CSE 21: Discussion 5, Solutions to Practice Midterm (May 7, 2007)
1.
The number of onto functions is 4!
S
(5
,
4) = 24
·
10 = 240. The total number of functions is 4
5
. Then
the probability is 240
/
4
5
= 60
/
4
4
= 15
/
64.
2.
Without any restriction, we have
(
2+3+4
3
)
= 84 choices. To get three balls with diﬀerent colors, we
have
(
2
1
)(
3
1
)(
4
1
)
= 24 choices. Then the probability is 24
/
84 = 2
/
7.
3.
We could have
k
= 0
,
1
,
2
,
3 or 4 Aces in the four cards. For each
k
(
k
Aces in the four cards), the
number of possibilities can can calculated as follows:
stage 1: choose
k
Aces. Number of choices:
(
4
k
)
.
stage 2: choose 4

k
more cards from the 48 cards which are not Aces. Number of choices:
(
48
4

k
)
.
Then for each
k
, the number of possibilities is
(
4
k
)(
48
4

k
)
. Then the corresponding probability is
(
4
k
)(
48
4

k
)
(
52
4
)
.
Finally, the expected value is
E
(
X
)
=
4
X
k
=0
k
·
(
4
k
)(
48
4

k
)
(
52
4
)
=
(
4
1
)(
48
3
)
(
52
4
)
+ 2
(
4
2
)(
48
2
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This note was uploaded on 04/23/2008 for the course CSE 21 taught by Professor Graham during the Spring '07 term at UCSD.
 Spring '07
 Graham

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