HW3 ans - STAT 400 Spring 2011 Homework #3 (10 points) (due...

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STAT 400 Spring 2011 Homework #3 (10 points) (due Friday, February 11, by 3:00 p.m.) 1. During two-and-a-half years of research, bio-psychologist Onur Güntürkün discovered that when people kiss, they turn their heads to the right roughly twice as often as to the left. (Güntürkün, O. Human behaviour: Adult persistence of head-turning asymmetry. Nature , 421, 711 , (2003).) Suppose the probability that a person would turn his/her head to the right is 2 / 3 , and the probability that a person would turn his/her head to the left is 1 / 3 . A couple is planning a kiss on Valentine’s day. Assume that their choice of which way to turn their heads is independent of each other. a) What is the probability that they would both turn their heads to the right (and kiss)? Person 1 Person 2 Right Left Right 2 / 3 × 2 / 3 4 / 9 kiss 1 / 3 × 2 / 3 2 / 9 bump noses 2 / 3 Left 2 / 3 × 1 / 3 2 / 9 bump noses 1 / 3 × 1 / 3 1 / 9 kiss 1 / 3 2 / 3 1 / 3 1.00 P( both turn their heads to the right ) = 4 / 9 . b) What is the probability that they would bump noses (i.e., choose the opposite direction to turn their heads)? P( bump noses ) = 2 / 9 + 2 / 9 = 4 / 9 . Similar arguments with 3 / 4 and 1 / 4 predict phenotypic ratios of 9:3:3:1 in F 2 offspring in Mendel's dihybrid crosses.
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2. From a group of 16 male and 9 female armadillos, Noah must choose two to travel on his ark. Unable to distinguish between male and female armadillos, Noah must choose at random. a) Noah chooses the two armadillos at random. Compute the probability that Noah gets two armadillos of the opposite sex (i.e., one male and one female armadillo). Do not choose the same armadillo twice without replacement. Does not matter whether a female is chosen first and a male is chosen second, or a male is chosen first and a female is chosen second. P( F M ) + P( M F ) = 9 / 25 16 / 24 + 16 / 25 9 / 24 = 288 / 600 = 12 / 25 = 0.48 . OR P( 1 F and 1 M ) = 25 12 300 144 300 16 9 2 25 1 16 1 9 C C C = = = = 0.48 . b) In order to improve his chances of selecting at least one male and one female armadillo, Noah decides to "cheat" and select three armadillos to travel on his ark. Compute the probability that Noah gets at least one male and one female armadillo. 8 possible outcomes: M M M F F F "bad" outcomes M M F F F M M F M F M F "good" outcomes M F F F M M 1 – [ P( M M M ) + P( F F F ) ] = 1 – [ 16 / 25 15 / 24 14 / 23 + 9 / 25 8 / 24 7 / 23 ] = 1 – [ 3360 / 13800 + 504 / 13800 ] = 1 – 3864 / 13800 = 1 – 0.28 = 0.72 . OR
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P( M M F ) + P( M F M ) + P( M F F ) + P( F M M ) + P( F M F ) + P( F F M ) = 16 / 25 15 / 24 9 / 23 + 16 / 25 9 / 24 15 / 23 + 16 / 25 9 / 24 8 / 23 + 9 / 25 16 / 24 15 / 23 + 9 / 25 16 / 24 8 / 23 + 9 / 25 8 / 24 16 / 23 = 2160 / 13800 + 2160 / 13800 + 1152 / 13800 + 2160 / 13800 + 1152 / 13800 + 1152 / 13800 = 9936 / 13800 = 0.72 . OR
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This note was uploaded on 05/31/2011 for the course SATS 400 taught by Professor . during the Spring '11 term at Moraine Valley Community College.

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HW3 ans - STAT 400 Spring 2011 Homework #3 (10 points) (due...

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