We open a 1500 page book to a random page. What is the probability that we open to?

a. page 23
b. page 987
c. page 0
d. page 1504
e. either page 1, 2, or 3
f. any page up to and including page 150
g. any page 12 through 17
h. any page after page 150
i. page 21.78
j. any page up to and including page 21.78

1 Answer
MathFact-orials.blogspot.com
Apr 8, 2017

See below:

Explanation:

With all of these questions, we're looking at a fraction:

#"the number of ways to meet the conditions"/"the total number of ways to pick"#

a, b

There is only 1 way to be on page 23 or 987, so the numerator is 1. There are 1500 pages in the book.

#:. P(Y=23)=P(Y=987)=1/1500#

c, d

There is no way to be on page 0 or 1504:

#:. P(Y=0)=P(Y=1504)=0/1500=0#

e

We can be on pages 1, 2, 3, and so we have 3 ways to meet the condition:

#:. P(Y<=3)=3/1500=1/500#

f

Similar reasoning here - there are 150 pages that satisfy the conditions:

#:. P(Y<=150)=150/1500=1/10#

g

Pages 12 through 17 satisfy the condition, making 6 pages:

#:. P(12<=Y<=17)=6/1500=1/250#

h

This is the same as #1-P(Y<=150)#

#:. 1-1/10=9/10#

i

There is no way to be on a fractional page, and so there is no way to meet the condition:

#:. P(Y=21.78)=0/1500=0#

j

There are 21 pages that meet this condition:

#:. P(Y<=21.78)=21/1500=7/500#

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