Another probability problem

titusland · 2010-09-21 09:18:45

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)

bobbym · 2010-09-21 09:57:40

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.

titusland · 2010-09-21 10:56:48

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

bobbym · 2010-09-21 11:14:50

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.

titusland · 2010-09-21 11:18:31

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

bobbym · 2010-09-21 11:28:25

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

titusland · 2010-09-21 11:30:41

so you had 3 events?

bobbym · 2010-09-21 11:41:20

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)

titusland · 2010-09-22 02:00:12

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.

bobbym · 2010-09-22 02:11:39

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.

mathsyperson · 2010-09-22 06:38:52

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%.

Math Is Fun Forum

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

Another probability problem

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

Re: Another probability problem

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

Re: Another probability problem

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

Re: Another probability problem

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

Re: Another probability problem

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

Re: Another probability problem

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

Re: Another probability problem

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

Re: Another probability problem

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

Re: Another probability problem

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

Re: Another probability problem

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

Re: Another probability problem

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