂C)= (0.42) (0.91) =0.3822
0.91
The probability that any Canadian I met in the month would have a smartphone
is 0.38 or 38%
Your Turn: Susan took two tests. The probability of her passing both tests is 0.6.
The probability of her passing the first test is 0.8. What is the probability of her passing
the second test given that she has passed the first test?
Do Daily problem for February 8th
HW: textbook pg. 350 #1-5, 7, 10
4
FD12
CH.5 Probability
Daily Problem
KIM
February 8th
/4
Name:
1. Two cards are drawn without replacement from a standard deck of 52 cards. Determine the
probability of the following events. Draw a tree diagram to represent each situation.
a) Both cards are red.
b) Neither card is a club.
2. A bag contains red and blue marbles. Two marbles are drawn without replacement. The probability
of selecting a red marble and then a blue marble is 0.28. The probability of selecting a red marble on
the first draw is 0.5. What is the probability of selecting a blue marble on the second draw, given that
the first marble drawn was red?
5
FD12
CH.5 Probability
KIM
5.4 Conditional Probability- Day Two
Ex. 4
Making predictions that involve dependent events.
Two machines A1 and A2 produce all the glass bottles made in a factory. The percentage of broke
bottles produced by these machines are 5% and 8% respectively. Machine A1 produces 60% of the
output.
Step 1: define two separate events
A=Machine
B=bottle broken
Step 2: draw the 2-steps tree diagram
Step 3: indicate the probability
For each branch
a) If a bottle is chosen at random, determine the probability that it is a broken bottle produced by
machine A1.
The first branch P(A1 and B) =0.6 x0.05 =0.03
b) If a bottle is chosen at random, determine the probability that the bottle is not broken.
The bottle is not broken Either ( A1 made and not broken) OR (A2 made and not broken)
2nd and 4th branches
= (0.6)(0.95)
+
6
(0.4)(0.92)
FD12
CH.5 Probability
KIM
Your Turn
Hillary is the coach of a junior ultimate team. Based on the team’s record, it has a 60% chance of
winning on clam days and a 70% chance of winning on windy days. Tomorrow, there is a 40%
chance of high winds. There are no ties in ultimate. What is the probability that Hillary’s team will win
tomorrow?
Daily Problem: Do #17, 19 pg.351
7
FD12
CH.5 Probability
KIM
HW: pg.351 # 8,9,13,14,15,18
Daily Problem
February 9th
/2
Name:
Textbook pg. 353 #17, 19
17. A computer manufacturer knows that, n a box of 100 chips, 3 will be defective. Jocelyn will draw
2 chips, at random, from a box of 100 chips.
a) Draw a tree diagram to represent this situation.
b) Determine the probability that exactly 1 chip will be defective.
19. Savannah’s soccer team is playing a game tomorrow. Based on the team’s record, it has a 50%
chance of winning on rainy days and a 60% chance of winning on sunny days. Tomorrow, there is a
30% chance of rain. Savannah’s soccer league does not allow ties.
Determine the probability that Savannah’s team will win tomorrow.
8