#### PlanetMaster

**XPRS Administrator**

- Messages
- 1,146

- Reaction score
- 2,014

- Points
- 273

There are several ways to tackle this kind of question.STATISTICSHELP NEEDED

I got the first two parts but I am so confused about solving (iii). Can anyone help me out by providing a good explanation so that I can, too, develop that way of thinking.

I prefer the plain probability way (as opposed to permutations).

So since Mrs Brown is a front row hogger, she could sit in any of the 3 seats from 14 total i.e. 3/14

Now for Mrs Lin to sit directly behind a student, she can sit anywhere except the front row so that's any of the seats highlighted in red.

So that's 11 possible seats but since Mrs Brown has occupied 1 of the 3 seats in front, Mrs Lin can't sit behind her so we have 10 possibilities here i.e. 10/13.

*(Its 13 because 1 seat is no longer an option and we have 13 to choose from)*

Finally, there are 5 students and 12 passengers so the probably of a student sitting in the seat in front of Mrs Lin is 5/12.

So,

\(p=\frac{3}{14}\times\frac{10}{13}\times\frac{5}{12}\)

Basically, we staged the entire scenario step by step here. First, we put Mrs Brown in one of the front seats. Then we put Mrs Lin in one of the possible seats where a student could be in front of her. Then, finally, we forced a student in that seat against their will!

Hope this helps!