1 probability question (boys, girls split between car, van)

[mcrae]

ten boys and 12 girls decide to rent a 16 passanger van and 6 passanger car. if the group is distributed randomly, what is the probability that
a) there are no boys in the car
b) there are no girls in the car
c) alan and margaret are both in the van
d) there are more girls than boys in the car?

a) 12C6 / 22C6 would be my answer
b) 10C6 / 22C6 would be my answer
c) 20C6 / 22C6 is the answer, and i don't quite get it, if someone could post an alternate method or reasoning that would be awsome ( im assuming theres a different way to do it, if not, oh well)
d)i figure you'd add up the cases when there are 4, 5, or 6 girls in the car, so {12C4 + 12C5 + 12C6 } / 22C6, but thats wrong.
so then i tried 14C6 + 13C6 +12C6, which is the same idea as before; 4 5 or 6 girls in the car, but thats wrong as well

thanks for your help

 

[pka]

d) \(\displaystyle \L
\frac{{\sum\limits_{k = 0}^2 {\left( \begin{array}{c}
12 \\
6 - k \\
\end{array} \right)\left( \begin{array}{c}
10 \\
k \\
\end{array} \right)} }}{{\left( \begin{array}{c}
22 \\
6 \\
\end{array} \right)}}\), no boys, 1 or 2 boys.

 

[mcrae]

what does the notation on the left signify? the 2, the sign similar to E and the k=0?
thanks for the help though, got the correct answers

 

[pka]

You do not know about summations?
The sum of k=0 to k=2 ?

 

[mcrae]

nope, we haven't heard of the word summations yet in the course. it seems to just be a more concise and 'proper' way to illustrate the solution than how we are being taught, which is just the same format as my attempts at d) show. i'm off to google summations now

 

[soroban]

Re: 1 probability question (boys, girls split between car, v

Hello, mcrae!

Ten boys and 12 girls decide to rent a 16-passanger van and 6-passanger car.
If the group is distributed randomly, what is the probability that:

c) Alan and Margaret are both in the van.
20C6 / 22C6 is the answer.

If Alan and Margaret are both in the van, then they are not in the car.
In how many ways can this happen?

The six car-passengers must be chosen from the other twenty people.
. . There are: .\(\displaystyle _{20}C_{_6}\) ways to do this.

Therefore, the probability is: .\(\displaystyle \L\,\frac{_{20}C_6}{_{22}C_6} \;=\;\frac{\frac{20!}{6!\cdot14!}}{\frac{22!}{6!\cdot16!}} \;=\;\frac{40}{77}\)


An alternate approach:
If Alan and Margaret are both in the van,
\(\displaystyle \;\;\)then the 14 other van-passengers are chosen from the other 20 people.
There are: .\(\displaystyle _{20}C_{14}\) ways.

Therefore, the probability is: .\(\displaystyle \L\,\frac{_{20}C_{14}}{_{22}C_{16}} \;= \;\frac{\frac{20!}{14!\cdot6!}}{\frac{22!}{16!\cdot6!}} \;=\;\frac{40}{77}\)

(d) there are more girls than boys in the car.

You had the right idea, but you didn't complete each case.

[1] There are 4 girls and 2 boys in the car.
\(\displaystyle \;\;\)There are: \(\displaystyle \,(_{12}C_4)(_{10}C_2)\) ways.

[2] There are 5 girls and 1 boy in the car.
\(\displaystyle \;\;\)There are: \(\displaystyle \,(_{12}C_5)(_{10}C_1)\) ways.

[3] There are 6 girls (and no boys) in the car.
\(\displaystyle \;\;\)There are: \(\displaystyle \,_{12}C_6\) ways.