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"You have a 16% chance of hitting the target on each attempt."
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what are your thoughts?There is a target.
You have a 16% chance to hit it each try.
You have 3 tries to hit it.
Once you hit it or you try 3 times you stop. (so if you hit it on the first try, you stop)
What is the chance of you hitting the target once.
Please share your work with us ...even if you know it is wrong.
If you are stuck at the beginning tell us and we'll start with the definitions.
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what are your thoughts?
Please share your work with us ...even if you know it is wrong.
If you are stuck at the beginning tell us and we'll start with the definitions.
You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:
http://www.freemathhelp.com/forum/announcement.php?f=33
Well I'm not sure if I'm right or wrong. I got 40.73% from P=1-(84/100)^3=1-(0.84)^3=1-0.5927=0.40729
I just wanted to see how somebody else would do it or if there was a better way.
I'm also uncertain about the "Once you hit it or you try 3 times you stop. (so if you hit it on the first try, you stop)" part
No one except the author of the exercise in question can say with any certainty, but my interpretation of that is that if your first throw hits the target, you stop and don't make the other two throws Likewise, if you miss your first throw but hit on the second throw, you stop and do not throw a third time. You would only throw a third time if you miss the first two throws. Working under this interpretation, I think the best approach to the problem is to think about what the odds of each individual throw are.
The problem tells you that the probability of hitting on any one throw is 16%. Accordingly, you know that the probability of missing are 84%. Being that you only get a second throw if you miss the first, what, then, is the probability of hitting the target on the second throw? Similarly, you only get a third throw if you miss the first two throws. So, what is the probability of hitting the target on the third throw? What does that make the total probability of hitting on any throw?
This looks good to me though there is another way shown by ksdhartWell I'm not sure if I'm right or wrong. I got 40.73% from P=1-(84/100)^3=1-(0.84)^3=1-0.5927=0.40729
I just wanted to see how somebody else would do it or if there was a better way.
\(\displaystyle 0.16 + 0.84 * 0.16 + 0.84 * 0.84 * 0.16 = 0.16 + 0.1344 + 0.112896 = 0.407296.\)
That is identical to your result.
\(\displaystyle 1 - 0.84^3 = 1 - 0.592704 = 0.407296.\)