Homework #1 - Probabilities

HOMEWORK 1 1.4 For events A and B, find formulas for the probabilities of the following events in terms of the quantities P(A), P(B), and P(A∩B). (a) either A or B or both (b) either A or B but not both (c) at least one of A or B (d) at most one of A or B a) P(A∪B)=P(A)+P(B)−P(A∩B) b) P(either but not both)=P(A∪B)−P(A∩B) c) P(A∪B)=P(A)+P(B)−P(A∩B) d) P(at most one of A or B)=1−P(A∩B) 1.6 Two pennies, one with P(head)=u and one with P(head)=w, are to be tossed together independently. Define P 0 ​ =P(0 heads occur) P 1 ​ =P(1 head occurs) P 2 ​ =P(2 heads occur) Can u and w be chosen such that P 0 ​ =P 1 ​ =P 2 ​ ? Prove your answer. P 0 ​ =P(0 heads)=(1−u)(1−w) P 1 ​ =P(1 head)=u(1−w)+w(1−u) P 2 ​ =P(2 heads)=uw P 0 ​ =P 1 ​ =P 2 ​ ⇒(1−u)(1−w)=u(1−w)+w(1−u)=uw 1 st :1−w−u+uw=u−uw+w−uw 1−w−u−u=w−uw−uw−uw 1−2w−2u=−3uw 1=2w+2u−3uw 2 nd :u−uw+w−uw=uw −2uw+u+w=0 −3uw+u+w=0 3 rd :replace w 1=2( −3u+1 −u ​ )+2u−3u( −3u+1 −u ​ ) w= −3u+1 −u ​ 1= −3u+1 −2u ​ +2u+ −3u+1 3u 2 ​ 1= −3u+1 −2u+2u(−3u+1)+3u 2 ​ 1= −3u+1 −2u−6u 2 +2u+3u 2 ​ 1= −3u+1 −3u 2 ​ −3u+1=−3u 2 3u 2 −3u+1=0 4 th :calculate w: w= −3u+1 −u ​ w= −3(1/2)+1 −(1/2) ​ w= −3/2+1 −(1/2) ​ w= −1/2 −(1/2) ​ =1 5 th :P 0 ​ =P 1 ​ =P 2 ​ P 0 ​ =(1−1/2)(1−1)=(1/2)(0)=0 P 1 ​ =1/2(1−1)+1(1−1/2)=1/2(0)+1(1/2)=1/2 P 2 ​ =1/2(1)=1/2 Therefore, u and w can’t be chosen to make that equation true. II. Additional Problems: A group of friends decide to go out to watch a movie. Fred and Tom are bringing dates, while their 2 friends are going alone. When the friends arrive at the movie theater, they find a row of six seats so they can all sit together. If Fred and Tom must sit next to their dates, how many different ways can the group of six sit down? Tom and date -> can be ordered in 2! ways Fred and date -> can be ordered in 2! ways 1 friend 2 friend Can be ordered in 4! ways 4! * 2! * 2! = 96 ways in which the group of 6 can sit down