Feb 7, 2008
NAME:
Math 425, SECTION 8
MIDTERM 1 – SOLUTIONS
Problem 1 [20 pts]
A poker hand consisting of 5 cards is dealt from a standard deck of 52 cards. Assume that
all combinations are equally likely to occur. What is the probability that the hand is:
(a) a flush (that is, all five cards are of the same suit)?
Answer:
P
(flush) =
4
(
13
5
)
(
52
5
)
(b) a pair (that is, there are two cards of the same denomination, and the rest are of different
denominations) ?
Answer:
P
(pair) =
13
(
4
2
)(
12
3
)
4
3
(
52
5
)
Our sample space
S
consists of
(
52
5
)
poker hands. Suppose we have a pair. There are 13
ways to choose the denomination in the pair,
(
4
2
)
ways to choose for the suits in the pair,
(
12
3
)
ways to choose the denominations of the remaining three cards and 4
3
ways to choose
their suits.
Note. It is tempting to write 12
×
11
×
10 instead of
(
12
3
)
in the formula above. By doing
so, we take into account the order of the remaining three cards, so we must divide by 3!.
Problem 2 [20 points]
(a) Ten students are randomly divided into Team A and Team B,
with 5 students each. How many divisions are possible?
Solution.
There are
(
10
5
)
ways to choose the five students in Team A. But now, we also
know who is in Team B. So, the answer is
(
10
5
)
= 252.
(b) Among the ten students, there are two brothers: John and Fred. What is the probability
that they are both in the same team? Please state clearly the assumptions you are making.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Winter '08
 Buckingham
 Probability, Probability theory, Alice, rob

Click to edit the document details