if first is hit then the probability will be 0.7....he missed 1st chance ,,
probability with which he missed is 0.3 ... same way he will try 2nd
chance.. hit chance is 0.6.. but he missed first already.. so total
probabilty of hit is 0.3*0.6... its the same way for all.




On Mon, Aug 29, 2011 at 6:06 PM, Naman Mahor <naman.ma...@gmail.com> wrote:

> abhishek's answer.
> plz  clear my confusion
> that i hv mentioned above.
>
> On Mon, Aug 29, 2011 at 6:04 PM, vishwa <vishwavam...@gmail.com> wrote:
>
>> @naman: whose answer man
>>
>>
>> On Mon, Aug 29, 2011 at 5:56 PM, Naman Mahor <naman.ma...@gmail.com>wrote:
>>
>>> ur answer is correct but i hv a confusion that all four shots fire at a
>>> time so there may be probability that all shots hits the craft. but ur
>>> assuming that first hit + first not hit * second hit+......
>>> plz clear my confustion
>>>
>>> On Mon, Aug 29, 2011 at 5:49 PM, Abhishek Yadav <
>>> algowithabhis...@gmail.com> wrote:
>>>
>>>> i guess it would be ....   0.7 + 0.3*0.6 + 0.3*0.4*0.5 + 0.3*0.4*0.5*0.4
>>>> =.964.....correct me if i am wrong.??
>>>>
>>>> On Mon, Aug 29, 2011 at 5:34 PM, Naman Mahor <naman.ma...@gmail.com>wrote:
>>>>
>>>>> An anti aircraft gun can fire four shots at a time. If the
>>>>> probabilities of the first, second, third and the last shot hitting the
>>>>> enemy aircraft are 0.7, 0.6, 0.5 and 0.4, what is the probability that 
>>>>> four
>>>>> shots aimed at an enemy aircraft will bring the aircraft down?
>>>>>
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