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# unsolved national maths contest-1995 question. watch

1. In order to get the sum of 6,it can also be 1+1+1+1+1+1(the probability is 1/5+1/5+1/5+....,2+2+2(2/5+2/5+.....),3+3(2/5+2/5+....),1+1+2+2 and many other ways ,not just the above two ways, right?

After that we add all the probability of possible ways together.......is it correct?

I do not understand how can we find a separated P(1), P(2) and P(3)?
2. (Original post by akiromuchi)
In order to get the sum of 6,it can also be 1+1+1+1+1+1(the probability is 1/5+1/5+1/5+....,2+2+2(2/5+2/5+.....),3+3(2/5+2/5+....),1+1+2+2 and many other ways ,not just the above two ways, right?
Did you read the question? It says if three marbles... that means you can't have 1+1+1+1+1+1, because that would require 6 marbles!

After that we add all the probability of possible ways together.......is it correct?
My thought was that 6(P(1,2,3))+1(P(2,2,2)) would be the answer, I'm not sure if that's what you mean.

I do not understand how can we find a separated P(1), P(2) and P(3)?
Have you drawn the tree diagram? If not, do it! If you have, well start off by noting that the probability is 1/2 to go on either side of a pin.
3. In order to get 6(P(1,2,3))+1(P(2,2,2)) ,the probability is 6(1/5 x 2/5 x 2/5) + (2/5 x 2/5 x 2/5) = 12/ 125

which is not the answer ?????
4. (Original post by akiromuchi)
In order to get 6(P(1,2,3))+1(P(2,2,2)) ,the probability is 6(1/5 x 2/5 x 2/5) + (2/5 x 2/5 x 2/5) = 12/ 125
No.

It is 6P(1)P(2)P(3)+1P(2)P(2)P(2).

You need to find P(1), p(2) and P(3)...
5. hmm i think he does not know how to use the tree diagram to find p(1),etc?

anyway , there is a chance of 1 to hit the first pin at the 1st level.
At the 2nd level (where they are two pins), each has a 0.5 chance of being hit. At the 3rd level (3 pins) what are the probabilities? the sum of the propbabilities of hitting each pin at each level must sum to 1.
6. (Original post by nota bene)
No.

It is 6P(1)P(2)P(3)+1P(2)P(2)P(2).

You need to find P(1), p(2) and P(3)...
Aren't P(1),P(2),P(3) are given ? P(1) is 1/5 ,P(2) is 2/5 ,P(3) is 2/5?I do not understand what is mean by 1st level,2nd level,3rd level? Seeing that the level related to number of pins,isn't that the number of pin is constant which is 4 ?
7. (Original post by akiromuchi)
Aren't P(1),P(2),P(3) are given ? P(1) is 1/5 ,P(2) is 2/5 ,P(3) is 2/5?
Certainly not! You have to take the pins into account...
I do not understand what is mean by 1st level,2nd level,3rd level? Seeing that the level related to number of pins,isn't that the number of pin is constant which is 4 ?
Look at the levels horizontally, not as slanted lines of the typ y=x or somesuch. First you have one pin, then two etc. equal probability(=1/2) to go right and left of the pin. Look at how many paths give you 1, 2 and 3. This approach will allow you to calculate P(1), P(2) and P(3).
9. (Original post by nota bene)
Do you have the answer to it? (If my answer is right I'll explain, otherwise I won't, because I'm likely wrong since I'm really bad at geometry)
The answer is A ......Is it just like you expect?
10. (Original post by akiromuchi)
The answer is A ......Is it just like you expect?
I don't think so. Show your working.
11. (Original post by DFranklin)
I don't think so. Show your working.
I think the answer is E......is it correct?
(root((root 200)^2 +10^2)))/6
Anyway,i have another two questions in the attached doc......
For the second question,is my working correct?
50 + x = 60 + 0.5 x
Attached Files
12. Maths.doc (70.0 KB, 92 views)
13. for qns 1, think of it this way, when u open up the magazine to its middle, there are half of the number of pages on each side.
further whenu open up the magazine in any way, the page on the left is even numbered. for example if there are 32 pages, 16 on the left, therefore the required answer will be 16 and 17.

for qns 2, you need to start by deriving 2 a;gebraic equation to model how the cost for hire varies with time... post yourworkings then=) if u still need help.
14. (Original post by OCC++)
for qns 1, think of it this way, when u open up the magazine to its middle, there are half of the number of pages on each side.
further whenu open up the magazine in any way, the page on the left is even numbered. for example if there are 32 pages, 16 on the left, therefore the required answer will be 16 and 17.

for qns 2, you need to start by deriving 2 a;gebraic equation to model how the cost for hire varies with time... post yourworkings then=) if u still need help.
For 1st question, i think like you as well.....But the answer given is E??
For second Q , I do not know what value of distance can be taken as the connection between the two situation since only inequality is given for the first situation that is more than 80.......
15. glad that u understand qns 1. thinkof my example and think through it again. you should get it.
qns 2, always use variables when in doubt. now try yto get the equation . let any unknown constanjt be represented byvariables
16. But,we do not know whether the car travels less than or more than 80 miles in the first place?OCC++,can you please work out the equation?
17. haizz indeed it is true that we do not know..but the answers are all bigger than 80. At this point knowing that the distance has to be greater than 80, i m lazy to think further.

edit: ok after thinking notice that the fixed cost incurred for the 2nd hire is 60, already greater than the fixed cost of 50 for first hire. in order to even catch up to 60 from 50, the distance travelled must be greater than 80.
18. After solving the equation 50 + x = 60 +0.5x , x= 20
Then,20 is added to the initial 80 results in 100.....
But the given answer is 180?Why?
19. your equation is wrong. for the 2ndhire, u must take into account the cost of 1st 80miles.
20. (Original post by OCC++)
your equation is wrong. for the 2ndhire, u must take into account the cost of 1st 80miles.
Do you mean like this 50 + x = 60 + 0.5 (80)x
obviously x= distance after 80 miles.
therefore for the 2nd hire, cost of travelling for first 80miles is itself a fixed cost, like the 60 \$

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