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(math144)[2009](s)final~wfli^_10459.pdf
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Math 144: (Solutions) Final examination 2009/2010,
Dec 16, 2009.

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All answers should be either exact or correct to 4 decimal places.
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1. [10 marks] Poker dice is played by simultaneously rolling 5 fair dice. Find the following probabilities:
(a) P{no two alike}, (b) P{one pair}, (c) P{two pair}, (d) P{.ve alike}.
(a) [2 marks]
6 5 4 3 2
P{ no two alike } =P{5 di.erent numbers} = =0.0926 (to 4 d.p.)
65
(b) [3 marks]
.5!
.6..5
132!
P{ one pair } =P{one number repeated twice} = =0.4630 (to 4 d.p.)
65
(c) [3 marks]
.6 . 5!
..4
212!2!
P{ two pair } =P{two di.erent numbers repeated twice} = =0.2315 (to 4 d.p.)
65
(d) [2 marks]
.6.
1
P{ .ve alike } =P{5 same numbers} = =0.0008 (to 4 d.p.)
65
2. The weight (in tons) of a vehicle crossing a certain bridge is found to have a density function shown below.
(a) [2 marks] Write down the equation for the increasing line in the picture and that for the decreasing line.
Let Y be the random variable of the weight of a vehicle, and f(y) be its density function. Then for the increasing line, we have
f(y)=0.1y, when 0 y 2,
while for the decreasing line, we have
1
f(y)= 1 . y, when 2 y 10.
4 40
(b) [2 marks] What proportion of a vehicle that crosses the bridge weighs more than 1.5 tons? The required proportion is
. 10
P (Y> 1.5) = f(y)dy,
1.5
where
.0.1y, if 0 y 2; f(y)= . 1/4 . y/40, if 2 y 10; . 0, otherwise.
Thus, we have
. 2 . 10
P (Y> 1.5)= 0.1ydy + (1/4 . y/40)dy =0.8875.
1.52
(c) [2 marks] Find the mean and standard deviation of the weight of a vehicle?
. 10 . 2 . 10 E[Y ]= yf(y)dy =0.1y 2dy +(y/4 . y 2/40)dy =4 0 02
Since
. 10 . 2 . 10
62
E[Y 2]= y 2f(y)dy =0.1y 3dy +(y 2/4 . y 3/40)dy = 20.66667 or ,
3
0 02
V ar(Y )= E[Y 2] . (E[Y ])2 =4.666667 or 14/3.
Thus, the standard deviation of Y is 2.1602 or .14/3.

2 (d) [2 marks] If there are 50 vehicles on the bridge, what is the normally-approximated probability
that their combined weight exceeds 200 tons?
Let Y1,...,Y50 be the weights of the 50 vehicles. So,

4 . E[Y ]
P (Y1 + Y> 4) P (Z>
+ Y50 > 200) = P ( =0)=0.5.
.V ar(Y )/50
(e) [4 marks] If there are 50 vehicles on the bridge, what is the probability that the heaviest ve-hicle exceeds 9.7 tons? [Hint: the weights of these 50 vehicles are independent and identically distributed.]
Let max{Y1,...,Y50} be the random variable of the weight of the heaviest vehicle. Then the required probability is
P (max{Y1,...,Y50} > 9.7)
=1 . P (max{Y1,...,Y50} 9.7)
=1 . P (all Y1,...,Y50 9.7)
=1 . P (Y1 9.7,...,Y50 9.7)
=1 . P (Y1 9.7) ... P (Y50 9.7) ( because of independence)
=1 . P (Y1 9.7)50 ( because of identical distribution)
. 9.7
=1 . [f(y)