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The probability of a target shooter hitting the bullseye on any one shot is 0.2. a If the shooter takes five shots at the target, find the probability of:
i) missing the bullseye every time
ii) hitting the bullseye at least once
b) What is the smallest number of shots the shooter should make to ensure a probability of more than 0.95 of hitting the bullseye at least once?

i) $n=5,p=0.2$
$(5)$($(0.2)$=1
$1(1-0.2)=0.8$
$(0.8)^5= 0.327$

I don't is this the right way to do but I got the right answer.
I don't know how to do ii) and B) , appreciate your help!

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But ii) is just the logical negation of i); ii) happens if and only if i) doesn't. So I'm sure you can work out the answer to ii) from the answer to i). –  Gerry Myerson Aug 22 '12 at 1:44
    
Question 1 correct. ii) in itself should be obvious as it is the opposite of i). [partially deleted] –  FrenzY DT. Aug 22 '12 at 1:45
    
@FrenzYDT.: It is the number of shots that is variable in b, not the probability of each shot hitting. –  Ross Millikan Aug 22 '12 at 1:47
    
@RossMillikan Oh. I misread... –  FrenzY DT. Aug 22 '12 at 1:49

2 Answers 2

up vote 4 down vote accepted

Hints: For ii, you have the probability of missing every time from i, so all the rest is the chance of hitting at least once.

For b, the same logic of deriving ii from i applies, but the number of shots is a variable. If you want 0.95 to hit at least once, what is the chance you miss them all? Then find a number of shots that gets the chance below that.

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Here's what I would do:

ii) The probability of hitting it once or more is opposite of the probability of missing it every time, so it's 1 - the probability of your result from i): $0.8^5 = 0.3276$`. So it's $1 - 0.32768 = 0.67232$.

b) I'm sure there's a pretty algorithm for this one, but sometimes I find it's easier just to experiment. Let's try 10: the probability of missing every time is $0.8^{10} = 0.10737$. The complement of this will be slightly under 0.9, so we're a bit off. Let's try 13: $0.8^{13} = 0.05498$. The complement of this will be slightly under 0.95, so we're extremely close. We confidently try 14 shots, which yields $0.8^{14} = 0.04398$. The complement of this is 0.95602, which is the smallest possible value over 0.95. So the answer is 14.

$Q.E.D.$

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You can set 0.8^x = 0.05, take natural log of both sides and solve for x, then round x up. –  Akavall Aug 22 '12 at 2:20
    
To get exponents of more than one character, put them in braces: use 0.8^{10} to get $0.8^{10}$ –  Ross Millikan Aug 22 '12 at 3:01
    
@RossMillikan thanks for editing that for me –  aopsfan Aug 22 '12 at 11:16

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