Statistics Probabilities Chapter 4 Example Problems

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

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

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

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

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?

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 1561 – 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.

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 = 0.238 (rounded to nearest thousandth) (1352/1561)^10 in calculator (find the ^ key)

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

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

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!

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

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

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”

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

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?

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

Permutations And you should get 1585080 and you can only win one way, so the probability is 1/1585080.  Long way would be P(37,4) = 37! / (37-4)! = 37! / 33! = 37*36*35*34*33!/33! = 37*36*35*34 = 1585080.

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 = 1048576

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 / 1048576 = 0.0001812 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.