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.
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calendarEducator since 2012
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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.
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calendarEducator since 2012
write609 answers
starTop subjects are Math, Science, and History
(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.