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

Question

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 )

Answer 1

$Increase =I$

$Decrease=D$

$Unchanged=U$

solution i)

sample spaces:

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

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

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

solution ii)

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

Therefore:

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

Since:

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

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

Similarly:

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

P(A) where:

$$A=\{\{x\in share1\}=\{y\in share2\}=\{z \in share3\}\}$$

$$=(\frac{1}{3}×\frac{1}{3}×\frac{1}{3})×3=\frac{1}{9}$$

answer: $\frac{1}{9}$

P(B) where:

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

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

answer: $\frac{2}{27}$

P(C) Where:

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

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

answer: $\frac{8}{27}$