The Student Room Group

Permutation and Combination

IMG-20160117-WA0004.jpg

I can't solve part (iii)
I did:
(3*11)/(14*14)
because probability of Mrs. Brown is 3/14
and Mrs. Li is 11/14
But this is not the answer
Original post by 123PC
IMG-20160117-WA0004.jpg

I can't solve part (iii)
I did:
(3*11)/(14*14)
because probability of Mrs. Brown is 3/14
and Mrs. Li is 11/14
But this is not the answer


This is a tricky little one! I think that the easiest approach is to think in terms of the number of configurations that work and then divide by the total number of configurations. So

(i) There are 3 seats that meet the criterion for Mrs Brown.
(ii) Once Mrs Brown is correctly seated, there are then 10 seats that allow for Mrs Lin to be correctly seated. (Why?)
(iii) Once Mrs Brown and Mrs Lin are seated, you need to count the number of possibilities for putting one student behind Mrs Lin.
(iv) Now how many possibilities are there for seating the remaining passengers?
(v) Finally, to turn this into a probability, divide through by the total number of seating configurations that you found in the first part.
Reply 2
Original post by Gregorius
This is a tricky little one! I think that the easiest approach is to think in terms of the number of configurations that work and then divide by the total number of configurations. So

(i) There are 3 seats that meet the criterion for Mrs Brown.
(ii) Once Mrs Brown is correctly seated, there are then 10 seats that allow for Mrs Lin to be correctly seated. (Why?)
(iii) Once Mrs Brown and Mrs Lin are seated, you need to count the number of possibilities for putting one student behind Mrs Lin.
(iv) Now how many possibilities are there for seating the remaining passengers?
(v) Finally, to turn this into a probability, divide through by the total number of seating configurations that you found in the first part.


Ohhhh. I get it. The answer matches now! Thank you!

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