(ii)

Let W be event of raining on Wednesday.

`P(A|B) = `probability of rain falling on thursday given it did not rain on tuesday

`P(A|B') = `probability of rain falling on thursday given it rained on tuesday

`P(A|B)-P(A|B')=`

`= P(A|W)P(W|B) + P(A|W')(1-P(W|B) - (P(A|W)P(W|B') +`

`P(A|W')(1-P(W|B')))=`

Everything before...

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(ii)

Let W be event of raining on Wednesday.

`P(A|B) = `probability of rain falling on thursday given it did not rain on tuesday

`P(A|B') = `probability of rain falling on thursday given it rained on tuesday

`P(A|B)-P(A|B')=`

`= P(A|W)P(W|B) + P(A|W')(1-P(W|B) - (P(A|W)P(W|B') +`

`P(A|W')(1-P(W|B')))=`

Everything before minus is `P(A|B)` and everything after minus is `P(A|B').`And since `P(A|W)=x` and `P(A|W)=y` we have:

`xP(W|B)+y (1-P(W|B) - x P(W|B') -y (1-P(W|B')) =`

`x^2 + y(1-x) - xy - y(1-y) = x^2 + y - xy - xy -y + y^2=`

`x^2 - 2xy + y^2 = (x-y)^2`

(iii)

If `x=y` then

`P(A|B)-P(A|B') = (x-x)^2 = 0 implies P(A|B) = P(A|B')`

Which means that probability of raining on thursday **if it rained** on tuesday is equal to probability of raining on thursady** if it did not rain** on tuesday meaning that `A` and `B` are independent.

(ii)

Let W be event of raining on Wednesday.

`P(A|B) = `probability of rain falling on thursday given it did not rain on tuesday

`P(A|B') = `probability of rain falling on thursday given it rained on tuesday

`P(A|B)-P(A|B')=`

`= P(A|W)P(W|B) + P(A|W')(1-P(W|B) - (P(A|W)P(W|B') + P(A|W')(1-P(W|B')))=`

Everything before minus is `P(A|B)` and everything after minus is `P(A|B').`And since `P(A|W)=x` and `P(A|W)=y` we have:

`xP(W|B)+y (1-P(W|B) - x P(W|B') -y (1-P(W|B')) =`

`x^2 + y(1-x) - xy - y(1-y) = x^2 + y - xy - xy -y + y^2=`

`x^2 - 2xy + y^2 = (x-y)^2`

(iii)

If `x=y` then

`P(A|B)-P(A|B') = (x-x)^2 = 0 implies P(A|B) = P(A|B')`

Which means that probability of raining on thursday **if it rained** on tuesday is equal to probability of raining on thursady** if it did not rain** on tuesday meaning that `A` and `B` are independent.