# Ch6.2 Solutions.pdf

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Seneca College

International Business

International Business QNM222

Nabil Ayoub

Winter

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Chapter 6.2
6.16 P(A 1) = .1 + .2 = .3, A(2) = .3 + .1 = .4, A(3) = .2 + .1 = .3.
P( B ) = .1 + .3 + .2 = .6,BP( ) = .2 + .1 + .1 = .4.
1 2
6.17 P(A 1) = .4 + .2 = .6, A(2) = .3 + .1 = .4. B1 ) = .4 + .3 = .7,BP2) = .2 + .1 = .3.
P(A and B )
6.18 a P(A | B ) 1 1 .4 .57
1 1 P(B ) .7
1
P(A a2d B ) 1 .3
b P(A |2B )1 P(B ) .7 .43
1
c Yes. It is not a coincidence. GiveB1 the eventsA 1and A 2 constitute the entire sample space.
P(A 1nd B )2 .2
6.19 a P(A1| B 2 .67
P(B 2 .3
P(A 1nd B )2 .2
b P(B 2 A 1) .33
P(A 1 .6
c One of the conditional probabilities would be greater than 1, which is not possible.
6.20 The events are not independent becauseP(A | B ) P(A ) .
1 2 1
6.21 a P(A 1 or B1) = P(A 1P(B )1(A and 1 ) 1 = .6 + .7 - .4 = .9
b P(A or B ) = P(A )P(B )P(A and B ) = .6 + .3 - .2 = .7
1 2 1 2 1 2
c P(A 1 or A 2) = P(A 1P(A ) 2 = .6 + .4 = 1
P(A and B ) .20
6.22 P(A 1 B 1 1 1 .25; P(A 1 .20.05 .25 ; the events are independent.
P(B 1 .20.60
P(A and B )
6.23 P(A | B ) 1 1 .20 .571;P(A ) .20.60 .80 ; the events are dependent.
1 1 P(B ) .20.15 1
1
6.24 P(A 1) = .15 + .25 = .40, PA2 ) = .20 + .25 = .45, A(3) = .10 + .05 = .15.
P( B ) = .15 + 20 + .10 = .45, PB ) = .25 + .25 + .05 = .55.
1 2
P(A 2nd B ) 2 .25
6.25 a P(A 2|B 2 ) = .455
P(B 2 .55
P(A 2nd B ) 2 .25
b P(B 2 A 2 ) = .556
P(A 2 .45 P(A and B ) .20
c P(B 1| 2) = 2 1 .444
P(A )2 .45
6.26 a P(A or A ) = P(A ) + P(A ) = .40 + .45 = .85
1 2 1 2
b P(A 2 or B 2) = P(A 2) + P(B 2) – P(A 2and B 2) = .45 + .55 - .25 = .75
c P(A 3 or B1) =P( A3 ) + PB 1) –

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