Statistics Probabilities Chapter 4 Example Problems
Slide 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
Slide 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 1st2nd3rd boy girl boygirlboy girl boy
girlboygirl boy girl
Slide 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 1st2nd3rd boy girl boygirlboy girl boy
girlboygirl boy girl
Slide 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 1st2nd3rd boy girl boygirlboy girl boy
girlboygirl boy girl
Slide 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?
Slide 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 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.
Slide 8
Acceptance Sampling Multiply (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)
Slide 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.
YesNo Positive12018 Negative21141
Slide 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 YesNo
Positive12018 Negative21141
Slide 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*98down to 1? Calculator
Time!
Slide 12
Factorials To find 6! Type 6 Press the MATH button Arrow over
to PRB Arrow down to ! Press Enter You should see 720
Slide 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
Slide 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
Slide 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
Slide 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?
Slide 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
Slide 18
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.
Slide 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 = 1048576
Slide 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 / 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.