EXAM P SAMPLE SOLUTIONS

Consequently,
1 2 1 2
( , ) 34
f t t
1 2
⎧ 1
⎪ , 0< t < 6 , 0< t < 6 , t + t < 10
= ⎨
⎪⎩ 0 elsewhere
and
ET+ T = ET + ET = 2 ET (due to symmetry)
[ ] [ ] [ ] [ ]
1 2 1 2 1
⎧ 4 6 1 6 10−t11 ⎫ ⎧ 4 ⎡t 6
2 6 ⎤ ⎡t2 10−t
⎤ ⎫
1
= 2⎨∫ t1 2 1 1 2 1 1 0 1 1 0
0 ∫ dt dt + 2
0 34 ∫ t
4 ∫ dt dt ⎬= ⎨ 0 34 ∫ t
0 ⎢
34
⎥dt + ∫ t
4 ⎢
34
⎥dt1⎬
⎩ ⎭ ⎩ ⎣ ⎦ ⎣ ⎦ ⎭
2
⎧ 43t 6
1 1 2 ⎫ ⎧3t1 4 1 ⎛ 2 1 3⎞6⎫
= 2⎨∫ dt1 + ( 10t1 − t1 ) dt
0 4
1⎬= 2 0+ ⎜5t1 − t1
⎩ 17 ∫
⎨ ⎟ 4 ⎬
34 ⎭ ⎩34 34 ⎝ 3 ⎠ ⎭
⎧24 1 ⎡ 64⎤⎫
= 2⎨ + 180 −72 − 80 + = 5.7
17 34 ⎢ ⎬
⎣ 3 ⎥
⎩ ⎦⎭
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95. Solution: E
tW 1 + t2Z t1( X+ Y) + t2( Y−X) ( t1− t2) X ( t1+ t2) Y
M ( t1, t2) = E⎡e E⎡e ⎤ E⎡e e ⎤
⎣
⎤
⎦ =
⎣ ⎦
=
⎣ ⎦
(
= E⎡ ⎣
e
) ⎤ (
E⎡ ⎦ ⎣
e
)
1 2 1 2 1 2 2 1 2 2
( t1− t2) ( t1+ t2) ( t1 − 2t1t2+ t2) ( t1 + 2t1t2+
t2)
⎤ 2 2 2 2
⎦
= e e = e e = e
2 2
t1− t2 X t1+ t2 Y t1 + t2
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96. Solution: E
Observe that the bus driver collect 21x50 = 1050 for the 21 tickets he sells. However, he
may be required to refund 100 to one passenger if all 21 ticket holders show up. Since
passengers show up or do not show up independently of one another, the probability that
21 21
all 21 passengers will show up is ( 1− 0.02) = ( 0.98) = 0.65 . Therefore, the tour
operator’s expected revenue is 1050 − ( 100)( 0.65) = 985 .
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