Intro to Possibility 2

By L.D.With the help of Ms. Becker

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

ForewordO Watch the slideshow “Intro to

Probability” on my blog before doing this.

Problem 1O 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

Problem 1: Question 1O What is the probability of getting an

odd numbered yellow ball?

Red Blue Yellow

Even # 6 8 10

Odd # 11 7 8

Problem 1: Question 1O 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

Problem 1: Question 1O 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

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

4/25

Problem 1: Question 2O What is the probability of getting an

even red ball?

Red Blue Yellow

Even # 6 8 10

Odd # 11 7 8

Problem 1: Question 2O 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

Problem 1: Question 3O What is the probability of getting an

odd number?

Red Blue Yellow

Even # 6 8 10

Odd # 11 7 8

Problem 1: Question 3O 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

Problem 1: Question 4O 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

Problem 1: Question 4O To find this we add up everything

that is not odd or blue placing it over the denominator.

O 6 + 10 = 16O 16/50 = 8/25Red Blue Yellow

Even # 6 8 10

Odd # 11 7 8

What are odds?O A ratio that compares the number of

variable outcomes to unfavorable outcomes.

Problem 2O There is a 25% chance of rain this

weekend. What are the odds in favor of rain?

Problem 2O 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 = 75O Since 75 is the left over, it is unfavorable

(what we don’t want), we will place both those numbers in a fraction.

Problem 2O 25/75 can also be read as 25:75

which can be reduced to 1:3.

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