Edward and his friend cannot go to school when it’s sunny outside because they a

Question

Edward and his friend cannot go to school when it’s sunny outside because they are vampires. Forks, Washington is experiencing a sunny period these days and the chance of being sunny each day is 85%. Assume the weather is independent from day to day. Let X be the number of days Edward has to wait through to go outside. (think about what kind of distribution this is. a) What is the probability Edward will have to wait between 10 to 12 days to go back to school? b) Given he has already waited five days, what is the probability he will have to wait between 10 and 12 days (inclusive) to go back to school? (note the given condition) c) What is the expected number of days that he will have to wait to go back to school? Edward and his friend cannot go to school when it’s sunny outside because they are vampires. Forks, Washington is experiencing a sunny period these days and the chance of being sunny each day is 85%. Assume the weather is independent from day to day. Let X be the number of days Edward has to wait through to go outside. (think about what kind of distribution this is. a) What is the probability Edward will have to wait between 10 to 12 days to go back to school? b) Given he has already waited five days, what is the probability he will have to wait between 10 and 12 days (inclusive) to go back to school? (note the given condition) c) What is the expected number of days that he will have to wait to go back to school? a) What is the probability Edward will have to wait between 10 to 12 days to go back to school? b) Given he has already waited five days, what is the probability he will have to wait between 10 and 12 days (inclusive) to go back to school? (note the given condition) c) What is the expected number of days that he will have to wait to go back to school?

Explanation / Answer

X - number of days it takes to go to school first time

p = 1-0.85 = 0.15

X follow geometric distribution

P(X = k) = (1-p)^(k-1)p

= 0.85^(k-1)*0.15

P(10<=X<=12) = P(X =10)+P(X =11)+P(X =12)

= 0.85^(9)*0.15 + 0.85^(10)*0.15 + 0.85^(11)*0.15

=0.089375

b)

geometric distribution has memoryless property

hence P(10<=X<=12|X>=5)

=P(5<= X<=7)

= 0.85^(4)*0.15 + 0.85^(5)*0.15 + 0.85^(6)*0.15

=0.20142

c)

E(X)= 1/p = 1/0.15 = 6.6666666

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