1. By: Andrew, Oscar, Gladys, Aneesa Mentor: Patrick
Probability
By: Andrew, Oscar, Gladys, Aneesa
Mentor: Patrick

2. Definition of Probability :
The possibilities of interest over the number of total possibilities.
Possibility of interest
Total Possibilities 1

3. The Problem?
Monty Hall
There are 4 cards in front of you and one has the ace underneath it. Your goal is to correctly guess the location of the ace. You make your initial guess, and the host flips over two cards he knows are not the ace. Now there are only two cards left: the one you chose and the one you didn’t choose. He asks you to either stay or switch. Which is the better choice?

4. MONTY MAUL PROBLEM
The Monty Maul Problem is like the Monty Hall problem but the host doesn’t know where the ace is, and flips 2 random cards.
If the host flips an ace then they redo the whole thing.

5. Experimental Data Host Doesn't Know Host Knows Win Lose Switch 7 2
Stay 1 10
Host Doesn't Know
Win
Lose
Switch 6 3
Stay 5

6. Experimental Graph: Win

7. P(B|A)= P(A|B) X P(A) P(B)
Bayes Law states that the probability of something happening (B) given an event before (A) is expressed by the equation:
P(B|A)= P(A|B) X P(A)
P(B)
This equation is very useful in finding the percentage rate of events happening in games, including Monty Hall, Crawl, and Maul.
Bayes Law

8. Bayes’ Law How to read P= probability of | = given ( this| that)
In P(B|A), the first variable is the probability of what you wan to know and the second is the probability of what you know.

9. The Problem?
Monty Hall
There are 4 cards in front of you and one has the ace underneath it. Your goal is to correctly guess the location of the ace. You make your initial guess, and the host flips over two cards he knows are not the ace. Now there are only two cards left: the one you chose and the one you didn’t choose. He asks you to either stay or switch. Which is the better choice?

10. Monty Hall P(A|B)= 1x¼ 1 P(A|B)=¼ OR… The Solution P(A|B)= P(B|A)xP(A)
A= the probability of picking the ace initially
B= the host flipping over two cards and not hitting the ace
P(A|B)=
1x¼ 1
P(A|B)=¼
OR…
Following Bayes Law, we combined the equation p(A|B) x p(B) and p(B|A) x p(A), then isolated p(A|B), the probability of flipping two cards and not hitting the ace given the possibility of finding the ace on the first try

11. MONTY MAUL PROBLEM
The Monty Maul Problem is like the Monty Hall problem but the host doesn’t know where the ace is, and flips 2 random cards.
If the host flips an ace then they redo the whole thing.

12. THE SOLUTION A= the contestant picked the ace
B= the host flips and ace
What we want to know: p(A| not B)
What we already know:p(B | A)=0
p(B | not A)= 2/3
p(A)= ¼
p(B)= ½
p(not B | A)= 1
p(not B | A)p(A)= 1 x ¼ = ½
p(B) ½
There is a ½ chance that staying with your original card and winning. T

13. Probability Zombie Flu
OH NO!! You just found out you have a fever and you heard that the first symptom of the Zombie Flu is FEVER!!
Now you have to find out what the probability of you having the Zombie Flu is since you have A Fever!!!
The chance of having Zombie Flu is 1 in 10,000.
During the winter a person WITH Zombie Flu has a 90% chance of having a fever
A person WITHOUT Zombie Flu has a 5% chance of having a fever.

14. Solution P(Z|F)= P(Z) P(F|Z) P(F)
So in conclusion, the chance of you becoming a Zombie because you have a fever is a 9/5009.
P(f)= possibility of having a fever
P(z)= possibility of being a zombie
P(F)= possibility of having a fever
P(Z)= possibility of becoming a Zombie

15. The Family Probability Game!
You ask a mother of two children if one of her children are a girl. She responds yes. What is the probability that both of her children are girls?
P(B|A)= P(A) X P(A|B) 1X¼ = ¼x4/3=⅓
P(B) ¾
A= the probability of one child being a girl
B= the probability of both children being girls