Probablity questions (1 Viewer)

[Sammy_b]

These are from Maths in Focus (1999 edition)

12.9
4. b) A 4 digit number is to be made from the digits 1 to 9, with no digit allowed more than once in the same number. In how many ways can an odd number be made?

12.10
14. Thirty people apply for a special housing package, but only 12 packages are available, and are located randomly.
a) What is the probability that a particular person applying will get a house?
(answer in back is 1/86 493 225)
b) Out of the 12 packages, 5 are in Sydney and 7 are in the Blue Mountains. If 19 people apply for the Sydney houses and the rest apply for the Blue Mountains, in how many ways can the houses be allocated?
(answer is 3 837 240)

Can anyone help, please?
Thanks

 

[cj_bridle]

4b)

By choosing combinations for the 4 digits where by u specify how many of the first 3 are odd or not.. thus leaving the last digit just choosing from the remaining odd numbers.. i.e. out of 5.

this is what i got.. ?

choose 0 odd:
9x8x7x5 = 2520
choose 1 odd:
9x8x7x4 = 2016
choose 2 odd:
9x8x7x3 = 1512
choose 3 odd:
9x8x7x2 = 1008

sum these together = 7056 possibilities

ill work on the next 2 now..

[edit] nop im stuck on the other 2..

 

[withoutaface]

14 a) 12/30=6/15 I dunno what the answer is going on about, either the question is ambiguous or it is wrong.

b)19P5 + 11P7

 

[coca cola]

4. b) i think cj_bridleis solution is wrong

it should be 8x7x6x5 = 1680, i think.

think of it in this way. an odd number occurs when the last digit is odd, i.e. 1, 3 ,5 ,7 ,9.

therefore there is 5 possibilities or choices of making up the last digit, i am assuming you know the rest as it is just standard.

 

[coca cola]

14 a) 12/30=6/15 I dunno what the answer is going on about, either the question is ambiguous or it is wrong.

b)19P5 + 11P7

yeah i think its very ambiguous indeed.

soln is 1/30C12 = 1/86 493 225

but this is essentially the probability that the any particular group of 12 people will get a house out of the 30 people.

 

[Sirius Black]

for the 1st one, we can use "box-filling" method so if we do not consider the condition applied (ie. odd number) there are 9x8x7x6 ways to do it. and the prob of an odd # occur is 5/9. so it should be 8x7x6x5 ways to do it(=1680? a bit weird)

for the 2nd question 1st part, prob=1/30C12

 

[Sirius Black]

Q14 (b)=19c5x11c7=...
hint: in Permutation and Combination questions, ppl choosing houses and houses choosing ppl are the same thing. don't think human are supior than houses/seats or whatever