MATH 105 101
Assignment 5 Solutions
Due date: November 13, 2014
MATH 105 101 Assignment 5 Solutions
All work must be shown for full marks.
1. (7 marks) In a deck of 30 distinct cards, there is exactly one legendary card called
Deathwing. If Artosis draws a hand of 3 cards randomly from the deck (without
replacing the cards drawn):
• How many possible hands of 3 cards are there?
Solution: To form a hand, we can think of the number of choices for each
time choosing a card. Note that we have 30 choices for the first card, 29
for the second (excluding the one we already drew), and 28 for the third.
However, in a hand of cards, the order of the cards does not matter, so we
are, in fact, over-counting. For example, for a hand including 3 cards A,B,C,
we can obtain this hand by drawing A first, B second, C third, or by drawing
A first, C second, B third. Given 3 cards, we can arrange them in order in 6
different ways, so the total number of hands are:
30.29.28
= 4060.
6
• What is the probability that he will draw the legendary card in his hand?
Solution: First, we can find the total number of hands including Deathwing.
Since we fix one of the card to be Deathwing, the remaining 2 cards can be
chosen freely from the 29 cards left. By the same reasoning in the first part,
we get the total number of hands of 3 cards including Deathwing is:
29.28
= 406.
2
Therefore, the probability of Artosis drawing the legendary card in his hand
is:
406
1
= .
4060
10
2. (9 marks) Let X be the duration of a telephone call in minutes and suppose that X
has the following probability density function:
f (x) = ce−x/10
for x ≥ 0, and c is some constant.
• Determine the value of c.
Page 1 of 3
MATH 105 101
Assignment 5 Solutions
Due date: November 13, 2014
Z
∞
Solution: In order to be a PDF, f (x) needs to satisfy the condition
f (x) dx =
−∞
1, that is:
Z
∞
1=
Z
f (x) dx =
−∞
0
ce−x/10 dx = lim −10ce−x/10 |b0
= lim
b→∞
0
= lim −10ce
−b/10
b→∞
Thus, c =
ce−x/10 dx
b
Z
b→∞
∞
+ 10c = 10c.
1
.
10
• What is the probability that a call lasts less than 5 minutes?
Solution: The probability that a call lasts less than 5 minutes is:
Z 5
Z 5
1 −x/10
Pr(X < 5) =
f (x) dx =
e
dx = −e−x/10 |50 = 1 − e−1/2 .
−∞
0 10
• Compute the expected value E(X).
Solution: We have that:
Z
Z ∞
xf (x) dx =
E(X) =
−∞
∞
0
1 −x/10
xe
dx = lim
b→∞
10
Z
b
0
1 −x/10
xe
dx.
10
1 −x/10
e
dx, du = dx and
Using integration by parts, we have u = x, dv = 10
−x/10
v = −e
. We get:
Z
Z
1 −x/10
−x/10
xe
dx = −xe
+ e−x/10 dx = −xe−x/10 − 10e−x/10 + C.
10
Then,
E(X) = lim (−xe−x/10 − 10e−x/10 ) |b0
b→∞
= lim −be−b/10 − 10e−b/10 + 10
b→∞
⇒ E(X) = 10.
Page 2 of 3
MATH 105 101
Assignment 5 Solutions
Due date: November 13, 2014
3. (9 marks) Let X be a continuous random variable, with the following probability
density function:
(
a + bx2 , 0 ≤ x ≤ 1
f (x) =
0, else.
• Suppose that the expected value E(X) = 3/5, determine the values of a and b.
Z
∞
Solution: Since f is a PDF, it has to satisfy
f (x) dx = 1, that is:
−∞
Z
∞
1=
Z
f (x) dx =
−∞
0
1
b 3 1
b
(a + bx ) dx = ax + x |0 = a + .
3
3
2
So, b = 3 − 3a. On the other hand, since the expected value E(X) = 3/5, we
also get:
Z ∞
Z 1
a 2 b 4 1 a b
3
3/5 = E(X) =
xf (x) dx =
(ax + bx ) dx =
x + x |0 = + .
2
4
2 4
−∞
0
So, 12 = 10a + 5b = 10a + 5(3 − 3a) = 15 − 5a. Hence, a = 3/5, and
b = 3 − 3a = 6/5.
• Compute the variance Var(X).
Solution: We have that Var(X) = E(X 2 ) − (E(X))2 . So, first we compute
Var(X 2 ).
Z 1
Z ∞
3 2 6 4
2
2
x f (x) dx =
E(X ) =
x + x dx
5
5
0
−∞
1 3
6
1
6
11
=
x + x5 |10 = +
= .
5
25
5 25
25
Thus,
Var(X) = E(X 2 ) − (E(X))2 =
Total: 25 marks.
Page 3 of 3
11
9
2
−
= .
25 25
25