1. Intro to Possibility 2
By L.D.
With the help of Ms. Becker

2. Table of Contents
O Slide 3: Foreword
O Slide 4: Problem 1
O Slide 5: Problem 1: Question 1
O Slide 9: Problem 1: Question 2
O Slide 11: Problem 1: Question 3
O Slide 13: Problem 1: Question 4
O Slide 15: What are odds?
O Slide 16: Problem 2

3. Foreword
O Watch the slideshow “Intro to Probability”
on my blog before doing this.

4. Problem 1
O A bag contains numbered balls in red,
blue and yellow, the table below shows
the number of each type of ball.
Red Blue Yellow
Even # 6 8 10
Odd # 11 7 8

5. Problem 1: Question 1
O What is the probability of getting an odd
numbered yellow ball?
Red Blue Yellow
Even # 6 8 10
Odd # 11 7 8

6. Problem 1: Question 1
O Since we know that there are 8 odd yellow
balls we have to find out what the
denominator would be by adding up all the
other balls and putting it into the possibility
formula.
O 6 + 8 + 10 + 11 + 7 + 8 = 50
Red Blue Yellow
Even # 6 8 10
Odd # 11 7 8

7. Problem 1: Question 1
O Since we know that there are 8 odd yellow
balls we have to find out what the
denominator would be by adding up all the
other balls and putting it into the possibility
formula.
O 6 + 8 + 10 + 11 + 7 + 8 = 50
Red Blue Yellow
Even # 6 8 10
Odd # 11 7 8

8. Problem 1: Question 1
O Our last answer on this is 8/50 or 4/25

9. Problem 1: Question 2
O What is the probability of getting an even
red ball?
Red Blue Yellow
Even # 6 8 10
Odd # 11 7 8

10. Problem 1: Question 2
O To answer this we will do the same thing
we did with the last problem, the
denominator is the same so the answer is
6/50 or 3/25.
Red Blue Yellow
Even # 6 8 10
Odd # 11 7 8

11. Problem 1: Question 3
O What is the probability of getting an odd
number?
Red Blue Yellow
Even # 6 8 10
Odd # 11 7 8

12. Problem 1: Question 3
O We just need to add up all the “odds” and
put them over our denominator.
O 11 + 7 + 8 = 26
O 26/50 = 13/25
Red Blue Yellow
Even # 6 8 10
Odd # 11 7 8

13. Problem 1: Question 4
O What is the possibility of not getting a ball
that is odd and blue?
Red Blue Yellow
Even # 6 8 10
Odd # 11 7 8

14. Problem 1: Question 4
O To find this we add up everything that is
not odd or blue placing it over the
denominator.
O 6 + 10 = 16
O 16/50 = 8/25Red Blue Yellow
Even # 6 8 10
Odd # 11 7 8

15. What are odds?
O A ratio that compares the number of
variable outcomes to unfavorable
outcomes.

16. Problem 2
O There is a 25% chance of rain this
weekend. What are the odds in favor of
rain?

17. Problem 2
O There is a 25% chance of rain this
weekend. What are the odds in favor of
rain?
O Since we are comparing the odds we
must think of what is leftover if you take
25% out of 100% (remember that
possibility must be between 0 and 1).
O 100 – 25 = 75
O Since 75 is the left over, it is unfavorable
(what we don’t want), we will place both
those numbers in a fraction.

18. Problem 2
O 25/75 can also be read as 25:75 which
can be reduced to 1:3.