HSC Maths Q Help (1 Viewer)

[carpe_diem]

Hello!

I came across a questions from one of the past papers and needed help working it out.

In the Jackpot Lottery, the probability of the Jackpot prize being won in any draw is approximately 1 in 50.

i) What is the probability that the jackpot prize will be won in each of the three consecutive draws?

ii) How many consecutive draws must be made for it to be 99% certain that a Jackpot prize will have been won?

If all working out could be shown, that would be appreciated.

Thanks!

 

[RealiseNothing]

For i) you just do:



Since there is a 1/50 chance of winning, and you want the chance of winning in all 3.

For ii) this is what I would do:

There is a 49/50 chance that the jackpot won't be won in a draw. So multiply 49/50 by itself over and over again, until the answer is less than 1%. This is because we want there to be more than a 99% chance of there being one winner, so that must mean there is less than a 1% chance of there not being a winner.

So we make the equation:



Where 'n' is the amount of draws. Now we just solve for 'n' and we have the amount of consecutive draws needed for there to be a 99% chance of a winner.







So there needs to be 228 draws.

 

[iBibah]

(i) (1/50)^3

(ii) Its asking for how many draws would it take for there to be at least one winner. So:

1-(49/50)^n=0.99

n=227.948

Therefore at least 228 draws.

 

[carpe_diem]

@RealiseNothing and @iBibah

Thanks so much. I think I misinterpreted the question. I understand it now!

Just a question. Realise said that you want more than a 99% chance. Why? Why not just 99% Or is that irrelevant?

 

[RealiseNothing]

@RealiseNothing and @iBibah

Thanks so much. I think I misinterpreted the question. I understand it now!

Just a question. Realise said that you want more than a 99% chance. Why? Why not just 99% Or is that irrelevant?

It's irrelevant, you want a 99% chance, but that can't happen (unless you have 227.95 draws, which is impossible). So I just said more than 99% to show that you would go up to 228.

 

[iBibah]

@RealiseNothing and @iBibah

Thanks so much. I think I misinterpreted the question. I understand it now!

Just a question. Realise said that you want more than a 99% chance. Why? Why not just 99% Or is that irrelevant?

It has to be equal to 99%, but you solve you get a decimal. But you can't draw 227.9 times so when you round it up, the chance becomes greater than 99%.

EDIT: Beat me again

 

[carpe_diem]

Ok, thank you both!