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# A man can hit a target once in 4 shots. If he fires 4 shots

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Manager
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A man can hit a target once in 4 shots. If he fires 4 shots [#permalink]  10 Oct 2006, 10:35
A man can hit a target once in 4 shots. If he fires 4 shots in succession, what is the probability that he will hit his target atleast twice?
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cicerone - can you explain how you came up with that anser. I went about this problem with trying to figure out the probability that he will not hit the target twice. Thanks
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Matrix02 wrote:
cicerone - can you explain how you came up with that anser. I went about this problem with trying to figure out the probability that he will not hit the target twice. Thanks

Prob of hitting atleast twice = 1- (prob of hitting 0 times + prob of hitting 1 time)

Prob of hitting 0 times = 3/4x3/4x3/4x3/4 = 81/256.
Prob of hitting 1 time out of 4 times:
We have 4 cases here i.e. he could hit once in anyone of the 4 hits.
i.e if he hits in the first shot the prob is 1/4x3/4x3/4x3/4 =27/256
if he hits in the second shot the prob is 3/4x1/4x3/4x3/4 =27/256

Continuing this way it would be 4x27/256 = 108/256.

So 1-81/256-108/256 = 67/256.

The other way is
prob of hitting 2 times + prob of hitting 3 times + prob of hitting 4 times.

Again hitting 2 out of 4 we will have 6 cases.
For each case the prob will be 1/4x1/4x3/4x3/4 = 9/256
So prob of hitting 2 out of 4 is 6x9/256 = 54/256

Prob of hitting 3 out of 4 we will have 4 cases
For each case the prob will be 1/4x1/4x1/4x3/4 = 3/256
So prob of hitting 3 out of 4 is 4x3/256=12/256

Prob of hitting 4 out of 4 we will have only 1 case
So prob of hitting 4 out of 4 is 1/4x1/4x1/4x1/4 = 1/256.

SO prob of hitting atleast twice = 54/256+12/256+1/256 = 67/256

So either way it is 67/256.

But i think the first way is faster
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77/256

at least 2 means that we have to account for probability of hitting the target 2,3,and 4 times, or we can calculate the probability that he hits the target 0, and 1 time and subtract it from 1, since

Probability of atleast 2 = 1 - Probability of hitting the target 0 and 1 time

so 1-[ ( C(4,0)*1/4^0*3/4^4 ) + ( C(4,1)*1/4^1*3/4^3 ) ]

= 1 - 189/256 = 77/256
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bui wrote:
77/256

at least 2 means that we have to account for probability of hitting the target 2,3,and 4 times, or we can calculate the probability that he hits the target 0, and 1 time and subtract it from 1, since

Probability of atleast 2 = 1 - Probability of hitting the target 0 and 1 time

so 1-[ ( C(4,0)*1/4^0*3/4^4 ) + ( C(4,1)*1/4^1*3/4^3 ) ]

= 1 - 189/256 = 77/256

Everything will be futile if u commit these silly mistakes in the exam

1-189/256 = 67/256.....................
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Joined: 05 Feb 2006
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Isn't it that the man will surely hit one time out of 4?

So the probability that he will not hit 2 times:

3/4*3/4*3/4*1=27/64

So the probability that he will hit the target at least twice should be:

1-27/64=37/64... or not ?
Manager
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OR

( C(4,2)*1/4^2*3/4^2 ) + ( C(4,3)*1/4^3*3/4^1 )+( C(4,4)*1/4^4*3/4^0 ) = 67/256
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