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#1 2010-09-21 09:18:45

titusland
Member
Registered: 2010-09-20
Posts: 9

Another probability problem

Hi All
Thanks in advance for answering my question.

I'm throwing 3 balls at a target.
I win if I can hit 2 in a row.

I must throw with my right hand first, then my left and finally my right again. (Note: Successive throws are independent events)
I estimate that my right hand throws hit the target 70% of the times, while my left hand hit for 40%.

If I hit the target with my first throw, what is the probability that I will win? (Basically hitting on the next throw)

Last edited by titusland (2010-09-21 09:20:02)

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#2 2010-09-21 09:57:40

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Another probability problem

Hi titusland;

I must throw with my right hand first, then my left and finally my right again. (Note: Successive throws are independent events)
I estimate that my right hand throws hit the target 70% of the times, while my left hand hit for 40%.

I am getting 36.4% just by treeing it.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#3 2010-09-21 10:56:48

titusland
Member
Registered: 2010-09-20
Posts: 9

Re: Another probability problem

yes, you are indeed correct. Do you know of another way of solving it?

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#4 2010-09-21 11:14:50

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Another probability problem

Hi;

This is how I enumerated it without using the tree.

(.7)(.4)(.7) + (.7)(.4)(.3) + (.3)(.4)(.7) = .364

In actuality it is the same thing.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#5 2010-09-21 11:18:31

titusland
Member
Registered: 2010-09-20
Posts: 9

Re: Another probability problem

I can understand (.7)(.4)(.4), but how did you come up with the other two?
what is your thought process?
Thanks.

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#6 2010-09-21 11:28:25

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Another probability problem

Hi;

Sorry for my poor typing. It should read as below. I have edited the top post. Funny because it is the sane principle it gets the same answer. Is this understandable to you?

(.7)(.4)(.7) + (.7)(.4)(.3) + (.3)(.4)(.7) = .364


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#7 2010-09-21 11:30:41

titusland
Member
Registered: 2010-09-20
Posts: 9

Re: Another probability problem

so you had 3 events?

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#8 2010-09-21 11:41:20

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Another probability problem

Yes, RL_ meaning R hit L  and anything after that.

(.7)(.4)(.7) + (.7)(.4)(.3)

And:

MLR meaning a miss on the first and RL

(.3)(.4)(.7)


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#9 2010-09-22 02:00:12

titusland
Member
Registered: 2010-09-20
Posts: 9

Re: Another probability problem

MLR meaning a miss on the first and RL

(.3)(.4)(.7)

but I'm assuming I hit on the first throw.
Sorry for all these replies. I'm not really good with probability.

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#10 2010-09-22 02:11:39

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Another probability problem

Hi;

I'm throwing 3 balls at a target.
I win if I can hit 2 in a row.

I must throw with my right hand first, then my left and finally my right again. (Note: Successive throws are independent events)
I estimate that my right hand throws hit the target 70% of the times, while my left hand hit for 40%.

That analysis is for the above question. Or at least I thought it was a question. I thought you were posing this question and then the one where you hit the first throw.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#11 2010-09-22 06:38:52

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Another probability problem

If you've already hit with the first throw, then the question is simple.

Hitting with the second throw means you win. Missing with it means you lose. The third throw doesn't matter either way.

Therefore, the probability is a nice round 40%.


Why did the vector cross the road?
It wanted to be normal.

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