Marco and a group of seven friends, four girls and four boys in all, meet for a dinner. The dinner table is round and the seats are numbered. The group decides to assign the seats at random, each drawing their own seat number. Calculate the following probabilities: a) PA: that Marco has two boys on either side; b) PB: that Marco has a boy and a girl on either side; c) PC: that Marco has two girls on either side.
For the first probability I obtained $ P_A = \frac{ \binom{3}{1} \frac{1}{3} \binom{2}{1} }{ \binom {4}{3} } = 0,5 $ For the second probability I obtained $ P_B = \frac{ \binom{4}{1} \frac{1}{4} \binom{3}{1} }{ \binom {15}{3} } = 0,00659 $ For the third probability I obtained $ P_c = 0,00659 $
But I don’t know if my result are correct.. thank you !
There are $$\binom{7}{2}$$ ways to select two people to sit next to Marco (without considering who sits to his left and who sits to his right). Since Marco is one of the four boys, three of the seven friends who could sit next to him are boys. Thus, there are $$\binom{3}{2}$$ ways to select two boys to sit next to Marco (again, without considering who sits to his left and who sits to his right). Hence, the probability that Marco is seated between two boys is $$P_A = \frac{\dbinom{3}{2}}{\dbinom{7}{2}}$$ If you prefer to think about order, note that there are seven ways to choose the person who sits to Marco's left and six ways to choose the person who sits to Marco's right. For the favorable cases, there are three ways to choose the boy who sits to Marco's left and two ways to choose the boy who sits to his right. Thus, $$P_A = \frac{3 \cdot 2}{7 \cdot 6}$$ which agrees with the answer obtained above.
For part (b), choose a boy and a girl who will sit next to Marco.
For part (c), choose two girls who will sit next to Marco.