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CS 70 Fall 2017 Discrete Mathematics and Probability Theory Kannan Ramchandran and Satish Rao
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CS 70 Discrete Mathematics and Probability TheoryFall 2017 Kannan Ramchandran and Satish Rao
HW 12
Sundry
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homework? List names and email addresses. (In case of homework party, you can also just describe
the group.) How did you work on this homework? Working in groups of 3-5 will earn credit for
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1 Darts with Friends
Michelle and Alex are playing darts. Being the better player, Michelle’s aim follows a uniform
distribution over a circle of radius r around the center. Alex’s aim follows a uniform distribution
over a circle of radius 2r around the center.
(a) Let the distance of Michelle’s throw be denoted by the random variable X and let the distance
of Alex’s throw be denoted by the random variable Y .
• What’s the cumulative distribution function of X?
• What’s the cumulative distribution function of Y ?
• What’s the probability density function of X?
• What’s the probability density function of Y ?
(b) What’s the probability that Michelle’s throw is closer to the center than Alex’s throw? What’s
the probability that Alex’s throw is closer to the center?
(c) What’s the cumulative distribution function of U = min{X,Y }?
(d) What’s the cumulative distribution function of V = max{X,Y }?
CS 70, Fall 2017, HW 12 1
, (e) What is the expectation of the absolute difference between Michelle’s and Alex’s distances
from the center, that is, what is E[|X −Y |]? [Hint: There are two ways of solving this part.]
Solution:
(a) • To get the cumulative distribution function of X, we’ll consider the ratio of the area where
the distance to the center is less than x, compared to the entire available area. This gives
us the following expression:
πx2 x2
P(X ≤ x) = = , x ∈ [0, r]
πr2 r2
• Using the same approach as the previous part:
πy2 y2
P(Y ≤ y) = = , y ∈ [0, 2r]
π · 4r2 4r2
• We’ll take the derivative of the CDF to get the following:
dP(X ≤ x) 2x
fX (x) = = 2, x ∈ [0, r]
dx r
• Using the same approach as the previous part:
dP(Y ≤ y) y
fY (y) = = 2, y ∈ [0, 2r]
dy 2r
(b) We’ll condition on Alex’s outcome and then integrate over all the possibilities to get the
marginal P(X ≤ Y ) as following:
Z 2r Z r 2 Z 2r
y y y
P(X ≤ Y ) = P(X ≤ Y | Y = y) fY (y) dy = 2
× 2 dy + 1× dy
0 0 r 2r r 2r2
r4 − 0 4r2 − r2 1 3 7
= + = + =
8r4 4r2 8 4 8
Note the range within which P(X ≤ Y ) = 1. This allowed us to separate the integral to simplify
our solution. Using this, we can get P(Y ≤ X) by the following:
1
P(Y ≤ X) = 1 − P(X ≤ Y ) =
8
A similar approach to the integral above could be used to verify this result.
Z r 2
r4
Z r Z r
x 2x 1 1
P(Y ≤ X) = P(Y ≤ X | X = x) fX (x) dx = 2 2
dx = 4 x3 dx = 4
=
0 0 4r r 2r 0 8r 8
CS 70, Fall 2017, HW 12 2
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