# Probability

• May 1st 2012, 04:54 AM
nazz
Probability
There are 4 girls and 5 boys would like to go on school trip but there are onlyy 2 seats available.

a) calculate the probability that one girl and one boy will be chosen
ans P (BG)+ P(GB)=( 5/9 x 4/8 ) +( 4/9 x 5/8 )

b) if Sarah is one of the girls who want to go on the trip, what is the probabily of not choosing Sarah?
How to solve it?
• May 1st 2012, 06:31 AM
Quant
Re: Probability
Hi!

Quote:

Originally Posted by nazz
There are 4 girls and 5 boys would like to go on school trip but there are onlyy 2 seats available.

a) calculate the probability that one girl and one boy will be chosen
ans P (BG)+ P(GB)=( 5/9 x 4/8 ) +( 4/9 x 5/8 )

I don't agree with that, because you don't care about who is chosen first.
So it's just P(one boy and one girl) = 5/9*4/8

Quote:

Originally Posted by nazz
b) if Sarah is one of the girls who want to go on the trip, what is the probabily of not choosing Sarah?
How to solve it?

There are 9 people: sarah and 8 other students.

P(sarah is not chosen to go on the trip) = 8/9 * 7/8

In this case you do not care about if two boys are chosen for example. You just want to know if sarah is one of the lucky ones getting the chance to go on this trip.
• May 1st 2012, 06:57 AM
Plato
Re: Probability
Quote:

Originally Posted by nazz
There are 4 girls and 5 boys would like to go on school trip but there are onlyy 2 seats available.
a) calculate the probability that one girl and one boy will be chosen
ans P (BG)+ P(GB)=( 5/9 x 4/8 ) +( 4/9 x 5/8 )

Quote:

Originally Posted by Quant
I don't agree with that, because you don't care about who is chosen first.
So it's just P(one boy and one girl) = 5/9*4/8

Actually the above answer is correct.

It is gotten by $\displaystyle \dfrac{\dbinom{5}{1}\dbinom{4}{1}}{\dbinom{9}{2}}= 2\cdot\frac{5}{9}\cdot\frac{4}{8}$
• May 1st 2012, 07:02 AM
Quant
Re: Probability
Hello Plato!

Quote:

Originally Posted by Plato
Actually the above answer is correct.

It is gotten by $\displaystyle \dfrac{\dbinom{5}{1}\dbinom{4}{1}}{\dbinom{9}{2}}= 2\cdot\frac{5}{9}\cdot\frac{4}{8}$

Where does the factor 2 come from?

I think

$\displaystyle \dfrac{\dbinom{5}{1}\dbinom{4}{1}}{\dbinom{9}{2}}$

is equal to
$\displaystyle =\frac{5}{9}\cdot\frac{4}{8}$

Am I wrong?
• May 1st 2012, 07:22 AM
biffboy
Re: Probability
9c2=36
• May 1st 2012, 07:24 AM
Plato
Re: Probability
Quote:

Originally Posted by Quant
$\displaystyle \dfrac{\dbinom{5}{1}\dbinom{4}{1}}{\dbinom{9}{2}}$
is equal to
$\displaystyle =\frac{5}{9}\cdot\frac{4}{8}$
Am I wrong?

Yes you are.
$\displaystyle \dfrac{\dbinom{5}{1}\dbinom{4}{1}}{\dbinom{9}{2}}= \dfrac{5\cdot 4}{\dfrac{9\cdot 8}{2\cdot 1}}$
• May 1st 2012, 07:41 AM
HallsofIvy
Re: Probability
Quote:

Originally Posted by Quant
Hi!

I don't agree with that, because you don't care about who is chosen first.
So it's just P(one boy and one girl) = 5/9*4/8

Actually, nazz is correct precisely because "you don't care about who is chosen first"! What you gives is the probability that a boy is chosen first and a girl second. The probability that as girl would be chosen first and then a boy would be (4/9)(5/8) and the probability of "one boy and one girl in either order" is the sum of those, as nazz gives.

Quote:

There are 9 people: sarah and 8 other students.

P(sarah is not chosen to go on the trip) = 8/9 * 7/8

In this case you do not care about if two boys are chosen for example. You just want to know if sarah is one of the lucky ones getting the chance to go on this trip.
• May 1st 2012, 08:30 AM
Quant
Re: Probability
Quote:

Originally Posted by Plato
Yes you are.
$\displaystyle \dfrac{\dbinom{5}{1}\dbinom{4}{1}}{\dbinom{9}{2}}= \dfrac{5\cdot 4}{\dfrac{9\cdot 8}{2\cdot 1}}$

whop, I'm terribly ashamed.
Well, thank you!

Quote:

Originally Posted by HallsofIvy
Actually, nazz is correct precisely because "you don't care about who is chosen first"! What you gives is the probability that a boy is chosen first and a girl second. The probability that as girl would be chosen first and then a boy would be (4/9)(5/8) and the probability of "one boy and one girl in either order" is the sum of those, as nazz gives.

I see, thanks.
• May 1st 2012, 02:37 PM
nazz
Re: Probability
What about part b) can someone shed more light on it please and thanks for all the replies
• May 1st 2012, 03:17 PM
Plato
Re: Probability
Quote:

Originally Posted by nazz
There are 4 girls and 5 boys would like to go on school trip but there are onlyy 2 seats available.
b) if Sarah is one of the girls who want to go on the trip, what is the probabily of not choosing Sarah?
How to solve it?

The probability that Sarah is chosen equals $\displaystyle \dfrac{\binom{8}{1}}{\binom{9}{2}}~.$
• May 1st 2012, 03:59 PM
nazz
Re: Probability
Thanks for for all the replies you all have been great.