Probability!! (1 Viewer)

[Premus]

Hi

here are some probability q's that i cant get my head around!

From a pack of 9 cards numbered 1,2,3.......9 , three cards are drawn at random and laid on a table from left to right.

a) how many combinations are possible?
My ans: is it just 9P3 ?

b)what is the probability that the number formed exceeds 400?
My ans: (6*8*7) / 9P3 = 2/3

c) what is the probability that the digits are drawn in descending order?
My ans: ??

d) how many of these numbers formed are such that the three digits do not occur in increasing order from left to right?

??

Thanks a lot!!

 

[ngai]

a) yup
b) yup
c) total ways = 9P3
descending orders = 9C3 (for each choice of any 3 #s, only one way to arrange in descending order)
P = 9C3/9P3
d) P(increasing) = P(decreasing), so P = 1-(9P3/9C3)

 

[Xayma]

Hi

here are some probability q's that i cant get my head around!

From a pack of 9 cards numbered 1,2,3.......9 , three cards are drawn at random and laid on a table from left to right.

b)what is the probability that the number formed exceeds 400?
My ans: (6*8*7) / 9P3 = 2/3

Easier to say that it will be >400 if a 4 or greater is drawn first. ie P(≥4)=6/9=2/3

 

[Premus]

sorry but i dont quite understand what the 'C' does??
how do u get 9C3?dont you use 'C' only if the order is not important?

thanks again!!

 

[ngai]

yeah, C when order isn't important
u 9C3 to choose any 3 numbers, disregarding order
then those three, theres only one way to arrange in descending/ascending order