5 boys and 10 girls are lined up randomly

[terrellc2002]

Okay, 5 boys and 10 girls are lined up randomly

1. What is the probability that the person in the 4th position is a boy?

2. What about the 12th position?

3. What is the probability that a particular boy is in the 3rd position?

I really need help with this one, or just to get it started. Thank you in advance.

 

[galactus]

For the first one:

There are \(\displaystyle \L\\\frac{15!}{10!5!}=3003\) ways to arrange the 15 boys and girls.

Put the boy in the 4th position and arrange the others:

\(\displaystyle \L\\\frac{14!}{10!4!}=1001\)

\(\displaystyle \L\\\frac{1001}{3003}=\frac{1}{3}\)

EDIT: expanding on this method, though Soroban's is easier.


#2. Same as #1

#3. \(\displaystyle \L\\\frac{14!}{15!}=\frac{1}{15}\)

 

[soroban]

Re: Boys and Girls

Hello, Terrell!

The problem is much simpler than you think . . .

5 boys and 10 girls are lined up randomly.

1. What is the probability that the person in the 4th position is a boy?

2. What about the 12th position?

3. What is the probability that a particular boy is in the 3rd position?

1. The 4th position can be occupied by any of the 15 children.
. . There are 5 possible boys.
. . Therefore: \(\displaystyle \,P(\text{4th is a boy}) \:=\:\frac{5}{15}\:=\:\frac{1}{3}\)

2. The same reasoning applies to the 12th position.
. . \(\displaystyle P(\text{12th is a boy})\:=\:\frac{1}{3}\)

3. The 3rd position can be occupied by any of the 15 children.
. . There is 1 way for it to be occupied by that particular boy.
. . Therefore: \(\displaystyle \,P(\text{3rd is a particular boy})\:=\:\frac{1}{15}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Think of it this way . . .

A deck of cards is randomly shuffled.
What is the probability that the top card is a Heart?

We could consider all 52! possible permutations of the cards,
. . but we don't have to.

We are concerned with the top card only.
. . We don't care about the other 51 cards.

The top card could be any of the 52 cards.
There are 13 available Hearts.
Therefore: \(\displaystyle \,P(\text{top card is a Heart}) \:=\:\frac{13}{52}\:=\:\frac{1}{4}\)
. . (This should make sense: one-fourth of the cards are Hearts.)

In fact, the probability that the \(\displaystyle n^{th}\) card is a Heart is \(\displaystyle \,\frac{1}{4}\)

 

[terrellc2002]

Thanks a lot Soroban, I don't know why I was making it more complicated than that. You sure are a big help in here and on that other board. Thank God for Soroban.

Edit: Much thanks to Galactus as well, sorry to leave you out at first