Probability Help (2 Viewers)

[Dragonmaster262]

I'll rep anyone who can solve these for me.

1) A group consisting of 3 men and 6 women attends a prizegiving ceremony.
i. If the members of the group sit down at random in a straight line, find the probability that the
three men sit next to each other.
ii. If 5 prizes are awarded at random to members of the group, find the probability that exactly 3
of the prizes are awarded to women if
a. there is a restriction of at most one prize per person.
b. there is no restriction on the number of prizes per person.

2) A die is loaded in such a way that in 8 throws of the die, the probability of getting 3 even numbers is
four times the probability of getting 2 even numbers. Find the probability that a single throw of the
die results in an even number.

3) Four families each have four children. What is the probability that exactly two of these families have
two boys and two girls? (Assume that each child is equally likely to be a boy or a girl)

 

[Tofuu]

gahh, permutations and combinations, never did this in so long, kinda forgot how to do it, well i'll attempt the first one LOL

i)

w=women, m=men

w w w w w w (3m)

so i think its 7! times (3 x 2 x1)
since the men can be withing their little group

=30245

total combination would be 362880


so the prob that the men sit together

=30245/362880


ii. If 5 prizes are awarded at random to members of the group, find the probability that exactly 3
of the prizes are awarded to women if
a. there is a restriction of at most one prize per person.

[3 women get it] = 6P3-------------------------(1)
[total possiblities] = 9P5-----------------------(2)

so i think its (1) / (2)

can you check that for me

 

[Dragonmaster262]

Part 1 seems to be right but part 2 doesnt. It's supposed to be 10/21.

 

[Luxxey]

For ii) a.

Total 5 prizes: 9C5
3W have prizes: 6C3
2M have prizes: 3C2

Therefore probability of 3 women receiving prizes w/ limit 1pp: (6C3 x 3C2) / 9C5 = 10/21.


Also, what's the answer to b? I want to check the answer I'm getting.

 

[Dragonmaster262]

The answer to b is 80/243.

The answer to 2 is 2/3 and the answer to 3 is 675/2048.

All I need is working out. Can anybody help?

 

[babysnow]

are you sure the answer for b is 80/243? because i got 8/243... :O

i have an answer for qn2 which is 2/3 - it involves binomials and i dont know if you have done them yet? let me know and ill post it up.. im not sure if its how you do it =/

 

[Dragonmaster262]

Well answers may be wrong but please post up any working out and answers that you have anyway.

 

[babysnow]

(b)
<a latex=6^3\times 3^2\\$since there are 6 women who can get 3 prizes and without restrictions$ \\9^5\quad$is the total number of possibilities\\\therefore \frac{6^3\times 3^2}{9^5}\quad=\frac{8}{243}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?6^3\times 3^2\\$since there are 6 women who can get 3 prizes and without restrictions$ \\9^5\quad$is the total number of possibilities\\\therefore \frac{6^3\times 3^2}{9^5}\quad=\frac{8}{243}" title="6^3\times 3^2\\$since there are 6 women who can get 3 prizes and without restrictions$ \\9^5\quad$is the total number of possibilities\\\therefore \frac{6^3\times 3^2}{9^5}\quad=\frac{8}{243}" /></a>


Q2:
<a latex=$Let$\quad\begin{matrix}p &=& P(\text{even}) \\ q &=& P(\text{odd}) \end{matrix} \qquad$where p@plus;q=1\\\\ \begin{matrix}P(\text{3 even)} &=& {8\choose3}p^3q^5 &=& 56p^3q^5 \\ \\ P(\text{2 even}) &=& {8\choose2}p^2q^6 &=& 28p^2q^6 \end{matrix} \\ \\ \therefore 56p^3q^5 \:=\:4\cdot28p^2q^6 \quad\Rightarrow\quad p \:=\:2q \\ \\ $Since\;p@plus;q\:=\:1,\; $we\:have\; p = \tfrac{2}{3},\;q =\tfrac{1}{3} \\ \\\therefore P(\text{even}) \:=\:\frac{2}{3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?$Let$\quad\begin{matrix}p &=& P(\text{even}) \\ q &=& P(\text{odd}) \end{matrix} \qquad$where p+q=1\\\\ \begin{matrix}P(\text{3 even)} &=& {8\choose3}p^3q^5 &=& 56p^3q^5 \\ \\ P(\text{2 even}) &=& {8\choose2}p^2q^6 &=& 28p^2q^6 \end{matrix} \\ \\ \therefore 56p^3q^5 \:=\:4\cdot28p^2q^6 \quad\Rightarrow\quad p \:=\:2q \\ \\ $Since\;p+q\:=\:1,\; $we\:have\; p = \tfrac{2}{3},\;q =\tfrac{1}{3} \\ \\\therefore P(\text{even}) \:=\:\frac{2}{3}" title="$Let$\quad\begin{matrix}p &=& P(\text{even}) \\ q &=& P(\text{odd}) \end{matrix} \qquad$where p+q=1\\\\ \begin{matrix}P(\text{3 even)} &=& {8\choose3}p^3q^5 &=& 56p^3q^5 \\ \\ P(\text{2 even}) &=& {8\choose2}p^2q^6 &=& 28p^2q^6 \end{matrix} \\ \\ \therefore 56p^3q^5 \:=\:4\cdot28p^2q^6 \quad\Rightarrow\quad p \:=\:2q \\ \\ $Since\;p+q\:=\:1,\; $we\:have\; p = \tfrac{2}{3},\;q =\tfrac{1}{3} \\ \\\therefore P(\text{even}) \:=\:\frac{2}{3}" /></a>

 

[Dragonmaster262]

Got answers for quesion 3?

 

[frenzal_dude]

This is what I got for the first one.

1)i)
There are 7 ways for 3 men to sit next to each other:
1. M M M W W W W W W
2. W M M M W W W W W
3. W W M M M W W W W
4. W W W M M M W W W
5. W W W W M M M W W
6. W W W W W M M M W
7. W W W W W W M M M

For the first scenario:
Imagine the first person goes to get a seat, the chance that it is a man=3/9, the chance that he chooses the first seat = 1/9. Therefore the chance that a man sits in the first seat = 3/9 * 1/9 = 1/27.

The 2nd peson goes to get a seat, the chance that it is a man is now 2/8, the chance that he chooses the 2nd seat is 1/8, therefore the chance that he is a man and chooses the 2nd seat is 2/8 * 1/8 = 1/32.

The 3rd person goes to get a seat, the chance that it is a man is 1/7, the chance that he sits in the 3rd seat is 1/7, therefore the chance that a man sits in the 3rd seat is 1/49.

The chance of those three things happening is (1/27) * (1/32) * (1/49) = 1/42236.

That is one way that 3 men could sit next to eachother, since there are 7 ways, multiply 1/42236 by 7.

The answer is 1/6048
_____________________________________________________________________________________________________
1)ii)a

The answer is 10/21

Draw up a tree diagram, remember to minus 1 person from the denominator as you move to the next prize.

1)ii)b

The answer is 80/243.
some people thought it was 8/243, that's the probability of getting 3 prizes for the women in ONE scenario, but there are 10 different scenarios, so you need to multiply 8/243 by 10 which gives you 80/243.

If you PM me your email I'll email you a copy of my working out.
_____________________________________________________________________________________________________
2)
(no idea)
_____________________________________________________________________________________________________
3)
There are four families:
Family 1
Family 2
Family 3
Family 4

There are 6 ways to have 2 families together: 1,2 1,3 1,4 2,3 2,4 3,4

Draw a tree diagram for one family having 4 kids, the chance of having 2 boys and 2 girls is 6 * 1/16 = 3/8.

The chance of family 1 and family 2 having 2 boys and 2 girls is 3/8 * 2 = 9/64

There are 6 combinations of families, therefore 6 * 9/64 = 27/32 is the answer

I'm not 100% sure I've done this right however.

 

[cutemouse]

I'll rep anyone who can solve these for me.

1) A group consisting of 3 men and 6 women attends a prizegiving ceremony.
i. If the members of the group sit down at random in a straight line, find the probability that the
three men sit next to each other.

Consider the 3 men as a unit. This is done in 3! ways. This leaves (9-3)+1=7 people units.



I'll do the rest tomorrow. I need some sleep...

 

[cutemouse]

I've done all questions. All the answers you had provided seemed correct.

BTW, I really liked these questions

 

[frenzal_dude]

Dragonmaster, do you have the answer for 1i? I got 1/6048 but noone else seems to be getting that answer.

 

[Dragonmaster262]

Dragonmaster, do you have the answer for 1i? I got 1/6048 but noone else seems to be getting that answer.

It's what Jim said 1/12.

 

[Dragonmaster262]

I've done all questions. All the answers you had provided seemed correct.

BTW, I really liked these questions

Thanks I'll rep you again as soon as I can.

 

[cutemouse]

You can rep me if you want I don't really need/care about it, as you can see

In (1)(ii)(b) you can also use 5C3 instead of 5C2, but it's the same thing really. ie. arranging 5 prizes into groups of 3 or 2 prizes respectively.