HSC 2015 MX1 Marathon (archive) (3 Viewers)

[Kaido]

Re: HSC 2015 3U Marathon

(a) MJ's chance of being selected is 5/9. HS's chance of being selected IF MJ is in is 4/8. Multiply to give 5/18.

(b) Not clear whether you mean exactly one or at least one. Assuming the first:
(MJ in / HS out) OR (MJ out / HS in): 5/9 times 4/8 + 4/9 times 4/8 = 1/2

(c) I'm stumped ...

shouldn't this be 5/8?

wow such a simple method, i was doing combs and stuff:
i) 7C3/9C5 etc...

 

[braintic]

Re: HSC 2015 3U Marathon

shouldn't this be 5/8?

Isn't half of 5/9 equal to 5/18 ?

 

[Drsoccerball]

Re: HSC 2015 3U Marathon

There are 9 identical coils that are to be put in 3 boxes. How many arrangements are there such that no box is left empty?

 

[braintic]

Re: HSC 2015 3U Marathon

There are 9 identical coils that are to be put in 3 boxes. How many arrangements are there such that no box is left empty?

This has been asked so many times (including by yourself only 6 days ago), how about we spice it up a bit.

In how many ways can 16 identical coins be placed into four boxes labelled 1, 2, 3 and 4, if the kth box is to contain no more than 2k coins? (A box CAN be empty)

[Without brutishly writing out all the possibilities]

 

[Drsoccerball]

Re: HSC 2015 3U Marathon

This has been asked so many times (including by yourself only 6 days ago), how about we spice it up a bit.

In how many ways can 16 identical coins be placed into four boxes labelled 1, 2, 3 and 4, if the kth box is to contain no more than 2k coins? (A box CAN be empty)

[Without brutishly writing out all the possibilities]

its not simply 19!/16! right?

 

[FrankXie]

Re: HSC 2015 3U Marathon

This has been asked so many times (including by yourself only 6 days ago), how about we spice it up a bit.

In how many ways can 16 identical coins be placed into four boxes labelled 1, 2, 3 and 4, if the kth box is to contain no more than 2k coins? (A box CAN be empty)

[Without brutishly writing out all the possibilities]

I got 6!/4!+5!/3!+4!/2!=62

 

[braintic]

Re: HSC 2015 3U Marathon

I got 6!/4!+5!/3!+4!/2!=62

The answer is 31 (which I verified by writing out all the outcomes). As you're out by a factor of 2, your answer is likely on the right track. Not sure about your logic - can you explain it. My method was much more cumbersome.

 

[FrankXie]

Re: HSC 2015 3U Marathon

The answer is 31 (which I verified by writing out all the outcomes). As you're out by a factor of 2, your answer is likely on the right track. Not sure about your logic - can you explain it. My method was much more cumbersome.

lol I simply used the wrong formula for the method of separator ( i used (m+n)!/m! instead of (m+n)!/m!/n! )

what i did is, suppose box 1 contain 2 coins, box 2 contains 4 coins, box 3 contains 6 coins and cup 4 contains 8 coins. now i need to take out four coins from thses four cups. because cups 2, 3, 4 have no restriction ( we can take out from 0 to 4 from any of them) and cup 1 does have restrictions, i use cases: case (i) from cup 1 take out 0 coin, we have 6!/4!/2!; case (ii) from cup 1 take out 1 coin, we have 5!/3!/2!; case (iii) from cup 1 take out 2 coins, we have 4!/2!/2!. finally add up.

 

[Ekman]

Re: HSC 2015 3U Marathon

Tbh I just considered all cases when box 4 had either 4, 5, 6, 7 or 8 coins in it. Then I just found the arrangements of coins in the boxes and got 31. Is there a shorter way?

 

[Drsoccerball]

Re: HSC 2015 3U Marathon

A fair dice is thrown 5 times. Find the probability that:
I) the 5 scores are all different numbers
Ii) the 5 scores are consecutive numbers
Iii) the 5 scored include exaclty 3 different nunbers,one of which occurs 3 times.
Iv) the product of the 5 scoree is an even number
Show the thought process behind it

 

[Kaido]

Re: HSC 2015 3U Marathon

Tbh I just considered all cases when box 4 had either 4, 5, 6, 7 or 8 coins in it. Then I just found the arrangements of coins in the boxes and got 31. Is there a shorter way?

i think for this case, it was faster to use the rigorous method :O

 

[Drsoccerball]

Re: HSC 2015 3U Marathon

A fair dice is thrown 5 times. Find the probability that:
I) the 5 scores are all different numbers
Ii) the 5 scores are consecutive numbers
Iii) the 5 scored include exaclty 3 different nunbers,one of which occurs 3 times.
Iv) the product of the 5 scoree is an even number
Show the thought process behind it

Bump

 

[braintic]

Re: HSC 2015 3U Marathon

lol I simply used the wrong formula for the method of separator ( i used (m+n)!/m! instead of (m+n)!/m!/n! )

what i did is, suppose box 1 contain 2 coins, box 2 contains 4 coins, box 3 contains 6 coins and cup 4 contains 8 coins. now i need to take out four coins from thses four cups. because cups 2, 3, 4 have no restriction ( we can take out from 0 to 4 from any of them) and cup 1 does have restrictions, i use cases: case (i) from cup 1 take out 0 coin, we have 6!/4!/2!; case (ii) from cup 1 take out 1 coin, we have 5!/3!/2!; case (iii) from cup 1 take out 2 coins, we have 4!/2!/2!. finally add up.

I've been trying to generalise this to n² matches and n boxes, with the same restriction, but I can't get a finite form. Any suggestions?

 

[braintic]

Re: HSC 2015 3U Marathon

Bump

(I) 5/6 × 4/6 × 3/6 × 2/6 = 5/54

(II) They have to be different, where the missing number is 1 or 6, ie. 1/3 of previous answer = 5/162

(III) P(1,1,1,2,3) = (1/6)⁵ [for getting precisely those numbers in that order] × 5C2 ways of picking the position of 2 and 3.
Then multiply by 6 ways of picking the triple and another 5C2 ways of picking the singles: 0.077

(IV) Need 5 odds: (1/2)⁵ = 1/32

 

[Drsoccerball]

Re: HSC 2015 3U Marathon

(I) 5/6 × 4/6 × 3/6 × 2/6 = 5/54

(II) They have to be different, where the missing number is 1 or 6, ie. 1/3 of previous answer = 5/162

(III) P(1,1,1,2,3) = (1/6)⁵ [for getting precisely those numbers in that order] × 5C2 ways of picking the position of 2 and 3.
Then multiply by 6 ways of picking the triple and another 5C2 ways of picking the singles: 0.077

(IV) Need 5 odds: (1/2)⁵ = 1/32

according to the answers iii and iv are wrong. can you explain your method for i and ii

 

[Kaido]

Re: HSC 2015 3U Marathon

for i) 1st number doesnt matter, so 1/1 chance lol, 2nd number has to be different, so 5/6 chance until 5th number which is 2/6

iii) (i think) 1st throw can be any number, 2nd throw is 5/6, 3rd is 4/6 then another 1/6 followed by 1/6 (which is the 3 recurring numbers) so 5/6x4/6x1/6x1/6

iv) even(e) occur when 1e,2e,3e,4e or 5e
for 1e 4odds(o) : 5C4 1/2^4 x 1/2
etc...
then add all of them

 

[Drsoccerball]

Re: HSC 2015 3U Marathon

for i) 1st number doesnt matter, so 1/1 chance lol, 2nd number has to be different, so 5/6 chance until 5th number which is 2/6
for ii) only combinations are 12345 or 23456 and you can follow braintic's way or just do some multiplication

iii) (i think) 1st throw can be any number, 2nd throw is 5/6, 3rd is 4/6 then another 1/6 followed by 1/6 (which is the 3 recurring numbers) so 5/6x4/6x1/6x1/6

i suck ass at permutations...

 

[Drsoccerball]

Re: HSC 2015 3U Marathon

for i) 1st number doesnt matter, so 1/1 chance lol, 2nd number has to be different, so 5/6 chance until 5th number which is 2/6
for ii) only combinations are 12345 or 23456 and you can follow braintic's way or just do some multiplication

iii) (i think) 1st throw can be any number, 2nd throw is 5/6, 3rd is 4/6 then another 1/6 followed by 1/6 (which is the 3 recurring numbers) so 5/6x4/6x1/6x1/6

iv) even(e) occur when 1e,2e,3e,4e or 5e
for 1e 4odds(o) : 5C4 1/2^4 x 1/2
etc...
then add all of them

for iii the answers

 

[Kaido]

Re: HSC 2015 3U Marathon

i suck ass at permutations...

i feel you, lost 4 marks in last exam, gg

 

[Drsoccerball]

Re: HSC 2015 3U Marathon

i feel you, lost 4 marks in last exam, gg

My teacher probably puts on a troll face when putting the permutation questions in