# Huntington’s Disease

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Order Now#12. Which of the following numbers could be the probability of an event? 1.5, 0, = ,0

#34 More Genetics In Problem 33, we learned that for some diseases, such as sickle-cell anemia, an individual will get the disease only if he or she receives both recessive alleles. This is not always the case. For example, Huntington’s disease only requires one dominant gene for an individual to contract the disease. Suppose that a husband and wife, who both have a dominant Huntington’s disease allele (S) and a normal recessive allele (s), decide to have a child. (a) List the possible genotypes of their offspring. (a) Sample space is {SS,Ss,sS,ss} where S=dominant disease allele and s=normal recessive allele (b) What is the probability that the offspring will not have Huntington’s disease? In other words, what is the probability that the offspring will have genotype ss? Interpret this probability(b) Since P(S)=P(s)=1/2 and S,s are independent, P(offspring will not have Huntington’s disease)=P(SS)=P(S)*P(S)=1/4

#40. Which of the assignments of probabilities should be used if the coin is known to be fair? If coin is fair, then assignment A is used because P(H)=P(T)=1/2 #48. Classifying Probability Determine whether the probabilities on the following page are computed using classical methods, empirical methods, or subjective methods. a) The probability of having eight girls in an eight-child family is 0.390625%. Empirical method

(b) On the basis of a survey of 1000 families with eight children, the probability of a family having eight girls is 0.54%. Classical method

(c) According to a sports analyst, the probability that the Chicago Bears will win their next game is about 30%. Subjective method

(d) On the basis of clinical trials, the probability of efﬁcacy of a new drug is 75%. Empirical method

5.2 #26 a-d #32

#26 . Doctorates Conferred The following probability model shows the distribution of doctoral degrees from U.S. universities in 2009 by area of study. Area of Study Probability

Engineering 0.154

Physical sciences0.087

Life sciences 0.203

Mathematics 0.031

Computer sciences0.033

Social sciences0.168

Humanities0.094

Education0.132

Professional and other ﬁelds0.056

Health0.042

(a) Verify that this is a probability model. ) First, all probabilities range from 0 to 1 inclusive. Second, the sum of probabilities=0.154+…+0.042=1. So it is a probability model.

(b) What is the probability that a randomly selected doctoral candidate who earned a degree in 2009 studied physical science or life science? Interpret this probability. P=0.087+0.203=0.29. Interpretation: if there are 100 candidates, 29 of them studies physical science or life science. (c) What is the probability that a randomly selected doctoral candidate who earned a degree in 2009 studied physical science, life science, mathematics, or computer science? Interpret this probability. P=0.087+0.203+0.031+0.033=0.354 Interpretation: if there are 1000 candidates, 354 of them studies physical science, life science, mathematics or compute science (d) What is the probability that a randomly selected doctoral candidate who earned a degree in 2009 did not study mathematics? Interpret this probability. d) P=1-0.031=0.969.

Interpretation: if there are 1000 candidates, 969 of them do not study mathematics #32. A Deck of Cards A standard deck of cards contains 52 cards, as shown in Figure 9. One card is randomly selected from the deck. (a) Compute the probability of randomly selecting a two or three from a deck of cards. Since there are 4 “two’s” and 4 “three’s” in a deck of 52 cards, P( a two or a three)=P(a two)+P(a three)=4/52+4/52=2/13 (b) Compute the probability of randomly selecting a two or three or four from a deck of cards. Since there are 4 “two’s”, 4 “three’s” and 4 “four’s”in a deck of 52 cards, P( a two or a three or a four)=P(a two)+P(a three)+P(a four)=4/52+4/52+4/52=3/13 (c) Compute the probability of randomly selecting a two or club from a deck of cards. Since there are 4 “two’s” and 13 “clubs” in a deck of 52 cards but there are 1 card which is both two and club, P( a two or a club)=P(a two)+P(a club)-P(both two and club)=4/52+13/52-1/52=4/13

5.3 #8, #18

#8. Determine whether the events E and F are independent or dependent. Justify your answer. (a) E: The battery in your cell phone is dead. The events are independent because the batteries have no connect to each other (different device uses different batteries). F: The batteries in your calculator are dead.

(b) E: Your favorite color is blue. The events are independent because my choice of color does not affect the other’s hobby. F: Your friend’s favorite hobby is ﬁshing.

(c) E: You are late for school. The events are dependent because if you run out of gas, then you are likely to be late for school F: Your car runs out of gas

#18. Life Expectancy The probability that a randomly selected 40-year-old female will live to be 41 years old is 0.99855 according to the National Vital Statistics Report, Vol. 56, No. 9. (a) What is the probability that two randomly selected 40-year-old females will live to be 41 years old? Since two females are independent, P(both 40-year-old females will live to be 41 years old)=0.99855*0.99855=0.997 (b) What is the probability that ﬁve randomly selected 40-year-old females will live to be 41 years old? Since five females are independent, P(all five 40-year-old females will live to be 41 years old)=0.99855*0.99855*0.99855*0.99855*0.99855=0.993

(c) What is the probability that at least one of ﬁve randomly selected 40-year-old females will not live to be 41 years old? Would it be unusual if at least one of ﬁve randomly selected 40-year-old females did not live to be 41 years old? P(at least one of five 40-year-old females will live to be 41 years old)=1-P(all five 40-year-old females will live to be 41 years old)=1-0.99855*0.99855*0.99855*0.99855*0.99855=0.007. Yes, the probability is only 0.007 which is less than 0.05 cut-off point. So the result is unusual and surprises me

5.5 #6, #12, #20, #32, #34, #46, #52, #56,#62

#6. 7!=7*6*5*4*3*2*1=5040

#12 7P2=7!/(7-2)!=5040/120=42 where ! is sign of factorial

#20 9C2=9!/[2!*(9-2)!]=362880/(2*5040)=36

#32 Clothing Options A woman has ﬁve blouses and three skirts. Assuming that they all match, how many different outﬁts can she wear?15 different #34 Arranging Students In how many ways can 15 students be lined up? The number of ways 15 students can be lined up is 15!=1307674368000 #46. Betting on the Perfecta In how many ways can the top 2 horses ﬁnish in a 10-horse race? Since the order matters, the top “2” horses can be selected out of 10 horses in 10P2 ways. So 10P2=10!/8!=90. #52.

Simple Random Sample How many different simple random samples of size 7 can be obtained from a population whose size is 100? Since the order does not matter, the sample of size 7 can be selected out of population of size 100 in 100C7 ways. So 100C7=100!/[7!93!]=16007560800 #56. DNA Sequences (See Example 10.) How many distinguishable DNA sequences can be formed using one A, four Cs, three Gs, and four Ts? This is the case of permutations with non-distinct objects. By the multiplication rule, the number of possible sequences is 12!/[1!4!3!4!]=138600

#62. Selecting a Committee Suppose that there are 55 Democrats and 45 Republicans in the U.S. Senate. A committee of seven senators is to be formed by selecting members of the Senate randomly. 55 democrats and 45 republicans, we have to form committee of 7 no. of ways of selecting 7 from total 100 = 100C7

(a) What is the probability that the committee is composed of all Democrats? no. of ways of selecting 7 democrats from 55 = 55C7 the probability that the committee is composed of all democrats= 55C7/ 100C7 (b) What is the probability that the committee is composed of all Republicans? no. of ways of selecting 7 republicans from 45 = 45C7 the probility the committe is composed of all republicans= 45C7 / 100C7 (c) What is the probability that the committee is composed of three Democrats and four Republicans. no. of ways of selecting 3 democrats from 55 and 4 republicans from 45 = 55C3 * 45C4 probability that the committee is composed of three democrats and four republicans = (55C3 * 45C4) / 100C7