# Please help me to solve this question. I'm facing difficulty in solving the question below:In Camelot, the rain does not fall on Fridays, Saturdays, Sundays, and Mondays. The probability that rain...

Please help me to solve this question. I'm facing difficulty in solving the question below:

In Camelot, the rain **does not** fall on **Fridays, Saturdays, Sundays, and Mondays**. The probability that **rain falls** on a **Tuesday** is **1/5**. For the other days of the week, **Wednesdays and Thursdays**, the conditional probability that there will be rain given that it has rained the previous night is **x**, and the conditional probability that rain will fall given that it did not rain the previous night is **y**.

(i) Show that the probability (unconditional) that it will rain on Wednesday is (x+4y)/5 and find the probability that it will rain on Thursday.**ANSWER PART (ii) and part (iii)**(ii) If A is the event that, in a randomly selected week, rain falls on Thursday, B is the event that rain does not fall on Tuesday,

**show**that

**P(A|B) - P(A|B') = (x-y)^2**

(iii) **Explain** the implications of the case **x=y**.

*print*Print*list*Cite

### 2 Answers

(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.