1
ECE 3020A Homework 2, Spring 2018
Solution
Problem 1.
a)
(
6
2
) + (
7
2
) + (
4
2
) = 42
possibilities.
b)
There are
6 × 7
choices of Impala and Malibu,
6 × 4
choices of Impala and Corvette, and
7 × 4
choices of Malibu and Corvette. That is, there are 94 possibilities.
Problem 2.
Let us call
r
for right and
u
for up. Then, each path is a linear arrangement of 4
r’
s and 3
u
’s. For
example:
r,r,u,u,r,r,u
. As such, the number of possible different paths from A to B is
:
7!
3!×4!
=
35
.
Problem 3.
Let G denote the event that th
e suspect “S” is guilty and C the event that he possesses the
characteristic of the criminal, we have:
𝑃(𝐺|𝐶) =
𝑃(𝐺 ∩ 𝐶)
𝑃𝐶
=
𝑃(𝐶|𝐺)𝑃(𝐺)
𝑃(𝐶|𝐺)𝑃𝐺 + 𝑃(𝐶|𝐺
𝑐
)𝑃(𝐺
𝑐
)
=
1 × (0.6)
1 × (0.6) + (0.2) × .04
≈ 0.882
Problem 4.
We’ll compute each measure by its definition, using the fact that each coin flip is fair and
independent: First, observe that Player 1 wins a dollar if at least 1 of the coins lands on heads.

** Subscribe** to view the full document.