course 152
Yale did not do the text assignment 11.5 and 12.1 is this okay?----Katye B
It's always best to do the assignments, but the bottom line is preparation for the tests. If that's OK, then all's well.
If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution.
If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
006. ``q Query 6
Question: **** **** `q Query 12.1.6 8 girls 5 boys
What is the probability that the first chosen is a girl?
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Your solution:
8 girls + 5 boys = 13 possibilities
Girls =8/13=0.6153846
Boys=5/13=0.384615
Confidence Assessment: 3
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Given Solution:
`a ** Assuming the choice is completely random there are 13 possible choices, 8 of which are female so we have
P(female) = 8 / 13 = .6154, approx. **
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Self-critique (if necessary):
Self-critique Rating:
Question: **** **** `q Query 12.1.12 3 fair coins: Probability and odds of 3 Heads.
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Your solution:
HHH, HHT, HTH, HTT
TTT, TTH, THT, THH
2 Possibilities of each
2*2*2=8
Confidence Assessment:
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Given Solution:
`aThere are 8 equally likely possible outcomes when flipping 3 fair coins. You can list them: hhh, hht, hth, htt, thh, tht, tth, ttt. Or you can use the fact that there are 2 possibilities on each flip, therefore 2*2*2 = 2^3 = 8 possible outcomes.
Only one of these outcomes, hhh, consists of 3 heads.
The probability is therefore
P(3 heads) = # of outcomes favorable/total number of possible outcomes = 1 / 8.
The odds in favor of three heads are
Odds ( 3 heads ) = # favorable to # unfavorable = 1 to 7. **
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Self-critique (if necessary):
&#Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the given solution, and if necessary asking specific questions (to which I will respond).
&#
Self-critique Rating:3
Question: **** **** `q Query 12.1.20 P(pink) from two pink parents (Rr and Rr)
What is the probability of a pink offspring.
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Your solution:
R and r = Red and white genes
Rr or rR- pink offspring
RR=red
Rr=white
R r
R RR Rr
r Rr rr
2/4 possibilities which is the same as 1/2
Confidence Assessment: 3
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Given Solution:
`a The genes R and r stand for the red and white genes.
A pink offspring is either Rr or rR. RR will be red, rr white.
R r
R RR Rr
r rR rr
shows that {RR, Rr, rR, rr} is the set of equally likely outcomes. We season two of the four possible outcomes, rR and Rr, will be pink.
So the probability of pink offspring is 2/4 = 1 / 2. **
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Self-critique (if necessary):
Self-critique Rating:
Question: **** **** `q Query 12.1.33 cystic fibrosis in 1 of 2K cauc, 1 in 250k noncauc
What is the empirical probability, to 6 places, that a randomly chosen non-Caucasian newborn will have cystic fibrosis?
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Your solution:
1/250000= 0.000004
Confidence Assessment: 2
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Given Solution:
`aThere is 1 chance in 250,000 so the probability is 1 / 250,000 = 4 * 10^-6, or .000004. **
12.1.40 Cc genes carrier, cc has disease; 2 carriers first child has disease **** What is the probability that the first child has the disease?
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Your solution:
C c
C CC Cc
c Cc Cc
1/4
CC,Cc,Cc,cc
Confidence Assessment:
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Given Solution:
`aIf cc has the disease, then the probability that the first child will have the disease is 1/4. **
What is the sample space for this problem?
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Your solution:
CC, Cc, Cc, cc
Confidence Assessment:
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Given Solution:
`aThe sample space is {CC, Cc. cC, cc}. **
12.1.60 36 in class, 3 chosen **** What is the probability that the choice will be the given three people in any order?
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Your solution:
P(36,3)
36*35*34=42840
Confidence Assessment:
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Given Solution:
`aThere are P(36,3) possible ordered choices of 3 people out of the 36.
P(36,3) = 36! / (36-3)! = 36! / 33! = (36*35*34*33*32*31*...*1) / (33*32*31*...*1) = 36*35*34=40,000 or so.
The probability of any given choice is therefore 1 / P(36,3) = 1/40,000 = .000025, approx..
For any given set of three people there are six possible orders in which they can be chosen. So the probability of the three given people, in any order, is 6 * probability of a given order = 6 / P(36,3) = 6/40,000 = .00015.
Alternatively we can say that we are choosing 3 of 36 people without regard for order, so there are C(36,3) possibilities and the probability of any one of them is 1 / C(36,3). **
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Self-critique (if necessary):
1/P(36,3)
Because there is a regard for order.
Self-critique Rating:
Question: **** **** `q Query 12.1.75 digits 1, 2, ..., 5 rand arranged; prob even, prob digits 1 and 5 even What is the probability that the resulting number is even and how did you obtain your answer?
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Your solution:
Did not do this one, was studying for test
Confidence Assessment:
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Given Solution:
`aThe number will be even if it ends in 2 or 4. There are 5 possible ending numbers. So 2 of the 5 possible ending numbers are even and the probability of an even number is 2/5.
We analyze in two ways the number of ways to choose a number with digits 1 and 5 even.
First way:
There are 5! = 120 possible arrangements of the 5 digits.
There are only two possible even digits, from which we will choose digit 1 and digit 5. The order of our choice certainly matters, since a different choice will give us a different 5-digit number. So we are choosing 2 digits from a set of 2 digits, where order matters. We therefore have P(2, 2) = 2*1 / 0! = 2 ways to choose these digits.
The remaining 3 digits will comprise digits 2, 3 and 4. We are therefore choosing 3 digits from a set of 3, in order. There are P(3, 3) = 3*2*1/0! = 6 ways to do so.
To obtain our number we can choose digits 1 and 5, then digits 2, 3 and 4. There are P(2, 2) * P(3, 3) = 2 * 6 = 12 ways to do this.
So the probability that digits 1 and 5 are even is 12 / 120 = 1/10.
Second way:
A simpler solution looks at just the possibilities for digits 1 and 5. There are P(2, 2) = 2 choices for which these digits are even, and P(5, 2) = 20 total choices for these two digits. The probability that both will be even is therefore 2/20 = 1/10, the same as before. **
&#Good work. See my notes and let me know if you have questions. &#