7. Four boys and three girls want to sit on the same bench. Is it more likely that the people at both ends of the bench will be of the same or opposite gender? Determine both probabilities.

The use of a pocket calculator or any manuals is not allowed.
First method:

First, calculate the total number of all possible seating arrangements: Any of the 7 people can sit in the first seat, any of the remaining 6 in the second seat, any of the remaining 5 in the third seat, and so on... The last remaining person can sit in the last seat. In total, there are $7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=5040$ arrangements for how the children can sit on the bench.

2 POINTS

Let's assume that two boys sit at the ends. The four boys can be arranged in these two seats in $4 \cdot 3=12$ ways, and then the remaining 5 people can sit anywhere, which they can do in $5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=120$ ways.

In total, there are $12 \cdot 120=1440$ arrangements in which two boys are at the ends.

2 POINTS

The people at the ends can also be girls. The three girls can be arranged in these two seats in $3 \cdot 2=6$ ways, which together with the 120 arrangements of the remaining 5 people gives $6 \cdot 120=720$ possibilities for how the children can sit on the bench, with two girls at the ends.

Together, there are $1440+720=2160$ arrangements of the children in which people of the same gender sit at the ends. The probability that this happens is $2160: 5040=\frac{3}{7}=0.428571 \ldots \approx 43 \%$

2 POINTS

The probability that people of different genders will sit at the ends is $\frac{4}{7}=0.571428 \ldots \approx 57 \%$

1 POINT

It is more likely that people of different genders will sit at the ends of the bench.

1 POINT

TOTAL 10 POINTS