A teacher wants to randomly form two teams of 5 students from a group of 5 girls and 5 boys for a sports activity. Two of the girls, Ann and Alice, are selected as team leaders. The ten students are seated at a round table of 10. Find the probability that no two of the remaining 3 girls are next to each other given that Ann and Alice are not seated together.
So we see that the conditional probability is $$P(A|B) = \dfrac{P(A \cap B)}{P(B)}$$
We define $B$ as the event that Ann and Alice are not seated together. Which is $$\dfrac{7! \times {8 \choose 2} \times 2!}{9!} = \dfrac{7}{9}$$
However i have trouble computing out the probability of $A \cap B$. So i fix the 5 guys as $4!$ . And i insert in Ann and Alice in their interval which is ${5 \choose 2} \times 2!$. Lastly i insert the 3 remaining girls into the rest of the intervals? Which is ${7 \choose 3} \times 3!$ ; Apparently this is wrong logic, can anyone help?
And it will be helpful that the given answer is $\frac{85}{196}$. Note this is not a duplicate question, it is asking a very different question.
Your answer is correct.
Seat Ann. If we seat everyone other than Alice at the table, there are nine spaces to the right of a person in which Alice could be seated. Two of these are next to Ann. Hence, the probability that Ann and Alice are not seated together is $7/9$, as you found.
For the favorable cases, seat Ann. As we proceed clockwise around the table from Ann, the five boys can be seated in $5!$ ways. This creates six spaces to the right of a person at the table, two of which are adjacent to Ann, so Alice can be seated in $4$ ways. We now have seven people at the table, so there are seven spaces to the right of a person at the table. To ensure that no two of the remaining girls are seated in adjacent seats, we must choose three of these spaces for those girls. Once the spaces have been selected, there are $3!$ ways of arranging the girls in the selected spaces. Since there are $9!$ ways of arranging the girls, the probability that no two of the three girls other than Ann and Alice sit in consecutive seats is $$\frac{5! \cdot 4 \cdot \binom{7}{3} \cdot 3!}{9!} = \frac{5}{18}$$ again as you found.
Therefore, the desired probability is $$\frac{\frac{5}{18}}{\frac{7}{9}} = \frac{5}{14}$$