questions - help appreciated (1 Viewer)

[Timothy.Siu]

Q. A total of 300 tickets are sold in a raffle which has three prizes. There are 100 red, 100 green and 100 blue tickets.

What is the probability that at least one of the winning tickets is not red.

i wud find the probability that they're all red and subtract from 1
well its not that confusing
1-1/3 x 99/299 x 98/298 = 0.963704518
i hope thats correct

 

[tommykins]

回复: Re: questions - help appreciated

P(not red) = 1 - P(all red) ?

 

[lyounamu]

Q. A total of 300 tickets are sold in a raffle which has three prizes. There are 100 red, 100 green and 100 blue tickets.

What is the probability that at least one of the winning tickets is not red.

i wud find the probability that they're all red and subtract from 1
well its not that confusing
1-1/3 x 99/299 x 98/298 = 0.963704518
i hope thats correct

Yeah, but I don't get the wording...

How did you instantly know that? It took me 5 minutes of thinking to get it right.

EDIT: Got it, lol. My brain is truly dead today.

 

[Timothy.Siu]

Yeah, but I don't get the wording...

How did you instantly know that? It took me 5 minutes of thinking to get it right.

well, "What is the probability that at least one of the winning tickets is not red."
they want u to find the probably that out of the 3 winning tickets drawn that at least one of them will b either green or blue.
The easiest way is to find the probably of all red and subject it from 1 because the only case where at least one of the tickets is NOT red is when they're all red.

Alternatively u cud do the slower way
P(at least one not red) = 2/3 + 1/3x200/299+1/3x99/299x200/298

edit: at ur edit i say lol

 

[lyounamu]

well, "What is the probability that at least one of the winning tickets is not red."
they want u to find the probably that out of the 3 winning tickets drawn that at least one of them will b either green or blue.
The easiest way is to find the probably of all red and subject it from 1 because the only case where at least one of the tickets is NOT red is when they're all red.

Alternatively u cud do the slower way
P(at least one not red) = 2/3 + 1/3x200/299+1/3x99/299x200/298

And I got it now. I took it from only the red perspective which was wrong. Thanks

 

[henry08]

That first quetion is hard.

 

[Timothy.Siu]

That first quetion is hard.

its pretty hard if u've never done one like that b4, but after that its easy

 

[lyounamu]

the above post is correct

let P = $10000 R= 1.03 Q=450 S=1.02

1st withdrawal- PR²-Q
2nd withdrawal- PR³-QR-QS
3rd withdrawal - PR⁴-QR²-QRS-QS²

nth withdrawal - PR⁽ⁿ⁺¹⁾ -QR^(n-1) -QR^(n-2) S-QR^(n-3) S²-......-QR(n-n)S^(n-1)

= PR⁽ⁿ⁺¹⁾ - Q(R^(n-1) -R^(n-2) S-.....-RS^(n-2) -S^(n-1) )
using the formula sum Nth = PR⁽ⁿ⁺¹⁾ - QR^(n-1) [(S/R)^{n -1})]/((S/R)-1)
therefore sum of 25th withdrawal = $456.41
around there

I just got an access to the answer and the answer is apparently $545 to nearest dollar...

EDIT: I keep getting $561 to nearest dollar. Any help?

 

[Timothy.Siu]

I just got an access to the answer and the answer is apparently $545 to nearest dollar...

EDIT: I keep getting $561 to nearest dollar. Any help?

well i'm off as well then, i'll do it again

"let P = $10000 R= 1.03 Q=450 S=1.02

1st withdrawal- PR²-Q
2nd withdrawal- PR³-QR-QS
3rd withdrawal - PR⁴-QR²-QRS-QS²

nth withdrawal - PR⁽ⁿ⁺¹⁾ -QR^(n-1) -QR^(n-2) S-QR^(n-3) S²-......-QR(n-n)S^(n-1)

= PR⁽ⁿ⁺¹⁾ - Q(R^(n-1) -R^(n-2) S-.....-RS^(n-2) -S^(n-1) )
using the formula sum Nth = PR⁽ⁿ⁺¹⁾ - QR^(n-1) [(S/R)ⁿ-1)]/((S/R)-1)
therefore sum of 25th withdrawal = $456.41"

i dont even know how i got 456.41 because when i used my formula again i got like $1k+
but i figured how to get that answer, instead of the beginning of the formula being PR⁽ⁿ⁺¹⁾ , if its just PRⁿ then u get $545 to the nearest dollar

edit: i see how u got $561 but i still dont understand how to get $545 yet. u get 561 if u assume the money is withdrawn before the interest is calculated that year.

 

[Ali92l]

Yeah, it should be undefined or 1 (as x-> 0, sinx/x = 1)

I sought out help from a buddy doing ext 2

Basically 1/1 = 1 , x / x = 1, anything/anything = 1 , nothing/nothing = 1 ;

sinx / x for x = 0 is 0/0 so its 1

 

[Timothy.Siu]

I sought out help from a buddy doing ext 2

Basically 1/1 = 1 , x / x = 1, anything/anything = 1 , nothing/nothing = 1 ;

sinx / x for x = 0 is 0/0 so its 1

thats one interpretation

 

[Ali92l]

If you use 0; You get an incorrect answer ...

only interpretation =)

 

[lyounamu]

well i'm off as well then, i'll do it again



i dont even know how i got 456.41 because when i used my formula again i got like $1k+
but i figured how to get that answer, instead of the beginning of the formula being PR⁽ⁿ⁺¹⁾ , if its just PRⁿ then u get $545 to the nearest dollar

edit: i see how u got $561 but i still dont understand how to get $545 yet. u get 561 if u assume the money is withdrawn before the interest is calculated that year.

I don't think

we do 10000 x 1.03^(n+1) ....rest is you know, follows the first one.

Shouldn't it just be 10000 x 1.03^(n)....because whenever you get interest you pay once to the award night?

But still, my answer is wrong...so I cannot really prove the credibility of my logic..

 

[lyounamu]

I sought out help from a buddy doing ext 2

Basically 1/1 = 1 , x / x = 1, anything/anything = 1 , nothing/nothing = 1 ;

sinx / x for x = 0 is 0/0 so its 1

I knew that. I knew that as x -> 0, sinx/x becomes 1. But here, they didn't say that x -> 0. They only said that x = 0. In the solution, they just told us to "assume" that x -> 0.

And 0/0 is nothing, it undefined. What your mate would have said is that, as x -> 0, it becomes like 0.whatever/0.whatever = 1

 

[Timothy.Siu]

I don't think

we do 10000 x 1.03^(n+1) ....rest is you know, follows the first one.

Shouldn't it just be 10000 x 1.03^(n)....because whenever you get interest you pay once to the award night?

But still, my answer is wrong...so I cannot really prove the credibility of my logic..

but isn't the fund in for one year b4 u take the money out for award night

 

[lyounamu]

but isn't the fund in for one year b4 u take the money out for award night

Well, yes but the frequency must be same.

If you see, for every award night, you times 1.03

So for 1st award night, it's 10000 x 1.03 - P (there is one)
Then

2nd award night, it's 10000 x 1.03^2 - P (1+1.03) (there are two here)

3rd award...rest should be conformed.

I used that method for part 1 and 2 of the questions and I was correct. (The first question is actually 3 -part question)

 

[Timothy.Siu]

i'm not sure then, just hope a question like that isn't in the exam then? lol

 

[Ali92l]

I knew that. I knew that as x -> 0, sinx/x becomes 1. But here, they didn't say that x -> 0. They only said that x = 0. In the solution, they just told us to "assume" that x -> 0.

And 0/0 is nothing, it undefined. What your mate would have said is that, as x -> 0, it becomes like 0.whatever/0.whatever = 1

Make a rough sketch and see what would give you a more valid answer; Answers aren't always right and if you do use 0; your estimate using the trapezoidal rule would be way off.