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1.3 Conditional Probability - Coggle Diagram
1.3 Conditional Probability
Motivation Example
Example 1: Tulip bulbs
Settings
Divided into 2 colors
Divided by blooming time
20 tulip bulbs
Experiment 1:
Select 1 randomly
S = {all bulbs}
Assumptions: all bulbs are "equally likely".
Event R: {the selected bulb is red}
P(R) = N(R) / N(S)
Experiment 2: Select one bulb randomly from the bulbs that will bloom early
Sample space: Reduced Sample Space = {all bulbs that bloom early}
Assumption: all bulbs are equally likely
Event R associated with Experiment 2 != Event R associated with Experiment 1
Different Sample Space
P(R | E)
Probability of R given E
N ( R intersection E) / N (E)
New probability function associated with a reduced sammple space E.
E is a subspace of S
Link the probability function with the original subspaces
P(R|E) = N(R|E) / N(E) = N(R intersection E) / N(S) / N(E)/N(S)
P( R intersection E) / P(E) -> Associaated with S
P(R|E) -> Associated with E
Conditional probability of A, given that B has occured
P(A | B) = P(A intersection B) / P(B)
Provided B is larger than 0
:!!:
Conditional Probability is a probability function
assoociated with the
reduced sample space
P( A | B) >= 0
P (B | B) = 1
Like saying N(S)/N(S)
If A1, A2, ... are countable and mutually exclusive events then
P(A1 U A2 U...| B) = P(A1| B) + P(A2 | B)....
Example 3: Shooting Game
Settings
10 Yellow
8 Red
25 Ballons
7 Green
A {the first is Yellow}
B {the second ballon shot is yellow}
Sample space contains finite number of outcomes
"equally likely"
P(the first two balloons are yellow)
Process:
First 2 are yellow = A intersection B
P(A) = 10 /25
P(B | A) = 9/24
Their product gies the answer
1 DIFFICULT PATH: N{desirable} / N{sample space} :hot_pepper:
P(A intersection B) = P(B|A) * P(A)
Key: Multiplication Rules
The probability that 2 events, A and B both occur
Multiplication rule = P(A intersection B) = P({A and B both occur})
P(A intersection B) = P(A)
P(B|A) provided P(A) > 0
P(A intersection B) = P(B)
P(A|B) provided P(B) > 0
Example 4: Bowl of chips
Settings
110 chips
7 blue
3 red
Drawn 2 chips successively at random without replacement, what is the probability that the 1st draw is red and the 2nd draw is blue?
3/10 * 7/9
P(A) * P(B|A)
3/10 * 7/9
"7/9" is the settings after event A has occurred
Multiplication rule for 3 events
A, B C happen at the same time
P(A intersection B intersection C) = P((A intersection B) intersection C)
P(A intersection B) * P(C | (A intersection B))
Example 5 4-sided dice and observ hte sum of hte dice
A = {a sum of 3 is rolled}
B = {a summ of 3 oor a sum of 5 is rolled}
C = {a sum of 3 is rolled before a sum of 5 is rolled}
S = {every pair of outcomes}
Assumption: Each outcome is likely
P(A) = N(A) / N(S) = 2/ 16
2, 1 or 1, 2
P(B) = 6/16
P(C)
Method 1
A. Figure out the simplified random experiment
B. Figure the corresponding sample space and the event
P({a sum of 3 is rolled before a sum of 5 is rolled})
Method 2 (CP)
P(C) = P(A | B)
P(A intersection B)/P(B)
2/16 / 6/16
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