Chapter 3
Calculate the probability that the sum of the two dice is equal to 7 or that one of the dice shows a 3. i.e., P(A or B). Calculate the probability that the sum of the two dice is equal to 7, given that one of the dice shows a 3. i.e., P(A | B). Calculate the probability that one of the dice shows a 3, given that the sum of the two dice is equal to 7. i.e., P(B|A). The probability that the sum of the two dice is equal to 7 is 0.167. Let C be the event that the rst die shows a 2 or less. Then P(C | A) = 2/6 = 1/3 = 0.333. Find P(A and C).
1. First, A∩B = {(3; 4); (4; 3)} P(A∩B) = 2/36 P(AUB) = P(A) + P(B) - P(A∩B) = 15/36 B/c 6/36+11/36-2/36 2. P(A|B) = P(A∩B) / P(B) = (2/36) / (11/36) = 2/11 3. P(B|A) = P(B∩A) / P(A) = (2/36) / (6/36) = 1/3 4. P(A and C) / P(A) = P(C|A) P(A and C) = P(C|A) x P(A) P(A and C) =(2/6) x (0.167) = 0.5566 (1/18)
One bag contains 4 white marbles and 3 blue marbles, and a second bag contains 3 white marbles and 5 blue marbles. One marble is drawn from the rst bag and then placed unseen in the second bag. Let B1 be the event that a BLUE marble is drawn from bag 1. Let B2 be the event that a BLUE marble is drawn from bag 2. Let W1 be the event that a WHITE marble is drawn from bag 1. Let W2 be the event that a WHITE marble is drawn from bag 2 1. What is the chance of drawing a blue marble from the second bag given a blue marble was drawn from the rst bag? 2. What is the chance of drawing a blue marble from the second bag given a white marble was drawn from the rst bag? 3. What is the probability or drawing a BLUE marble from bag 1? 4. If you are told that you win $ 50 if you draw a BLUE marble from the second bag, then what are your chances of winning? Based on this scenario what are your chances of losing? 5. What is the probability that a WHITE marble was drawn from bag 1 and then a BLUE marble is drawn from bag 2?
1. The question implies finding (B2|B1), which from the figure is 6/9 =2/3 = 0.6667 2. The question implies finding P(B2|W1), which from the future is 5/9 = 0.5556 3. From the above figure, P(B1) = 3/7 =0.4286 4. This can go Branch 1 or Branch 3 -Branch1: P(B2 and B1) = P(B2|B1) x P(B1) = (6/9)(3/7) =2/7 =0.2857 -Branch3: P(B2 and W1) = p(B2|W1) x P(W1) = (5/9)(4/7) =20/63 =0.3175 P(B2) = P((B2andB1) or (B2 and W1)) = P(B2 and B1) +P(B2 and W1) =0.6032 -- P(losing) = 1-0.6023 = 0.3968 5. p(B2 and W1) = 45/79 = 20/63 = 0.3175
Of 10 births, what is the probability of getting 8 females and 2 males?
10! / (8!)(2!) = 45 ^^ n! / (n-x)!x!
Based on T(108) blood of 100 people 13) Prob of randomly selecting someone not in group A? 14) Prob of randomly getting someone is is type Rh-? 15) Prob of randomly getting someone who is group A or Rh-? 16) Prob of randomly getting someone in group A or group B? 17) Prob of randomly getting someone not type Rh+? 18) Prob of randomly getting someone in group B or Rh+ 19) Prob of randomly getting someone in group AB or type Rh+ 20) Prob of randomly getting someone in group A or O or Rh+
13) P(A) = 40 / 100 = 0.4 P(A') = 1-0.4 = 0.6 14) P(Rh-) = 14/100 = 0.14 15)P(A or Rh-) = P(A) + P(Rh-) P(A or Rh-) = 40 + 9 = 49 = 49/100 = 0.49 16) P( A or B) = P(A) + P(B) P( A or B) = 40 + 10 = 50 = 50/100 = 0.5 17) P(Rh+) = 86 = 0.86 P(Rh+') = 1 - 0.86 = 0.14 18) P(B or Rh+) = P(B) + P(Rh+) P(B or Rh+) = 10 + (39+35+4) = 88 = 0.88 19) P(AB or Rh+) = P(AB) + P(Rh+) P(AB or Rh+) = 5 + (39+35+8) = 87 = 0.87 20) P(A or O or Rh+) = P(A) + P(O) + P(Rh+) P(A or O or Rh+) = 40 + 48 + 12 = 100 = 1
Five couples give birth to five children. How many gender differences are possible?
2C1 Each box is independent so = 2C1 x 2C1 x 2C1 x 2C1 x 2C1 = (2C1)^5 = (2)^5 = 32
Calculate 3! , 5!, 6!
3! = 3x2x1 = 6 5! = 5x4x3x2x1 = 120 6! = 6x5x4x3x2x1 = 720
Want to do a drug test patient by patient to evaluate effects, 30 people volunteer but you only have 20 samples of the drug, how many different sequences of the volunteers are possible? nPr or nCr?
30! / (30-20)! = 30!/10! = 7.31x10^25 nPr
You are assigned to travel to all 50 states, but can only travel to 4 in the timeframe given, how many possible routes can you take? Is this nCr or nPr?
50! / 50-4! = 50! / 46! =5,527,200 This divides the number of possible ways by the number we want to elect of them nPr
Based on T(91), what is the probability of selecting a subject who is pregnant or had a positive rest result
- P(Pregnant) = 85; P(positive result) = 83 P(A)+P(B)-P(AandB) = 85+83-80 - 88/99 = 8/9
Based on T(108), if one of the titanic passengers is randomly selected, find the probability of getting a woman or something who did not survive
- P(woman) = 422 (318 + 104) - P(dying) = 1413 (1360 + 35 + 18) P(WorNS) = P(woman) + P(dying) = 1413 + 422 = `1835 / 2223 = 0.825
Mendel had pea experiments, 428 green peas and 152 yellow pea. Probability of getting an offspring that is green? Is it close to the 3/4 that was expected?
- S = 428 + 152 = 580 G = 428/580 = 0.738 Yes this is close to the 3/4 (its almost 0.75)
Describe the types of compound events: - Union - Intersection
-Union is a collection of two elements of each event without repeats. - It uses the vocab "or" - "AUB" or "A or " are valid. - A or B or both events occur so it goes wider e.g "I'm looking for the number of students who are girls or received an A grade" (Much larger pool) - Intersection is a collection containing common elements to each of the events - It uses the vocal "and" - "A∩B" or "A and B" are valid. - It includes an additional sample to be considered so it goes narrower (specific) e.g "I'm looking for the number of students who are female *and* received an A grade" (Much smaller pool)
P20 Two fair dice are tossed, and the following events are defined: A: Sum of numbers showing is odd B: Sum of the numbers showing is 9, 11, or 12 Then, which of the following statements will hold true? A) events A and B are independent B) events A and B are dependent C) events A and B are mutually exclusive D) events A and B are neither mutually exclusive nor independent
ME is an event where outcomes cannot simultaneously happen Ind means two events do not influence eachother (vv for dependent) Not ME because being odd and 9,11 can happen, so both outcomes can occur. Dependent because outcome of one sum does impact the outcome of another sum
If operation 1 can be performed N1 ways, and operation 2 can be performed N2 ways, how do you calculate the number of ways both operations can be performed?
N1 x N2
Describe odds and odds ratio Interpretation of these values? Do these values indicate probability or not? Interpret Odds ratio of 1.571
Odds (in favour) of an event is defined as the ratio of the probability of event to the probability not observing an event. Let: - P(t) = Risk of disease (Treat) = a / (a+b) - P(c) = Risk of disease (Placebo = c / (c+d) Then: - Odds of disease (Treat) = p(t) / [1-p(t)] - Odds of disease (Placebo) = p(c) / [1-p(c)] ^^ These aren't impactful by themselves, but when used to compare in an odds ratio they are Also: - Odds ratio can be calculated in multiple ways 1. Odds of Disease(Treat) / Odds of disease (Plac) 2. (a x d) / (b x c) ^^ Odds of observing a positive test result in a person with the disease They do not indicate probability, and only occur under binary events (win/lose) Odds in favour of ac column are 1.57 times higher than in bd row when compared to cd row
Describe the more theoretical relationship between sample space , events and compliments What is a null set?
Sample space = 1 (100% that something in it will occur) Probability ranges from not occurring (0) to certainly occurring (1) The event probability is impossible to go beyond 1 (can't be more likely than the sample space A null set is an empty set, it is the compliment of S, (S') as it is the probability of nothing occurring (is therefore 0) Finding a compliment is merely 1-P(X) and vice versa for event Everything together should add up to 1
What is the sample space? Event? Compliment? Compound event?
The collection of all the possible outcomes in an experiment. It is unconditional. A specific outcome of interest that occurs within the sample space (like rolling a 6 with a single fair die).An event is a collection of ways an outcome can occur, such as having two F and one M (FFM, FMF, MFF) since multiple ways can occur it is an event. The opposite of the event (not getting a 6 with a die - getting a 1,2,3,4,5) The probability after mixing two events together ( flipping a coin and rolling a fair die) Simple event is an event that cannot be broken down further, like giving birth (M or F) Sample space is all simple events The #A/#S requires that each simple event has equal chance of occurring (like in die, birth, coin tosses) ^^ If they are not equal must make estimations and educated guesses (subjective probability) like guessing if it's going to rain tomorrow, can just divide interest group by total
What are conditional events? Changes to the sample space?
The equation makes sense given ven diagrams, you find overlap and divide by B overlap to get whats left in A ^^Get image where two ven circles overlap then keep the entire one circle with overlap and kill other one with chunk out
What would an odds ratio of 0.3 indicate?
The odds of polio in the treatment group is 1/3 of the odds of polio in the placebo group The odds of polio in the placebo group are x3 higher than in the treatment group
Describe the theory behind disease and screening Perfect test gives (+) result, reliability? 0.1 Error test gives (+) result, reliability?
There is one truth, a patient either has the disease or the patient does not have the disease We never know this truth, but use diagnostic tests to see. These tests have binary outcomes: positive and negative. We can know this truth, and use these to know the first truth on whether there is a disease or not. We want to go from test to the truth, but there is often error in the test itself - 100% reliability -99% reliability The line where we deem something unreliable is very grey
Rules nCr?
Totally n different items are available Can select row the n items (without replacement) Rearrangements of the same items must be the same (ABC = CNA) where 0-<r-<n and and both r and n being whole numbers
What is Bayes Theorem? When do you use it?
When given P(A|B) but asked about P(B|A) So you can be given P(A|B), P(A), and may even know P(A) = 0 P(B|A) = [P(A|B) x P(B)] / ([P(A|B) x P(B)] + [P(A|B') x P(B')])
Given below is data corresponding to viewers of the evening news from a certain county in Texas a) What is the probability of selecting a Republican viewer? b) What is the probability of selecting a viewer who watches the CBS evening news? c) What is the probability of selecting a Democrat who watches the NBC evening news? d) What is the probability of selecting a Republican who watches the CBS evening news? e) What is the probability of selecting a Republican viewer or a CBS evening news viewer?
a) 800/1700 =0.47 b) 675/1700 = 0.4 c) 350/1700 = 0.21 d)225/1700 = 0.13 e) A=Republican; B=CBS viewer AUB =P(A)+P(B)- P(A∩B) P(AUB) = 800+674-225 = 25/34 f) B=democrat; A=NBC viewer = 350/900 (=350/1700 / 900/1700)<<Cancel 1700 out
The probability of getting a multiple choice question wrong is 0.8. If a weekly quiz has three multiple choice questions (assume independence) then a) What is the probability that you will get a perfect score in the quiz? b) What is the probability that you will get a zero in the quiz?
a) Pw = 0.8; Pc = 0.2 Pperf=(0.2)(0.2)(0.2) = (0.2)^3 = 0.008 = 8e-3 b) Pzero=(0.8)(0.8)(0.8)=(0.8)^3 = 0.512
A single fair coin is tossed: a) what is the sample space? Elements in the sample space? b) number of elements in observing a head? c) probability of observing a head? d) number of elements in observing a tail? e) probability of observing a tail?
a) S = {H,T}, #S = 2 = 2^1 (*one* coin flip, a coin with *two* total possible outcomes) b) #A = 1 (only one way to only see a head, no combinations) c)P(A) = #A / #S = 1 / 2 = 0.5 d) Same as (b) e) Same as (c)
A single fair toss of two coins a) what is the sample space? Elements in the sample space? b) number of elements in observing two tails? c) probability of observing two tails? d) number of elements in observing one head? e) probability of observing one head? f) number of1 elements in observing at most one tail? g) probability of observing at most one tail? how can this be reworded?
a) S = {HH, TT, HT, TH}; #S = 4 = 2^2 = 2^1 x 2^1 b) {TT}; #A = 1 (theres only one way to get this combo) c) P(A) = #A / #S = 1 / 4 = 0.25 d) {TH, HT}; #B = 2 (two different ways to get this combo) e) P(B) = #B / #S = 2 / 4 = 0.5 f) {TH, HT, HH}, #C = 3 (three different ways to get this combo) g) P(C) = #C / #S = 3 / 4 = 0.75 ^^ This is pretty much the compliment of event A, the prob of not observing event A
A single toss of two fair die a) What is the sample space? elements in the sample space? b) Elements and probability in observing two ones? c) Elements and probability in observing a 1 in at least one of the die d) Elements and probability in observing a 1 in one die only
a) { (1,1)...(1,6), (6,1)...(6,6)}; #S = 36 = 6^2 (two die, with six outcomes on each die) b) {1(,1)}; #A = 1 (only one way to get this combo) P(A) = #A / #S = 1 / 36 c) {(1,1)...(1,6), (6,1)...(6,6)}; #B = 11 (one fewer than 12 because (1,1) only appears once in context, while the others - (1,2)//(2,1) appear twice; it just falls 1 short) P(B) = #B / #S = 11 / 36 d) {(1,2)...(1,6) (6,1)...(6,6)}; #C = 10 (one fewer than in the previous question, as now we don't count the (1,1) as it has a 1 from both die. -1 gives us 10) P(C) = #C / #S = 10 / 36
Given the table on the other side, find: a) P(D) b) P(D') c) P(T+) d) P(T-) e) P(T+|D) f) P(T-|D) g) P(T+|D') h) P(T-|D') Interpret - P(D) - P(T+|D) - P(T-|D')
a. 200 / 100,000 = 0.002 b. 99,800 / 100,000 = () or 1-P(D) c. 8000 / 100,000 = 0.08 d. 92000 / 100,000 () or 1-P(T+) e. 170 / 200 f. 30 / 200 g. 7830 / 99800 h. 9170 / 99800 P(D) is the prevalence of the disease P(T+|D) is called the sensitivity of the screening test P(T-|D') is called the specificity of the screening test
A box contains five $25, three $50, two $100 gift cards from Best Buy, and 10 blank cards. If you are given the chance of drawing one card from this box then a) what is the probability that you draw the $25 gift card? b) what is the probability that you draw the $50 gift card? c) what is the probability that you draw the $100 gift card? d) what is the probability that you draw a blank card? e) what is the probability that you draw a gift card greater than or equal to $25? f) what is the probability that you draw a gift card greater than $25?
a. 5/20 = 1/4/ 0.25 b 3/20 = 0.15. c. 2/20 = 1/10 = 0.1 d. 10/20 = 1/2 = 0.5 e. 5+3+2 /20 = 10/20 =1/2 = 0.5 f. 3+2 /20 = 5/20 = 1/4 = 0.25
Company A makes 80% of product (4% defect), Company B makes 15% of product (6% defect), and Company C makes 5% of product (9% defect). a) What is the probability of picking a random product and it being made by Company A? b) If randomly pick a product and it was defective, what is the probability that it was made by Company A?
answer in notes
Detection device is claimed to be 2% lower defect that competing brands. Makes 5000 of the devices and randomly takes 15 of them. None of them had defects. What is this probability. Is there strong evidence to support the claim?
0.739; no
What is number needed to treat? Interpret a value of 2458
1 / | Pt - Pc | Need to vaccinate 2458 people to prevent one from getting the disease
What is the probability that all the babies are female?
(1/2)x(1/2)x(1/2)x(1/2)x(1/2)x(1/2) = 0.015625 or 6C6 = 1 therefore 0.015625 x (1) = 0.015625
Based on P(120), What is the probability a negative test result given that the subject is not pregnant? Is this value trustable?
(=11/14; 0.786; get another test)
Odds ratio for facial injuries in no helmet compared to group that had helmets? Does wearing a helmet appear to decrease the risk of facial injuries? Explain
(=2.13, The risk of facial injuries for those not wearing helmets is 2.13 times the risk for those not wearing helmets) Yes because of the answer in 15
Given that a couple has three children (equal prob of either sex), find: a) Among three, there is exactly one girl b) Among three, there are exactly two girls c) Among three, all are girls
- #S=8, #A = {bbg, bgb, gbb}, = 3/8 - #S=8, #B = {bgg, gbg, ggb} = 3/8 - #S=8, #C = {ggg} = 1/8
a) Find P(A') given that P(A) = 0.0175 b) A poll showed that 61% of Americans they believe in aliens. What is the probability of picking an American who does not believe in aliens?
- 1 - 0.0175 = 0.9825 - P(A) = 0.61 - P(A') = 1 - 0.61 = 0.39
a) If P(A) = 0.05, find P(A') b) Women have 0.25% rate of colourblindness. If a woman is randomly selected, what is the probability that she does not have it?
- 1 - 0.05 = 0.95 - P(blind) = 0.0025 - P(blind') = 1 - 0.0025 = 0.9975
Q. Prob of a woman having three kids, 2 of them being boys, and one girl?
- 8 combinations (just in general for 3 kids) - only 3 of the 8 meet the 2:1 parameter so 3/8
Based on T(91) a) Find the probability of a wrong test conclusion for a woman who is pregnant b) Is it unusual for the test conclusion to be wrong for women who are not pregnant?
- False negative = 5/85 = 0.0588 - False positive = 3/14 = 0.214. This value is larger that 0.05, so it is not unusual.
Difference between absolute and relative risk?
Absolute risk of a disease is your risk of developing the disease over a time period. We all have absolute risks of developing various diseases such as heart disease, cancer, stroke, etc. The same absolute risk can be expressed in different ways. For example, say you have a 1 in 10 risk of developing a certain disease in your life. This can also be said to be a 10% risk, or a 0.1 risk - depending on whether you use percentages or decimals. Relative risk is used to compare the risk in two different groups of people. For example, the groups could be smokers and non-smokers. All sorts of groups are compared to others in medical research to see if belonging to a group increases or decreases your risk of developing certain diseases. For example, research has shown that smokers have a higher risk of developing heart disease compared to (relative to) non-smokers.
Disjoint/Non disjoint: P(AorB) = ?
Disjoint (mutually exclusive) is when two events cannot occur at the same time If disjoint then P(A or B) = P(A) + P(B) If not disjoint then P(A or B) = AUB = P(A) + P(B) - P(A∩B) ^^ In the equation above the P(A∩B) is the overlap in the van diagrams, and recall since definitions you don't want anything to be repeated, you take them away
Formal multiplication rule?
For (A∩B) 1. If independent = P(A) x P(B) 2. If dependent = P(A) x P(B|A)
Company B is looking to hire technicians for two departments. Suppose 50 candidates applied for the position at both departments a. How many different choices for IT-techs for the two departments are possible? nCr or nPr?
Having Natasha in D1 and Anthony in D2 is different from having Anthony in D1 and Natasha in D2 So it's nPr = 50P2 = 2450
The number of ways of getting a three-digit number using 1,5,7,9. A number used once cannot be used again
If our first digit is 1, the second cane 5,7, or 9 (three ways), leaving 2 ways for the third box With 1 fixed, there are 6 ways Since 1 was arbitrary, we could have used 1,5,7, or 9. (four different starts) So 4 x 6 = 24
P(A|B)?
If something already happened, whats the chance of x happening? A|B means "A given B", meaning that with the requirement that B already occurred, whats the prob of getting A? e.g. " Given that I failed 2 of 3 midterms, what is the probability I will be able to pass the course?"
Based on the table on the other side, calculate the risk of polio Define the variables Find P(Polio|Salk Vaccine); P(Polio|Placebo) Limitations of risk? Absolute and Relative risk?
In what? Needs to specify in what group. E.g. "What is the risk of polio in the treatment group p(t) = Risk of polio in salk vaccine group p(c) = Risk of polio in the placebo group This is a risk: P(Polio|Salk Vaccine) = 33 / 200,745 = 0.0001644 This is a risk: P(Polio|Placebo) = 115/ 201229 = 0.0005615 AR = | Pt - Pc | = 0.000407 RR = Pt / Pc = 0.287
Independent: P(AandB) = ? P(A|B) = ? Disjoint P(AandB) = ? P(A|B) = ?
Independent = one event has no impact on another (rolling two fair die) Second bullet should be P(A|B) = P(A) Mutually = Should say "aka disjoint events" No common ground between two sets so prob of intersection is 0 - Final bullet should have P(A|B) = P(B|A) = 0 ^^Two ven diagram circles that are completely separate
Rules nPr?
Must have all items be different No replacement Order is taken into account
Testing new drug, we have 30 volunteers and 20 samples of the drug. If 20 subjects are selected from the 30 that are available, and the 20 selected are all treated the same, how many different treatment groups are possible?
Order does not count so we want combinations n = 30 available r = 20 we are interested inn nCr = 30! / (30-20)!20! = 30,045,015
P20 For two events, A and B, we have that P(A) = 0.8 , P(B) = 0.7 , P(A∩B) = 0.6. What is P(B|A)?
P(B|A) = P(B and A) / P(A) = 0.6 / 0.8 = 0.75
P20 Home modification for wheelchair users: The table classifying a sample of 306 wheelchairs users according to type of features installed in the home and whether or not they had an injurious fall follows. Suppose we select, at random, one of the 306 surveyed wheel chair users, then given that the wheelchair user had none of the features installed, what is the probability that the user had an injurious fall?
P(F|0) = P(F∩0) / P(0) P(F|0) = (20) / (109) P(F|0) = 0.183
Based on P(111), what is the probability of picking a green pod? What if you did not put the pod back and went to pull a yellow pod?
P(green pod) = 8/14 P(yellow pod) = 6/13 Both occurring back to back = 8/14 x 6/13 = 0.264
What types of ratios would a prospective study use? A retrospective study?
Pro = Risk and odd Retro = odds
Describe the theoretics behind a disease tree diagram, draw one (Tree diagram is from disease example) How to calculate branch 1? How do you calculate branch 3? What is P(D|T+)? Interpret the value ^^ Can you do this with specificity instead of prev?
Refers to the stage 1 being truth and stage two being outcomes from (+) test and (-) test Use Bayes theorem when going from S2 --> S1 (right to left) Branch 1 = multiple down the branch = P(D) x (P(T+|D) = 170 / 1000 Branch 3 = P(D') x P(T+|D') = (7830 / 9980) x (9980 / 100,000) = 7830 / 100,000 (similar cancels out) P(D|T+) is a reverse condition so use Bayes! - B1 = P(D) x (P(T+|D) - B3 = P(D') x P(T+|D') () P(D|T+) = B1 / (B1 + B3) = 170 / (170+8730) = 0.02125 ^^ Given a person's test returns positive for the disease, there is a 2.125% chance that the person has the disease. This is the positive predictive value (PPV)
Describe risk, absolute risk, and relative risk Interpretation of these values? Do these values indicate probability or not? Difference between absolute and relative risk? What would a RR of 0.287 mean (inverse = 3.48)
Risk of disease (Treat) = a / (a+b) = P(T+|Treated) Risk of disease (Placebo = c / (c+d) = P(T+ | Placebo) ^^ These aren't impactful by themselves so need to compare between them to have a better understanding Absolute risk = | [a / (a+b)] - [c / (c+d)] | Relative risk = [a / (a+b)] / [c / (c+d)] The basic risk assessments indicate probabilities, they measures the chance of observing a condition in a group under a certain treatment Absolute and relative are not probabilities; relative is a ratio RR=0.287 means that 0.287 less likely to get disease in Pt than in Pc. Inversely, people without the vaccine are 3.48 times more likely to get the disease
Consider the experiment of tossing two fair dice. a) Give number of elements in the sample space of this experiment. b) Which of the following is not a possible outcome for this experiment: (6,6), (0,6), (5,5), (2,3) c) In how many ways can you observe the event of a pair whose sum is greater than or equal to 10? d) What is probability of observing a pair whose sum is greater than 10? e) What is the probability of observing a pair whose product is less than 12?
a) #S = 36 = 6^1 x 6^1 b) (0,6) as can't have a "0" value on a die c) {(5,5), (5,6), (6,6), (6,5), (6,4),4,6)} = 6 ways d) P(A) = #A/#S = {(5,6), (6,6), 6,5)} / 36 = 3/36 = 1/12 e) P12 = {6,6} = 1/36 P12' = 36/36-1/36 = 35/36 = 0.97
Sample of 20 Babies 10 Girls 10 Boys a) If twenty newborn babies are randomly selected, how many different gender sequences are possible? b) How many ways can 10 boys and 10 girls be arranged in a sequence? c) What is the probability of getting 10 boys and 10 girls?
a) (2C1)^20 = 2^20 = 1048576 b) 20C10 = 184756 c) # ways 10b and 10g / # gender sequences = 0.176197
Based on T(116),Two of one hundred people are randomly selected. Find the probability that they both have group O blood a) Assume the first selected subject is replaced b) Assume that the first selected subjects is not replaced
a) (= 0.203) b)(=0.2)
20% HIV rate for "at risk" population. Screening is correct 95% of the time. a) Probability of selecting a person with HIV and tested positive? b) Probability that the person tests positive given they have HIV
a) (=0.679) b) P(T+|D) = P(T+ and D) / P(D) P(T+|D) = (0.19) / 0.2 = 0.95
There are 10 RPG games and 5 indie games. If you are allowed to pick 5 games, then how many ays are there to select 3 RPG's and 2 indie games? nPr or nCr?
nCr # ways of choose 3 RPG games is 10C3 = 120 #ways of choosing 2 indie games is 5C2 = 10 Total number of ways is 120x10 = 1200
nCr vs nPr?
nCr = Combination of 'r' objects chosen from 'n' objects. Disregarding arrangement or order nCr = n! / r!(n-r)! nPr = Permutation of 'r' objects from 'n' objects. Arranging "r" from "n" objects nPr = n! / (n-r)! = (r!)(nCr)
How to distinguish nPr and nCr?
nPr = Different orderings of the same item when counted separately nCr = Different orderings of the same items when not counted sperately
