QUESTION based on COMBINATIONS!!! (1 Viewer)

[slix_88]

I am having difficulties with 1 particular, yet simple question which i cannot solve. I have done half of it, but will still post the whole thing.

Q: "The digits 1, 2, 3 and 4 are used to form numbers that may have 1, 2, 3 or 4 digits in them. If one of the numbers is selected at random, find the probability that:
a) It has three digits,
b) it is even,
c) it is greater than 200,
d) it is odd and greater than 3000,
e) it is divisible by 3."

BTW, this question is from the yr12 Cambridge Mathematics book, pg 437.

Hope you have fun!

 

[withoutaface]

a)4P3/(4P1+4P2+4P3+4P4)
b) 1/2
c) (3*(4P3)+4P4)/(4P1+4P2+4P3+4P4)
d) 1/2*(1+4P4/(2(4P1+4P2+4P3+4P4)))
e) Can't think of another way but to list them...

 

[Slidey]

Q: "The digits 1, 2, 3 and 4 are used to form numbers that may have 1, 2, 3 or 4 digits in them. If one of the numbers is selected at random, find the probability that:
e) it is divisible by 3."

How many of the digits, up to 4 add up to a number divisible by 3?
3 = 1P1 = 1 number
1 and 2 = 2P2 = 2 nums
2 and 4 = 2P2 = 2 nums
1, 2 and 3 = 3P3 = 6 nums
2, 3 and 4 = 3P3 = 6 nums
No 4-digit numbers are divisible by 3 (since 1+2+3+4 is not divisible by 3). So that's a total of:
1+2*2!+2*3!=17 numbers. Hope that's right. I'm assuming no numbers repeat.

As for probability... well, you'll have to do that bit on your own... though at a guess: 17/4P4=17/24?

 

[slix_88]

These are the answers:
a)3/8
b)1/2
c)21/32?????????
d)3/32??????????
e)17/64 (close) - lolz

 

[slix_88]

Few more on circle!

I have 1 thing that really annoys me! How do you solve alternating things in a circle. Ok, i don't think this sounded right, but i'll give a demo Q.

Q: The letters A, E, I, P, Q and R are arranged in a Circle. Find the probability that:
c) the vowels and consonants alternate.

As well 1 more........

part d) at least 2 vowels are next to one another.

ANSWERS:
C)1/10
D)9/10

THNX a lot for your time and effort!

 

[Pace_T]

ok in the 1st one, the vowels and the consonents can be arranged in 2! * 3!
the reason why it's not a further *2! is because it doesn't matter if it is AB or BA since it is a circle. (similarly how the 1st position doesn't matter when poeple sit around the table, hence (n-1)!)
the reason why its not 3! * 3! is because it is a circle once again
then divide by all ways (6-1)! = 5!
= 12/120
= 1/10

 

[Pace_T]

atleast 2 vowels next to each other = 1 - (no vowels next to each other)
= 1 - 1/10 (as previous question)
= 9/10

*lets hope im right lol*