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# Urgent help S2 watch

1. the following question please. I just simply cant get the jist of it . any help appreciated. thank you

A child has a pair of scissors and a piece of string 8cm long, which has a mark on it at one end. The child cuts the string, randomly at a chosen point, into two pieces. Let X represent the length of the piece of string with the mark on it.

Evaluate the probability that the shorter piece of string is at least 3cm long.
2. well this is a continuous uniform distribution..
so the probability of getting any length is 1/8

P(X≥3) = 1 - P(X<3)

= 1 - 3/8 = 5/8

(i hope!)
i think about it in terms of a rectangular graph for the pdf..

good luck!
3. yes thats what i got too.

its a continuous uniform distribution (rectangular)
you can double check the above method using integration:

P(X>>3) = ∫1/8 dx (limits are 8 and 3)
= 1/8 (8-3) = 5/8
4. (Original post by hajira)
well this is a continuous uniform distribution..
so the probability of getting any length is 1/8

P(X≥3) = 1 - P(X<3)

= 1 - 3/8 = 5/8

(i hope!)
i think about it in terms of a rectangular graph for the pdf..

good luck!
thanks but the answer is 1//4 or .25 that is a quarter. there is a chance of either cutting it where X is more than three or just three. we have to combine the two probabilities. but i can't work them out.
5. (Original post by KingAS)
yes thats what i got too.

its a continuous uniform distribution (rectangular)
you can double check the above method using integration:

P(X>>3) = ∫1/8 dx (limits are 8 and 3)
= 1/8 (8-3) = 5/8
thanks but the answer is 1//4 or .25 that is a quarter. there is a chance of either cutting it where X is more than three or just three. we have to combine the two probabilities. but i can't work them out.
6. (Original post by ruzaika)
thanks but the answer is 1//4 or .25 that is a quarter. there is a chance of either cutting it where X is more than three or just three. we have to combine the two probabilities. but i can't work them out.
hmm i shudv thought that 'shorter' was implying something
and there a mark at the end
7. in that case wat it seems to me is that either P(X≥3) or P(X≤5)
a wild guess is to multiply the two?
tho i hav no clue y in the world that might work..
but that goes out as 5/8*3/8=15/64=0.2344

u'v got me worried now
8. (Original post by hajira)
in that case wat it seems to me is that either P(X≥3) or P(X≤5)
a wild guess is to multiply the two?
tho i hav no clue y in the world that might work..
but that goes out as 5/8*3/8=15/64=0.2344

u'v got me worried now
i will get back to you on it. i would call my master now. remember you can't multiply probabilities that are mutually exclusive, you add them. its the independent ones that you multiply.
9. (Original post by ruzaika)
i will get back to you on it. i would call my master now. remember you can't multiply probabilities that are mutually exclusive, you add them. its the independent ones that you multiply.
yup
maybe the ans is wrong in the first place

waiting to hear wat ur masters got to say abt this!
10. (Original post by hajira)
yup
maybe the ans is wrong in the first place

waiting to hear wat ur masters got to say abt this!

he is having a lesson in school...but the answer is right...since it is the same answer in two places..but there is an example in the text book modular s2 heinemann for as and a2. its the 2002 specification one. if it helps and clears the fog could u please post the answer thanks.
11. just outta curiosity, where did this question come from?
12. (Original post by hajira)
in that case wat it seems to me is that either P(X≥3) or P(X≤5)
a wild guess is to multiply the two?
tho i hav no clue y in the world that might work..
but that goes out as 5/8*3/8=15/64=0.2344

u'v got me worried now
13. (Original post by KingAS)
its an 1997 jun or jan paper i believe.
14. got it finally
if we let X be the length from the marked end of string,

we need 3≤X≤5
think about it as..either the bit with the marked end can be the shorter length, or it can be the longer end..making the other piece of string (lets call it Y) shorter
so if its the shorter end...to satisfy the question, we need it to be at least 3 i.e 3≤X
if its the longer length n Y is shorter..then it needs to be X≤5 in order to make Y at least 3

combining the conditions gives 3≤X≤5
as shown in graph...
probability is simply the shaded area, 2/8 = 1/4

Attached Images

15. (Original post by hajira)
got it finally
if we let X be the length from the marked end of string,

we need 3≤X≤5
think about it as..either the bit with the marked end can be the shorter length, or it can be the longer end..making the other piece of string (lets call it Y) shorter
so if its the shorter end...to satisfy the question, we need it to be at least 3 i.e 3≤X
if its the longer length n Y is shorter..then it needs to be X≤5 in order to make Y at least 3

combining the conditions gives 3≤X≤5
as shown in graph...
probability is simply the shaded area, 2/8 = 1/4

oh thank you thank you thank you.

:withstupi i had gotten the conditions and was stumped as to what to do with it next.
16. yep u gotta use both ends p(3>=x>=5)

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