Question 86613: I need help with these problems if someone has time to help me explain them to me step by steps thanks.
Problem #1
Breast Cancer To explain why the chance of a woman
getting breast cancer in the next year goes up each year,
while the chance of a woman getting breast cancer in her
lifetime goes down, Ruma Falk made the following analogy.†
Suppose you are looking for a letter that you may
have lost.You have 8 drawers in your desk. There is a probability
of .1 that the letter is in any one of the 8 drawers,
and a probability of .2 that the letter is not in any of the
drawers.
a. What is the probability that the letter is in drawer 1?
b. Given that the letter is not in drawer 1, what is the probability
that the letter is in drawer 2?
c. Given that the letter is not in drawer 1 or 2, what is the
probability that the letter is in drawer 3?
d. Given that the letter is not in drawers 1–7, what is the
probability that the letter is in drawer 8?
e. Based on your answers to parts a–d, what is happening
to the probability that the letter is in the next drawer?
f. What is the probability that the letter is in some drawer?
g. Given that the letter is not in drawer 1, what is the probability
that the letter is in some drawer?
h. Given that the letter is not in drawer 1 or 2, what is the
probability that the letter is in some drawer?
i. Given that the letter is not in drawers 1–7, what is the
probability that the letter is in some drawer?
j. Based on your answers to parts f–i, what is happening to
the probability that the letter is in some drawer?
Answer by Edwin McCravy(20055)   (Show Source):
You can put this solution on YOUR website!
Suppose you are looking for a letter that you may
have lost.You have 8 drawers in your desk. There is a probability
of .1 that the letter is in any one of the 8 drawers,
and a probability of .2 that the letter is not in any of the
drawers.
a. What is the probability that the letter is in drawer 1?
P(1 or 2 or ... or 8) = .1
P(1)+P(2)+ ... + P(8) = .1
P(1)=P(2)= ... =P(8) = .1/8 = 1/80
b. Given that the letter is not in drawer 1, what is the probability
that the letter is in drawer 2?
P(2 or 3 or ... or 8) = .1
P(2)+P(3)+ ... + P(8) = .1
P(2)=P(3)= ... =P(8) = .1/7 = 1/70
c. Given that the letter is not in drawer 1 or 2, what is the
probability that the letter is in drawer 3?
P(3 or 4 or ... or 8) = .1
P(3)+P(4)+ ... + P(8) = .1
P(3)=P(4)= ... =P(8) = .1/6 = 1/60
d. Given that the letter is not in drawers 1–7, what is the
probability that the letter is in drawer 8?
P(8) = .1
e. Based on your answers to parts a–d, what is happening
to the probability that the letter is in the next drawer?
It's denominator is decreasing by 10, which means the
probability is increasing.
f. What is the probability that the letter is in some drawer?
1 - P(not in any drawer) = 1 - .2 = .8
g. Given that the letter is not in drawer 1, what is the probability
that the letter is in some drawer?
It's still .8
h. Given that the letter is not in drawer 1 or 2, what is the
probability that the letter is in some drawer?
It's still .8
i. Given that the letter is not in drawers 1–7, what is the
probability that the letter is in some drawer?
It's still .8
j. Based on your answers to parts f–i, what is happening to
the probability that the letter is in some drawer?
It's not changing, but staying .8.
Edwin
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