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Probability-Bowman

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Manager

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Probability-Bowman [#permalink]

10 Mar 2005, 06:53

Here is a simple question

A bowman hits his target in 1/5 of his shots. What is the probability of him missing the target atleast once in 4 shots.

I know how to solve this...

I saw this Binomial Distribution explained by Hong in the following thread
http://www.gmatclub.com/phpbb/viewtopic.php?t=14706

Can anybody explain me how can I solve the above question using the concept explained in the thread...

Hong, Plz help..

Thanks,
Vijo.

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[#permalink]

10 Mar 2005, 08:42

Probability that one target is missed + probability that two targets are missed + ... + Probability that four targets are missed

4c1*4/5*(1/5)^3+4c2*(4/5)^2*(1/5)^2+4c3*(4/5)^3*1/5+4c4*(4/5)^4

or

1- Probability that no targets are missed (recommended)

1 - 4c4*(1/5)^4

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[#permalink]

10 Mar 2005, 08:51

P(Hitting) = 1/5
P(Missing) = 4/5

P(Atleast once in four shots) = 1 - P(None in four shots)

With four shots there are 16 ways out of which there is only one way in which No shots are hit. This is same as finding C(4,0) = 1

P(None in four shots) = (Number of ways this happens)(probability this happens)

= C(4,0) * (1/5)^0 * (4/5)^4
= (4/5)^4

Hence

P(Atleast once in four shots) = 1 - (4/5)^4

Ketan

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10 Mar 2005, 09:53

Great job both of you!
One correction, the question says at least one miss (not at least one hit), so it is equal to 1-no miss, or 1-all hit.

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10 Mar 2005, 10:45

Is 4/5 the chance of miss or the chance of hit, rupstar?

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10 Mar 2005, 13:08

of what i understood 1/5 is making the target and 4/5 is missing the target

hence
him missing the target atleast once is[ 1- him never missing the target]

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10 Mar 2005, 13:45

That's right. :b:

(I saw you corrected ketanm's C(4,0), thought you didn't correct the other part, which you did.

)

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[#permalink]

10 Mar 2005, 14:10

Rupstar wrote:

yea 1-4c4(1/5^4)

Hi Hong

I am just trying to understand the application of the concept in that thread.

Rupstar's expression is equivalent to
1- ( C(4,4) * (1/5)^4 * (4/5)^0) )

Am I right Hong???

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10 Mar 2005, 14:52

Yes, exactly.

[#permalink] 10 Mar 2005, 14:52

Probability-Bowman

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