Probability Question

**qwerty44** · 10 Jun 2012, 12:41 AM

Part (ii)

**RealiseNothing** · 10 Jun 2012, 12:48 AM

Find how many draws must be made for there to be a 1% certainty that a Jackpot prize isn't won.

This is much easier imo.

**qwerty44** · 10 Jun 2012, 12:49 AM

Originally Posted by RealiseNothing

Find how many draws must be made for there to be a 1% certainty that a Jackpot prize isn't won.

This is much easier imo.

How?

**RealiseNothing** · 10 Jun 2012, 12:51 AM

Originally Posted by qwerty44

How?

I think this is how you would do it:

There's a 49/50 chance that a Jackpot won't be won on any draw. You want this to be 1%, so find an integer 'n' such that:

$(\frac{49}{50})^n = \frac{1}{100}$

**qwerty44** · 10 Jun 2012, 1:02 AM

Originally Posted by RealiseNothing

I think this is how you would do it:

There's a 49/50 chance that a Jackpot won't be won on any draw. You want this to be 1%, so find an integer 'n' such that:

$(\frac{49}{50})^n = \frac{1}{100}$

yep that worked but how come $\left ( \frac{1}{50} \right )^{n}=\frac{99}{100}$ doesn't work. Obviously the answer it gets smaller and smaller as it reaches n but shouldn't it be the opposite of your way?

on a side note, for your bracket in latex to cover the whole fraction, ""\left ( *insert text here* \right )""

**bleakarcher** · 10 Jun 2012, 1:03 AM

Originally Posted by RealiseNothing

I think this is how you would do it:

There's a 49/50 chance that a Jackpot won't be won on any draw. You want this to be 1%, so find an integer 'n' such that:

$(\frac{49}{50})^n = \frac{1}{100}$

yeh this is correct.

**bleakarcher** · 10 Jun 2012, 1:05 AM

Originally Posted by qwerty44

yep that worked but how come $\left ( \frac{1}{50} \right )^{n}=\frac{99}{100}$ doesn't work. Obviously the answer it gets smaller and smaller as it reaches n but shouldn't it be the opposite of your way?

on a side note, for your bracket in latex to cover the whole fraction, ""\left ( *insert text here* \right )""

because (1/50)^n is not the probability that at least one draw will be won after n draws. it is actually the probability that the n draws will be won consecutively. so that doesnt work.

**RealiseNothing** · 10 Jun 2012, 1:07 AM

Originally Posted by bleakarcher

because (1/50)^n is not the probability that at least one draw will be won after n draws. it is actually the probability that the n draws will be won consecutively. so that doesnt work.

Pretty much this.

$(\frac{1}{50})^n$ implies that the draw is being won everytime.

**qwerty44** · 10 Jun 2012, 1:09 AM

Originally Posted by bleakarcher

because (1/50)^n is not the probability that at least one draw will be won after n draws. it is actually the probability that the n draws will be won consecutively. so that doesnt work.

Yeh that makes sense. I also got another way. Your is defs easier but....

**qwerty44** · 10 Jun 2012, 1:11 AM

$\frac{1}{50}\left ( \frac{1-\frac{49}{50}^{228}}{1-\frac{49}{50}} \right )$

Its trial and error but still works.

**qwerty44** · 10 Jun 2012, 1:12 AM

It's kind of the same as

"David has invented a game for one person. He throws two ordinary dice repeatedly until the sum of
the two numbers shown is either 7 or 9. If the sum is 9, David wins. If the sum is 7, David loses. If
the sum is any other number, he continues to throw until it is 7 or 9. What is the probability of winning?"

where geo series are used.

**qwerty44** · 10 Jun 2012, 1:26 AM

Also tried another way:

$1-\left ( \frac{49}{50} \right )^{n}=0.99$

which is actually realises way rearranged. I kinda did it accidently when bleakarcher said

Originally Posted by bleakarcher

because (1/50)^n is not the probability that at least one draw will be won after n draws. it is actually the probability that the n draws will be won consecutively. so that doesnt work.

which reminded me to try the compliment of no one winning.