Question

PLEASE HELP!! GRADE 12 DATA MANAGEMENT 
Twelve boys and fourteen girls decide to rent a 16 passenger van and a 10 passenger van to drive to a
Math Contest in Whitby. If the group is distributed randomly between the vehicles, what is the
probability that
a)There are no boys in smaller van
b)There are no girls in the smaller van
c)Sureba and Sana are both in the larger van
d)There are more boys than girls in the larger van

Answer 1

$C(n, r)= \frac{n!}{(n-r!)r!}$

is the formula which gives the total number of groups of r, that can be formed from n objects.

r! is "r factorial", which means 1*2*3*...*(r-1)*r, so multiplication of all integers from 1 to r.

In the problem there are 12 boys, 14 girls. The vans are 16 and 10 passengers vans.

In total, there are 26 students, so there are C(26, 16) ways to sit the students in the larger van. For each group we select for the larger van, the remaining 10, go automatically to the smaller van.

$C(26, 16)= \frac{26!}{10!16!}= \frac{26*25*24*23*22*21*20*19*18*17*16!}{10!*16!}$

$=\frac{26*25*24*23*22*21*20*19*18*17}{10!}= \frac{26*25*24*23*22*21*20*19*18*17}{10*9*8*7*6*5*4*3*2*1}$

simplifying, we have 5,311,735

which is the number of the sample space.

a) "no boys in the smaller van" means "only girls in the smaller van".
This can be done by C(14,10) ways. (so we consider only the 14 girls to sit in the 10 passenger van)

$C(14,10)= \frac{14!}{4!10!}= \frac{14*13*12*11}{4*3*2}=7*13=91$

P(no boys in the smaller van)=$\frac{91}{5,311,735}=0.000017$

b)

"no girls in the smaller van" means "only boys in the smaller van".
This can be done by C(12,10) ways.

$C(12,10)= \frac{12!}{2!10!}= \frac{12*11}{2}=6*11=66$

P(no girls in the smaller van)=$\frac{66}{5,311,735}=0.0000124$

c) assuming Sureba and Sana are, most likely, girls.

Sureba and Sana are already seated in the larger van. The remaining 14 seats, can be completed in C(24, 14) ways.

$C(24, 14)= \frac{24!}{14!10!}= [tex]\frac{24*23*22*21*20*19*18*17*16*15}{10*9*8*7*6*5*4*3*2*1}$=1,961,256

P(both Sureba and Sana are in the larger van)= $\frac{1,961,256}{5,311,735}=0.37$

d) In the larger Van there can be maximum 12 boys. we may have the following cases, with the given total number of groups:

12b 4g      C(12, 12)C(14, 4)=1*1001=1001
11b 5g      C(12, 11)C(14, 5)=12*2002=24,024
10b 6g      C(12, 10)C(14, 6)=66*3003=198,198
 9b  7g      C(12, 9)C(14, 7)=220*3432=755,040

since these are separate, disjoint possibilities, we add the probabilities. To calculate C(n,r) directly, there are many online software available.

P(There are more boys than girls in the larger van)

=$\frac{1001+24,024+198,198+755,040}{5,311,735}= \frac{978,263}{5,311,735}=0.184$

Answers:

a) 0.000017

b) 0.0000124

c) 0.37

d) 0.184