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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 293

Hi, I need someone to help me make sense of something :

We have three person shooting one arrow.

P(Thomas hits the target) = 0,4 >> 0,6 (He doesn't hit it)

P(Alicia hits the target) = 0,3 >> 0,7

P(Julie hits the target)= 0,8 >> 0,2

Now, the question is : What is the probability of atleast one person out of the three hits the target if they shoot one arrow each.

A' = No one hitting the target

0,6 * 0,7 * 0,2 = 0,084 or 8,4 %

Now, we know that :

A = Atleast one person hitting the target

1- 0,084 = 0,916 or 91,6%

There is one LITTLE thing which I do not understand ! It concerns the A, what does the 91,6 % stand for ? Yes, it's said to be " Atleast one person hitting the target" But I don't understand why. How can you substract the probability of no one getting the target and get the probability of atleast one person hitting it ? I feel it doesn't have any link together ? COuld someone enlighten me to see how it's possible ? Thank you ( If you do not understand what I mean, I will reformulate )

*Last edited by Al-Allo (2013-06-21 06:55:53)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

Hi;

How can you substract the probability of no one getting the target and get the probability of atleast one person hitting it ?

At least one person hitting the target means 1, 2 or all 3.

There are only these possibilities

1) No one hits

2) 1 person hits

3) 2 persons hit

4) 3 persons hit

At least one person hitting the target is 2) + 3) + 4)

From the laws of probability.

1) + 2) + 3) + 4) = 1

You calculated 1) to be .084

.084 +2) + 3) + 4) = 1

subtract .084 from both sides

2) + 3) + 4) = .916

And you know what 2) + 3) + 4) is by now.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 293

AHHHHHHH.. Thank you. THe problem was, I wasn't sure that atleast one person would also mean all the three like you showed me(It didn't make sense that it would mean ONLY one person, this is why I asked the question about it). A vocabulary problem ?

Anyway, thank you very much ! Im happy that I can count on this forum my problems

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

Hi;

You are welcome.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 293

Also, would this be corrrect :

1-0,084 = 0,916

0,916- 0,096 = 0,82 0,096(all the 3 hitting) = 0,4*0,3*0,8

Two person hitting = 0,4 * 0,3= 0,12 or 0,4*0,8= 0,32 or 0,3*0,8=0,24

0,12+0,32+0,24= 0,68

0,82-0,68=0,14

Is this any good ?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

You have .096 for all three that is correct.

You have .084 for none hitting.

What are you now looking for?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 293

I'm trying to get to 0 by substracting every probability. To be able to see in depth ^^

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

You want probability that only one hits and probability that just 2 hit?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 293

Yes, so we can get to 0.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

Two person hitting is .32

Is that what you got?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 293

Well, I got three of them

Two person hitting : 0,4 * 0,3= 0,12 or 0,4*0,8= 0,32 or 0,3*0,8=0,24

0,12+0,32+0,24= 0,68

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

Yes, you are correct?

Now how do you get the last one.Just one hit

Do you should see why that is wrong though?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 293

Here's what I did :

0,82-0,68=0,14

and 0,14-0,14= 0

0,14, the probability of only one person hitting the target

Is there any corrections to be made ? Thanks

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

Is this the answer you were expecting?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 293

Which one ? The 0,14 ?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

Yes. That will make a good choice.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 293

Well, I don't see what else it could be.... So...

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

Okay, I thought you had the answer already.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 293

Ah, no I don't, I only had to find the probability of atleast 1 person hitting the target. But would the 0,14 that only one person hits the target be good, would it be the answer you would have chosen ?

*Last edited by Al-Allo (2013-06-21 08:27:59)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

I got two different answers when I treed it.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 293

... What are those answers ?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

For any 2 hitting the target .392 and for 1 hitting the target .428

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 293

Well, here's how I see it :

If 1 is our omega, the we SHOULD get this :

14% (Only one person) + 68 % (Only two persons, with the different variations of the two persons, because this is omega) + 9,6 % (The three of them hitting) + 8,4 % (none hitting) = 100 %

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 293

bobbym wrote:

For any 2 hitting the target .392 and for 1 hitting the target .428

How did you get .392 ? Here's how I got for any two : 0,4 * 0,3= 0,12 or 0,4*0,8= 0,32 or 0,3*0,8=0,24

Could you show me ?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

That is what you are doing wrong you are only firing two arrows for each one. But the problem fires 3 arrows!

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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