• Congratulations to the Class of 2024 on your results!
    Let us know how you went here
    Got a question about your uni preferences? Ask us here

CSSA 2004 Trails Question (1 Viewer)

ressul

Mr
Joined
Mar 17, 2004
Messages
94
Gender
Male
HSC
2004
In a certain street, 40% of the households have at least 2 cars. If 4 households are chosen at random, find the peobability that:
b.at most of these households have at least 2 cars.
What does this question mean and how to do it?
 

Rorix

Active Member
Joined
Jun 29, 2003
Messages
1,818
Gender
Male
HSC
2005
its a binomial probability question
 

cranberries

show Pony
Joined
Aug 16, 2003
Messages
74
Location
totally gone with the wind
Gender
Undisclosed
HSC
N/A
you do it like:

P(At least 2 cars)= 0.4 = p
P(NOT at least 2 cars) =0.6= q

(p+q)^4 = whatever the binomial expansion is

i would write it but i'm not very good with typing maths stuff

and then you find the one with the p's and q's that you want and sub in the 0.4's and 0.6's !
 

Rorix

Active Member
Joined
Jun 29, 2003
Messages
1,818
Gender
Male
HSC
2005
ressul said:
In a certain street, 40% of the households have at least 2 cars. If 4 households are chosen at random, find the peobability that:
b.at most of these households have at least 2 cars.
What does this question mean and how to do it?
by the way, your question is missing a number. It should be "at most X of these" where x is a number.

hence it cant be answered.

it will have something to do with summing 4Cn (0.4)^n (0.6)^(4-n) thou
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top