# Allele and Probability

#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...

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