# Thread: Probability of boys and girls

1. ## Probability of boys and girls

We have 2 girls and 4 boys to be seated ina 6-seat row at the movie theater. What is the probability that the two people at each end of the row were both boys or both girls? Answer should be in common fraction.

My approach to this problem:
Total possibilities of boys and girls is 4! * 2!
No of success outcomes: for Girls: 2! * 2!
for Boys : 4!
p(both boys or both girls)= (4! + 2!*2!)/4! * 2!

Is my approach correct?

2. are you sure the total possibilities of boys and girls is 4!.2! (because it's like girls and boys are seated separately, i think it's 6!)

from the principle of counting

(numbers are printed according to the positions of the seats)

for girls;
2.(4.3.2.1).1 (mentioned within brackets are rest of the members)

for boys;
4.(4.3.2.1).3 (mentioned within brackets are rest of the members)

3. Originally Posted by Ka9
We have 2 girls and 4 boys to be seated ina 6-seat row at the movie theater. What is the probability that the two people at each end of the row were both boys or both girls? Answer should be in common fraction.

My approach to this problem:
Total possibilities of boys and girls is 4! * 2!
No of success outcomes: for Girls: 2! * 2!
for Boys : 4!
p(both boys or both girls)= (4! + 2!*2!)/4! * 2!

Is my approach correct?
Any boy or girl could be in any seat,
therefore the total number of seating arrangements is $6!$

since we do not distinguish between a boy or a girl to count all possibilities.

If the 2 girls are at each end, then either girl can be in position number 1.
Therefore, the number of ways the girls can be at either end is $2!$
The 4 boys are in between and the number of ways they can be arranged is $4!$

Hence, the number of ways the girls can be at both ends of the row with the 4 boys between is ?

There are 4C2 ways to choose 2 boys to be at the ends.
For every such pair, there are $2!$ ways to arrange them.
The 2 girls will be between them with the other 2 boys.
There are again $4!$ ways to arrange those 2 girls and remaining 2 boys.

Hence, the number of ways to have 2 boys at either end with the other 2 boys and 2 girls in between is ?

The sum of these two seating arrangements is ?

4. I could understand the first part i.e girls at the end. But, not the boys are the end.

4!*2!/6! + ?/6!

5. B----B
for boys at the end, 4c2
boys in the middle ----, since there are 4 spots, 4c2 and remaining 2 spots for girls i.e 2!

(4c2*4c2*2!)/6!

So, the final answer is p(boys or girls at the end) = (4!2!)/6! + (4c2*4c2*2!)/6!

May I know where I went wrong?

6. Originally Posted by Ka9
B----B
for boys at the end, 4c2
boys in the middle ----, since there are 4 spots, 4c2 and remaining 2 spots for girls i.e 2!

(4c2*4c2*2!)/6!

So, the final answer is p(boys or girls at the end) = (4!2!)/6! + (4c2*4c2*2!)/6!

May I know where I went wrong?
You must choose 2 boys to be at the ends only
(or choose 2 boys to be in the middle only but that's maybe more awkward).

This is because when you choose 2 boys to be at the ends,
then the "other" 2 boys automatically go with the 2 girls to be in between,
and we can just treat those in between as a group of 4.

So, do you see that you only need to choose 2 boys to be at the end
and every time you have 2 boys at the ends, there will be 2 boys and 2 girls
(4 teenagers or children) in between.

There are 4C2 ways to choose 2 boys (just a pair before arranging them).
Arranging them, there are 2!(4C2) arrangements of 2 boys at the ends
(in the case of the girls at the end, there were just 2! arrangements).
For each of these arrangements of 2 boys at the ends there are 4! arrangements
of the 2 boys and 2 girls in between.

If boys are at the end, 4!*4c2*2!/6!

p(2girls and 4boys at the end)= 2!*4!/6! + 4!*4c2*2!/6!

Thanks,

8. Originally Posted by Ka9

If boys are at the end, 4!*4c2*2!/6!

p(2girls or 2boys at the ends)= 2!*4!/6! + 4!*4c2*2!/6!

Thanks,
Yes,

though it's the probability of "2 girls or 2 boys" at the ends.
Then the probabilities of the two individual cases are summed.

As a fraction then, we have

$\displaystyle\frac{4!2!+4!\frac{4!}{2!2!}2!}{6!}=\ frac{4!(2+12)}{6!}=\frac{4!(14)}{6(5)4!}=\frac{14} {30}=\frac{7}{15}$

To see how it works out, we can list

$G1\;-\;-\;-\;-\;G2$

$G2\;-\;-\;-\;-\;G1$

$B1\;-\;-\;-\;-\;B2$

$B2\;-\;-\;-\;-\;B1$

$B1\;-\;-\;-\;-\;B3$

$B3\;-\;-\;-\;-\;B1$

$B1\;-\;-\;-\;-\;B4$

$B4\;-\;-\;-\;-\;B1$

$B2\;-\;-\;-\;-\;B3$

$B3\;-\;-\;-\;-\;B2$

$B2\;-\;-\;-\;-\;B4$

$B4\;-\;-\;-\;-\;B2$

$B3\;-\;-\;-\;-\;B4$

$B4\;-\;-\;-\;-\;B3$

In each case there are $4!$ arrangements in between.

Therefore the probability is

$\displaystyle\frac{(14)4!}{6(5)4!}$

Yes, it's kind of confusing in the beginning. But, thanks for the detailed reply.

10. For a question like this, if you think it through until you fully understand it,
you learn a lot.

Another way to calculate the arrangements with the boys at the end is....

Any of 4 boys can be in position 1.
For each of these, any of the 3 remaining boys can be in position 6.

Hence there are 4(3)=12 ways to have boys at either end.

Now all you have to do is arrange the 4 people in between for each case.
There are 4! arrangements of the people in between, hence there are 12(4!) arrangements with boys at the end.

11. The easiest way to answer a question like this is...

There are always 4 people in between,
hence we need only examine the 2 people at the ends.

Any of 6 can be in position 1.
For each of these, there are 5 remaining to choose from for position 6.

Therefore there are 30 possible ways to fill positions 1 and 6.

If these positions are occupied by the 2 girls, there are 2 ways to fill position 1
and the other girl fills position 6.
That's 2 ways.

If these positions are filled by boys, there are 4 ways to fill position 1.
For each of the boys in position 1, there are 3 ways to fill position 6.
That's 12 ways.

Therefore the probability is $\frac{2+12}{30}$