college prob + stat

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July 18th 2006, 09:09 AM
magicpunt

college prob + stat

2 Questions:

1) There are 5 men and 5 women in a group. A committee is to be formed of 5 people. What is the probability that the group is all of the same sex (all men or all women)?

2) There are 10 people in an elevator in the basement. They can get off at any of 5 floors above. The probability that they get off at each floor is equal and each person leaving is independant of each other. What is the probability that 2 people get off at each floor?
July 18th 2006, 10:08 AM
ThePerfectHacker

Quote:

Originally Posted by magicpunt

1) There are 5 men and 5 women in a group. A committee is to be formed of 5 people. What is the probability that the group is all of the same sex (all men or all women)?

$P(\mbox{men or woman})=P(\mbox{men})+P(\mbox{woman})$
Let us find,
$P(\mbox{men})$
There are 10 people and 5 are selected there are a total of $_{10}C_5=252$
From 5 men you take 5 men intotal the number of combinations is,
$_5C_5=1$
Thus,
$P(\mbox{men})=\frac{1}{252}$
Similarily ya have,
$P(\mbox{women})=\frac{1}{252}$
Thus, the probability is,
$\frac{1}{252}+\frac{1}{252}=\frac{2}{252}=\frac{1} {126}$
July 18th 2006, 11:45 AM
galactus

Quote:

2) There are 10 people in an elevator in the basement. They can get off at any of 5 floors above. The probability that they get off at each floor is equal and each person leaving is independant of each other. What is the probability that 2 people get off at each floor?

This is similar to the 'boxes and balls' problems.

There are $5^{10}=9765625$ ways to distribute the people to the 5 floors.

There are $\frac{5!}{2!}=60$ ways that no two people get off on the same floor.

EDIT:
Yes, I misinterpreted the problem. Cerebral flatulence. Soroban's method is solid.
My second part should be $\frac{10!}{2^{5}}$

Therefore, $\frac{5^{10}}{\frac{10!}{2^{5}}}$

This is a version of the classic 'assigning diplomats' problem.

For instance:

How many ways can 10 different diplomats be assigned to 5 different countries if 2 diplomats must be assigned to each country?.

See the similarity with your problem?. Yours is just elevators and people, instead.

This can be done in $\frac{10!}{(2!)^{5}}=113,400$ ways.

The method I used is just another variation on this problem. Keep it, you may need it.
July 18th 2006, 12:05 PM
magicpunt

Quote:

Originally Posted by galactus

This is similar to the 'boxes and balls' problems.

There are $5^{10}=9765625$ ways to distribute the people to the 5 floors.

There are $\frac{5!}{2!}=60$ ways that no two people get off on the same floor.

I think you might've misunderstood the problem

The question is: what is P(Exactly 2 people get off at each floor)
July 18th 2006, 12:33 PM
Soroban

Hello, magicpunt!

Here's my approach to #2 . . .

Quote:

2) There are 10 people in an elevator in the basement.
They can get off at any of 5 floors above.
The probability that they get off at each floor is equal
and each person leaving is independant of each other.
What is the probability that 2 people get off at each floor?

Each of the ten people has a choice of five floors.
. . There are: $5^{10}$ ways that they can leave the elevator.

If exactly two people get off at each floor,
. . the number of ways is: $\begin{pmatrix}10\\2,2,2,2,2\end{pmatrix} = 113,400$

Therefore, the probabilty is: . $\frac{113,400}{5^{10}}\:=\:\frac{4,536}{390,625}$
July 18th 2006, 07:53 PM
malaygoel

Quote:

Originally Posted by magicpunt

2 Questions:

1) There are 5 men and 5 women in a group. A committee is to be formed of 5 people. What is the probability that the group is all of the same sex (all men or all women)?

Another approach to #1
Find the total no. of ways of selecting 5 out of 10= $\frac{10!}{5!5!}=C_5^{10}$
No. of ways in which all are men=1
No. of ways in which all are women=1
Required probability= $\frac{1+1}{C_5^{10}}=\frac{1}{126}$

Malay