Binomial Probability (1 Viewer)

[karen88]

this was question 4 c iii from da 2004 maths ext 1 hsc paper...

Katie is one of ten members of a social club. Each week one member is selected at random to win a prize. 

(iii) For how many weeks must Katie participate in the prize drawing so that she has a greater chance of winning exactly 3 prizes than of winning exactly 2 prizes?

i had worked up to n!/((n-3)!3!) . p^3q^(n-3) > n!/((n-2)!2!) . p^2q^(n-2) where p=1/10 and q=9/10

looked at answers but i dun get how it is from ^ to :

1/3 . p > 1/3 . p > 1/(n-2) . q

other parts i get... just dun get how to simplify da factorial part.... could anyone explain?

 

[香港!]

this was question 4 c iii from da 2004 maths ext 1 hsc paper...

Katie is one of ten members of a social club. Each week one member is selected at random to win a prize. 

(iii) For how many weeks must Katie participate in the prize drawing so that she has a greater chance of winning exactly 3 prizes than of winning exactly 2 prizes?

i had worked up to n!/((n-3)!3!) . p^3q^(n-3) > n!/((n-2)!2!) . p^2q^(n-2) where p=1/10 and q=9/10

looked at answers but i dun get how it is from ^ to :

1/3 . p > 1/3 . p > 1/(n-2) . q

other parts i get... just dun get how to simplify da factorial part.... could anyone explain?

"n!/((n-3)!3!) . p^3q^(n-3) > n!/((n-2)!2!) . p^2q^(n-2) where p=1/10 and q=9/10"

n!/((n-3)!3!) . p^3q^(n-3) > n!/((n-2)!2!) . p^2q^(n-2)
the n! cancels so we get:
1\3!(n-3)! p^3 q^(n-3)>1\2!(n-2)! p^2 q^(n-2)
times the 2! on both sides and times (n-3)! on both sides:
2!\3! p^3 q^(n-3)>(n-3)!\(n-2)! p^2 q^(n-2)
(1\3) p^3 q^(n-3)>1\(n-2) p^2 q^(n-2)
divide p^2 on both sides and also divide q^(n-3) on both sides
(1\3) p>1\(n-2) q

N.B: 2!\3! means (1*2)\(1*2*3)=1\3
so a number on top that's one less than the bottom, u'll just get the bottom by itself... as seen in (n-3)!\(n-2)!=1\(n-2) since the top is one less than the bottom in the brackets....

 

[karen88]

thnx!!! i finally get it!!!