Answer to Question #128644 in Statistics and Probability for Muhammad Khizer Kamal

Question #128644
Consider a set of three shares in different companies. On any day the market price of each share can either increase (I), decrease (D) or remains unchanged (U).Assuming P(D) =P(I) =P(D)
(i) Write the sample space for the behaviour of the 3 shares
(ii) Compute the probabilities of the following events.
A = ( shares 1 , 2 , 3 behaves in the same way )
B = ( shares 1 and 2 increase in value )
C = ( none of the share price increases )
1
Expert's answer
2020-08-06T17:41:41-0400

Increase=IIncrease =I

Decrease=DDecrease=D

Unchanged=UUnchanged=U

solution i)


sample spaces:


Share1={I,D,U}Share 1=\{I,D,U\}

Share2={I,D,U}Share2=\{I,D,U\}

Share3={I,D,U}Share3=\{I,D,U\}


solution ii)


Let X be the movement from share1, Y be the movement from share2 and Z from share3


Therefore:



P(X=x)=P(D)orP(U)orP(I)P(X=x)=P(D)orP(U)or P(I)


Since:



P(D)=P(I)=P(U)P(D)=P(I)=P(U)


P(X=x)=13P(X=x)=\frac{1}{3}

Similarly:


P(Y=y)=P(Z=z)=13P(Y=y)=P(Z=z)=\frac{1}{3}


P(A) where:



A={{xshare1}={yshare2}={zshare3}}A=\{\{x\in share1\}=\{y\in share2\}=\{z \in share3\}\}


=(13×13×13)×3=19=(\frac{1}{3}×\frac{1}{3}×\frac{1}{3})×3=\frac{1}{9}

answer: 19\frac{1}{9}


P(B) where:



B={{X=I}and{Y=I}and{Z=DorU}}B=\{\{X=I\} and \{Y=I\} and\{Z=D or U\}\}


=13×13×(13+13)=227=\frac{1}{3}×\frac{1}{3}×(\frac{1}{3}+\frac{1}{3})=\frac{2}{27}



answer: 227\frac{2}{27}


P(C) Where:



C={{X=DorU}and{Y=DorU}and{Z=DorU}C=\{\{X=DorU\}and\{Y=DorU\}and\{Z=DorU\}

=(13+13)×(13+13)×(13+13)=827=(\frac{1}{3}+\frac{1}{3})×(\frac{1}{3}+\frac{1}{3})×(\frac{1}{3}+\frac{1}{3})=\frac{8}{27}

answer: 827\frac{8}{27}


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