Question

Assume that 4 digits are selected at random from the set { 1, 2, 4, 6, 7, 9 } and are arranged in random order.

What is the probability that the resulting 4-digit number is less than 7000?

Answer 1

lets say Event A= picking up 4 nos at rando from 6
which is done in 6C4 ways

Answer 2

event B= number is greater than 7000 ( yes we do this and then we can find its probability, and subtract the same from 1).

Answer 3

so if we want a number larger than 7000 we should have first digit as 7, or 9
lets look at 4 digit numbers beginnign with 7
fix 7 in first place, then second place you can have other 5 numbers, and in thired place other 4 and in fourth place other 3 nos.
similarly do it for 9

and then add these two events and find prob

Answer 4

So...
4*5*4*3=240
4*5*4*3=240
4*5*4*3=240
4*5*4*3=240
______________
240*4?

Answer 5

nope 5*4*3=60 , for numbers with 7 and 60 for numbers with 9, makes 120

Answer 6

are all numbers beggining >7000

Answer 7

I'm still kinda confused

Answer 8

it is simple. you must use your permutations formula.

Answer 9

I know how to use them, but I'm not sure how to use them here

Answer 10

okgive me a minte to work it out.

Answer 11

you have 720 possiple answers. but you cant use 7 or 9 to start the number. so if you take out all the options with 7 or 9 at the start, then you would do 720-x=y where x is the number of permutation starting with 9 and 7.

Answer 12

So the ones that start with 9 and 7 would be:
2*5*4*3?

Answer 13

wait i have that wrong
it is a total of 360 answers

Answer 14

why?

Answer 15

permutation formula would be 6^P^4 wich makes you have 6x5x4x3.

Answer 16

no.

Answer 17

Ahh! How would I find the ones that start with 7 and 9?!

Answer 18

when is this due?

Answer 19

Got it! 6P4 - 2*5P3=240
240/360=answer