Probability Question (1 Viewer)

[qwerty44]

Part (ii)

 

[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.

 

[qwerty44]

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]

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:

 

[qwerty44]

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:

yep that worked but how come 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]

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:

yeh this is correct.

 

[bleakarcher]

yep that worked but how come 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]

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.

implies that the draw is being won everytime.

 

[qwerty44]

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]

Its trial and error but still works.

 

[qwerty44]

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]

Also tried another way:



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

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.