Homework Set 4 Due Thursday February 10

Homework Set 4

Due Thursday February 10

1. Suppose there are 7 candidates in an election for President of the United

States. Each of 20 people votes for one person.

a. How many election outcomes are possible if an outcome includes not

just the totals, but also who voted for each candidate? (Example: If

only Laurie votes for Ralph Nader and 19 people vote for Carrot Top,

then that is a different outcome than if I only vote for Ralph Nader and

19 people {including Laurie} vote for Carrot Top.)

720 because each of the 20 people have 7 candidates to choose from.

b. How many election outcomes are possible if only one person votes for

Ralph Nader and only one person votes for Carrot Top?

20¡¤19¡¤518 because we need to choose who voted for Nader, who

voted for Carrot Top. The remaining 18 people only have 5

candidates to choose from since they cannot vote for Nader or

Carrot Top.

2. There are three women and seven men who will split up into two five-person

teams (to play against each other). How many ways are there to do this so

that there is (at least) one woman on each team?

The teams are 1 woman and 4 men vs 2 women and 3 men. We¡¯ll just

choose the team with the 1 woman and 4 men. The other team is

then determined.

? 3 ?? 7 ?

?? ???? ?? = 3 ? 35 = 105 ways

? 1 ?? 4 ?

3. There are 100 students at a school and three dormitories, A, B, and C, with

capacities 35, 45 and 20 respectively.

a. How many ways are there to fill the dormitories?

Choose which people go in dorm A. This leaves 65 people left to

choose from to put in dorm B. The remaining 20 go in dorm C.

?100 ?? 65 ?

??

???? ??

? 35 ?? 45 ?

b. Suppose that, of the 100 students, 50 are men and 50 are women and

that A is an all-men¡¯s dorm, B is an all-women¡¯s dorm, and C is co-ed.

How many ways are there to fill the dormitories?

Choose which men go in dorm A. Choose which women to put in

dorm B. The remaining 20 people (15 men and 5 women) go in dorm

C.

? 50 ?? 50 ?

?? ???? ??

? 35 ?? 45 ?

4. A box of 30 different iPads contains exactly 7 defective iPads. What is the

probability of finding:

a. Exactly 1 defective iPad in a random group of 5 iPads?

? 7 ?? 23 ?

?? ???? ??

? 1 ?? 4 ?

? 30 ?

?? ??

?5?

This is because we have 5 iPads to choose out of 30. There are 7

iPads that are bad and we are choosing one of them. We are also

choosing 4 good iPads out of the 23 available.

b. At least 2 defective iPads in a random group of 10 iPads?

? 7 ?? 23 ? ? 7 ?? 23 ?

?? ???? ?? ?? ???? ??

0 10

1 9

1 ? ? ?? ? ? ? ?? ?

? 30 ?

? 30 ?

?? ??

?? ??

? 10 ?

? 10 ?

This is because we don¡¯t want 0 or 1 defective iPads (out of 10

chosen).

c. The first defective iPad to be the 17th iPad taken out of the box (in a

random sample of 17 iPads)?

Choose 16 out of the 23 good ones initially.

After you choose the 16 good ones, there are 14 available. Now,

choose 1 bad one out of the 7 bad ones (and 14 total) available.

? 23 ? ? 7 ?

?? ?? ?? ??

? 16 ? ? ? 1 ? ?

? 30 ? ?14 ?

?? ?? ?? ??

? 16 ? ? 1 ?

5. How many arrangements of MATHMAJOR are there in which each consonant

is adjacent to a vowel?

All possible arrangements look CVCCVCCVC where C stands for

6!

consonant and V stands for vowel. There are = 360 ways to arrange

2!

the consonants because of the repeated M¡¯s. Similarly, there are

3!

= 3 ways to arrange the vowels because of the repeated A¡¯s.

2!

Our final answer is therefore 360*3-=1080 ways.

6. Consider the following modified Plinko! Game. This game models the

situation where you play on a smaller board (6 rows and 4 columns). Since

there is no ¡°center,¡± you start as close to the center as possible.

¡ñ

¡ñ ¡ñ

¡ñ ¡ñ

¡ñ ¡ñ ¡ñ

¡ñ ¡ñ ¡ñ

¡ñ ¡ñ ¡ñ ¡ñ

A B C D

Pr(A) = ___________ Pr(B) = ____________

Pr(C) = ____________ Pr(D) = ____________

Going left to right in each row, here are the probabilities of landing on each

dot.

Row 2: ? and ?

Row 3: ? and ?

Row 4: 3/8, ?, 1/8

Row 5: 5/8, 5/16, 1/16

Row 6: Pr(A) = 5/16, Pr(B) = 15/32, Pr(C) = 3/16, Pr(D) = 1/32

7. Consider the following blocks of Seattle.

C

B

A

a. How many routes are there directly from A to C?

? 4 + 5? ? 9 ?

?? = ?? ??

??

? 4 ? ? 4?

b. How many routes are there from A to C that do NOT go through B?

You want the total # of routes - # routes that go A to B to C

?9? ? 5? ? 4?

?? ?? ? ?? ?? ¡¤ ?? ??

? 4? ? 2? ? 2?

8. You want to seat 4 boys and 4 girls along 1 side of a long table with 8 seats.

a. If the boys and girls are distinguishable, how many ways are there to seat

these eight children?

8! because you are just arranging 8 named people.

b. If the boys and girls are indistinguishable (This means we don¡¯t know

names. We just know B and G.), how many ways are there to seat these

eight children?

8

? ?

?? ?? because placing where 4 B¡¯s go in 8 available spots.

? 4?

c. If the boys and girls are distinguishable and the boys and girls alternate

seats, how many ways are there to seat these eight children?

It either goes BGBGBGBG or GBGBGBGB where we know the names of

the boys/girls. Therefore it is (4!)2 + (4!)2 = 2(4!)2 = 1152

11. If 4 players are each dealt 13 cards from a (standard) 52-card deck, what is

the probability that each player gets one card of each value?

52

? ? 52 ?? 39 ?? 26 ??13 ?

?

?? = ?? ???? ???? ???? ??

The total # of ways to deal is: ??

?13,13,13,13 ? ? 13 ?? 13 ?? 13 ??13 ?

Total # of good outcomes for 1st person is 413 because she has 4 choices

for which ace she gets, 4 choices for which 2 she gets¡­¡­

Similarly, the # of good outcomes for 2nd person is 313 because she has 3

choices for which ace she gets, 3 choices for which 2 she gets¡­¡­This is

because the 1st person already took one of each face value

Similarly, the 3rd person has 213 choices for their hand.

The last person has 1 outcome or 113. This means the total # of good

outcomes is (4!)13. Our final answer for the probability is:

(4!)13

52

?

?

??

??

?13,13,13,13 ?

12. Suppose I choose 14 days from the month of February. What is the

probability that there are 2 of each day (of the week) in the subset?

There are 4 of each of the days of the week in February and want two of

? 28 ?

each of them. The total # of outcomes is ?? ??. The total # of good outcomes

? 14 ?

7

? 4?

is ?? ?? because we want 2 of each of the 7 days of the week. Thus our final

? 2?

7

? 4?

?? ??

2

answer is: ? ? .

? 28 ?

?? ??

? 14 ?

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download