1. Statistics Probabilities
Chapter 4
Example Problems

2. Probability Rules Probabilities must be between 0 and 1
Where 0 is impossible
Where 1 is certain
No negative probabilities
All probabilities must add to 1

3. Finding Probabilities
What is the probability of exactly 0 girls out of the 3?
I see only 1 way for this to happen out of a total of 8 outcomes or 1/8
1st
2nd
3rd
boy
girl

4. Finding Probabilities
What is the probability of exactly 3 girls out of the 3?
I see only 1 way for this to happen out of a total of 8 outcomes or 1/8
1st
2nd
3rd
boy
girl

5. Finding Probabilities
What is the probability of exactly 2 girls out of the 3?
I see 3 ways for this to happen out of a total of 8 outcomes or 3/8
1st
2nd
3rd
boy
girl

6. Acceptance Sampling
With one method of a procedure called acceptance sampling a sample of items is randomly selected without replacement and the entire batch is accepted if every item in the sample is okay.
A company has just manufactured 1561 CDs, and 209 are defective. If 10 of these CDs are randomly selected for testing, what is the probability that the entire batch will be accepted?

7. Acceptance Sampling
A company has just manufactured 1561 CDs, and 209 are defective. If 10 of these CDs are randomly selected for testing, what is the probability that the entire batch will be accepted?
P(entire batch accepted) = # of good CDs / total number of selecting CDs
If there are 1561 CDs and 209 are defective this means – 209 = 1352 are the number of good CDs.
In the denominator we have 1561 total CDs to chose from. This gives the probability that if we chose one CD it would be accepted.
P(choose one and it was good) = 1352/1561
But we want to choose 10, not one.

8. Acceptance Sampling Multiply Or
(1352/1561) (1352/1561) (1352/1561) (1352/1561) (1352/1561) (1352/1561) (1352/1561) (1352/1561) (1352/1561) (1352/1561) Or
(1352/1561)10 = (rounded to nearest thousandth)
(1352/1561)^10 in calculator (find the ^ key)

9. Conditional Probabilities
The data represents the results for a test for a certain disease.  Assume one individual from the group is randomly selected. 
Find the probability of getting someone who tests positive, given that he or she did not have the disease.
Yes No
Positive
120 18
Negative 21
141

10. Conditional Probabilities
Find the probability of getting someone who tests positive, given that he or she did not have the disease.
So you are looking for P(test positive | did not have the disease) = P (test positive AND did not have the disease) / P(did not have the disease) = 18/(18+141) = 18/159 = 0.113
Yes No
Positive
120 18
Negative 21
141

11. Factorials
The ! means factorial and to multiply from the value shown down to 1. For example,
6! = 6*5*4*3*2*1
But what if you had 100!
Do you want to multiply 100*99*98…down to 1?
Calculator Time!

12. Factorials To find 6! Type 6 Press the MATH button Arrow over to PRB
Arrow down to !
Press Enter
You should see 720

13. Permutations / Combinations
Permutations may be written as
46 P 3
Which means 46 objects taken 3 at a time where ORDER MATTERS
Combinations may be written as
46 C 3
Which means 46 objects taken 3 at a time where ORDER DOES NOT MATTER

14. Combinations
Find the probability of winning a lottery with the following rule.
Select the five winning numbers from 1, 2, …, 31 (in any order, no repeats)
Remember to find probabilities you find the number of ways the outcome you are looking for can occur / number of events. So in other words this would be the number of ways you can win the lottery / the number of combinations of numbers you can pick. I know this is a combinations versus permutations because in the question it says “any order”

15. Combinations Thus, I use the formula nCr = n! / (n-r)! r!
  This will give me the number of combinations you can pick, where n = 31 numbers, r = 5 winning numbers
So with your calculator you want 31C5 or 31! / (31-5)! 5!. With the TI calculator the steps would be
Type 31
Press Math
Arrow over to PRB
Arrow down to nCr and press enter
Type 5
Press Enter
You should get 169,911 which is the number of combinations of selecting 5 numbers, which is the denominator in our definition = the number of ways you can win the lottery / the number of combinations of numbers you can pick.
Then, there is only ONE way to win the lottery, thus our final answer would be 1/169,911

16. Permutations
A certain lottery is won by selecting the correct four numbers from 1, 2, …, 37. The probability of winning that game is 1/66,045. What is the probability of winning if the rules changed so that in addition to selecting the correct four numbers you must now select them in the same order as they are drawn?

17. Permutations
Ok, the key is that you are trying to find the number of ways you can win, that is, select the four numbers in order / number of total ways to select any numbers. Because order matters this makes it permutations. We have 37 total numbers and we want to take 4 at a time.
So on your calculator this is 37
Math
Arrow over to PRB
Down to nPr 4
Enter

18. Permutations
And you should get and you can only win one way, so the probability is 1/
 Long way would be
P(37,4) = 37! / (37-4)! = 37! / 33! = 37*36*35*34*33!/33! = 37*36*35*34 =

19. Permutations Given: 20 babies were born and 18 were boys.
Find the number of different possible sequences of genders that are possible when 20 babies are born.
Ok, there can only be two things (we hope) that can occur.  A boy or A girl. So the possible sequences then would be 2 ways for the 1st child AND 2 ways for the 2nd child......*2 ways for the 20th child or 2^20 =

20. Permutations
How many ways can 18 boys and 2 girls be arranged in sequence.
The key word is arranged.  Does order matter?  It didn't say so this is the use of Combinations. 
On the TI-83 (or 84) type 20 and then press the MATH button, arrow to the PRB and down to nCR and ENTER and then type 18 and ENTER.
If 20 babies are randomly selected, what is the probability that they consist of 18 boys and 2 girls?
P(18 boys and 2 girls) = # of ways to get 18 boys and 2 girls / # ways to get 20 kids which we just found 190 / =
Does the gender-selection method appear to yield results that is significantly different from a result that might be expected by random change? 
Certainly because this is a very small probability.