1995 HSC Q5b iii. (2 Viewers)

[tempco]

In a Jackpot Lottery, 1500 numbers are drawn from a barrel containing the 100 000 ticket numbers available.

The probability that the jackpot prize is won in a given game is 0.015.

The jackpot prize is initially $8000, and it increases by $8000 each time the prize is NOT won.

Calculate the probaibility that the jackpot prize will exceed $200 000 when it is finally won.

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The answer given is:

(1 - 0.015)²⁵

but that doesn't take into account whether it is won or not ("...when it is finally won").

I would've expected ²⁶C₁(0.015)(0.985)²⁵...

Bad wording? or am I reading the Q wrong?

 

[Bi0mic]

the way i see it is, the question actually does not say it _will_ be won, however if in fact it is won, that the prive money exceeds 200k .. it may be that nobody wins but if they do, they recieve >200,000

Well my explanation may make sense only to me so ... ya apologies if my point is vague

 

[Danny11]

I would've expected ²⁶C₁(0.015)(0.985)²⁵...

why? there are only one way that can happen... not ²⁶C₁ ways, eg L, L, L, L.... because you have to lose 25 times in a row for that to happen.

 

[Danny11]

the solution would be if you don't understand:

[(1- 0.015)^25 + (1- 0.015)^26 + (1- 0.015)^27 +.... + (1- 0.015)^infinity]*(0.015)

this is a geometric series. since (1- 0.015) <1 it has a limiting sum.

S = (1- 0.015)^(25)/ [1 - (1- 0.015)] = (1- 0.015)^(25)/(0.015)

now S*(0.015) = (1- 0.015)^25

 

[Danny11]

the way i see it is, the question actually does not say it _will_ be won, however if in fact it is won, that the prive money exceeds 200k .. it may be that nobody wins but if they do, they recieve >200,000

Well my explanation may make sense only to me so ... ya apologies if my point is vague

no thats not correct, you guys just don't understand the theory in limiting sums in an infinite series.

 

[tempco]

nyoh... I get it now...

the solution would be if you don't understand:

[(1- 0.015)^25 + (1- 0.015)^26 + (1- 0.015)^27 +.... + (1- 0.015)^infinity]*(0.015)

this is a geometric series. since (1- 0.015) <1 it has a limiting sum.

S = (1- 0.015)^25/ [1 - (1- 0.015)] = (1- 0.015)^25/(0.015)

now S*(0.015) = (1- 0.015)^25

Wows0r. Yeh I get wat you're saying... I'm getting mixed up with the "number of combinations" type of Qs :s

 

[Danny11]

hehe. then you should get some sleep, don't do any math the night before exam.

 

[BillyMak]

the solution would be if you don't understand:

[(1- 0.015)^25 + (1- 0.015)^26 + (1- 0.015)^27 +.... + (1- 0.015)^infinity]*(0.015)

this is a geometric series. since (1- 0.015) <1 it has a limiting sum.

S = (1- 0.015)^(25)/ [1 - (1- 0.015)] = (1- 0.015)^(25)/(0.015)

now S*(0.015) = (1- 0.015)^25

Err.... there is a much simpler way to do it than that.....

The probability of anyone winning is 0.015, so the probability of no one winning is (1 - 0.015). When someone wins, the prize is increased by $8000. The initial prize is $8000, so it must increase by $8000 another 24 times to reach $200,000.
Noone can win in this time, so there must be 25 games with noone winning.

:. probability of jackpot reaching $200,000 = (1 - 0.015)²⁵

 

[Bi0mic]

and do not eat butterscoth crunchy nut cornflakes for afternoon tea ... ... take my word for it

 

[Danny11]

Err.... there is a much simpler way to do it than that.....

The probability of anyone winning is 0.015, so the probability of no one winning is (1 - 0.015). When someone wins, the prize is increased by $8000. The initial prize is $8000, so it must increase by $8000 another 24 times to reach $200,000.
Noone can win in this time, so there must be 25 games with noone winning.

:. probability of jackpot reaching $200,000 = (1 - 0.015)²⁵

there is something really wrong with that logic, math teachers are trained to spot fudge solutions. i'll let someone else to point out the mistakes.

 

[BillyMak]

You'll find that no one will point out any mistakes because there are no mistakes. I'm sure if you were asked that question in the exam that would be the solution that the markers would expect, and if anyone did anything more it would just be a waste of time.

 

[Danny11]

ok the question says exceeding $200 000. so i can lose 100 times in row too. why 25? 25 is simply the minimum number for it to exceed.

don't worry about it though, its nothing personal. i just thought it was very wrong.

 

[jumb]

Because, we asume that its one as soon as it goes above. Otherwiset there would be things to the power of infinity.

 

[BillyMak]

Ok, for the prize to reach $200 000 before anyone wins, no one can win until it reaches this mark. The probability of no one winning in any one game is (1 - 0.015). The prize starts at $8000, and increases by $8000 each time someone loses. Therefore there must be 25 losses in a row for the prize to reach $200 000, since 25*8000 = 200 000.
Therefore the probability of prize reaching $200 000 is (1 - 0.015)²⁵.

Yes, you might not win for 100 turns, but when you do win, the prize will still be above $200 000, which is a requirement of the question. The probability that the prize will eventually be won is 1 (i.e. it will happen eventually), and the probability that it reaches $200 000 before it is won is (1 - 0.015)²⁵, from the above reasoning.

Got it?

 

[Danny11]

im not even gonna argue with you... you people are too smart

for those who wants to get the question right do it my method posted above, and use geometric series with limiting sums.

Because, we asume that its one as soon as it goes above. Otherwiset there would be things to the power of infinity.

i thought infinite series was in 3u, or am i wrong?

 

[tempco]

Never come across infinite series... but isnt a series like k¹+k²+...+k^n-1+kⁿ eventually to the power of infinity? meh... -___-

 

[mojako]

You'll find that no one will point out any mistakes because there are no mistakes. I'm sure if you were asked that question in the exam that would be the solution that the markers would expect, and if anyone did anything more it would just be a waste of time.

ok BillyMak, let me be a neutral third party...
I agree with Danny.

The Q says "Calculate the probaibility that the jackpot prize will exceed $200 000 when it is finally won."
So you need to NOT win for at least 25 times.
And you need to win for the 26th time or in any of the next draw after the 26th.
One way to interpret (1-0.015)^25 is as the probability that you don't win in the first 25 draws, and it doesn't matter whether you win on the next draws or not.
Another way is to say that it's the probability that you'll win only in one of the 26, 27, 28,... th draw (and after you win once you don't care what your next outcome is).

Why are they equal?
With the first interpretation of (1-0.015)^25, you don't care whether you win after the 25th draw. But if the draw keeps going, eventually you'll win anyway. So essentially the probability that you'll win after the 25th draw is 1. So the two interpretations agree.

Having said this... it gets confusing...
there should be only 1500 draws (doesn't get to infinity).
But from the series
S=[(1- 0.015)^25 + (1- 0.015)^26 + (1- 0.015)^27 +.... + (1- 0.015)^infinity]
you can see that when n=(1500-24) the term is really small and it's not significant.
If you do summation from n=1 to n=(1500-24) and ignore the small r^n term you might get the same answer anyway.

 

[Danny11]

actually majoko, it can get to infinity. the question says its 1500 numbers, not draws. this numbers are replaced, or else the Lose/Win probablity will change with every draw.

 

[Danny11]

With the first interpretation of (1-0.015)^25, you don't care whether you win after the 25th draw. But if the draw keeps going, eventually you'll win anyway. So essentially the probability that you'll win after the 25th draw is 1. So the two interpretations agree.

the difference is one intepretation has mathematics to back it up, the other just takes a stab at a solution which is given with some sought logic that agree with the solution.

 

[Managore]

at first that question looked like a question 7 toughy