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highhope
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devinawilliam83
is the answer E?

if yes, then this is an inequality cum probability question
assuming n missiles are fired before it hits target

probaility it hits is 0.4 or 2/5
probability it fails is 1-2/5 = 3/5

so if the missile is fired n times before it hits probability is

(3/5)(n-1)*2/5.. now this should be greater han or equal to 80%

therefore, (6n-6)/25>=80/100 or 4/5

solving we get n should be atleast 5

If the answer is 5 than the below expression shud be -
(3/5)*(3/5)*(3/5)*(3/5)*(2/5) >= 4/5 - which is not true.
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IMO B. Will explain the solution if that is correct
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The probablity of hitting the target =0.4
so,the probab of missing is = 0.6 i.e (1-0.4)
if the missile is fired n times the prob of missing is 0.6^n

we need to find the smallest n here.

since the least is 80 percent :

\(1-(0.6)^n > 0.8\\
\\
or (0.6)^n < 0.2\)

min value of n must be 4

SO, D :)
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Thank you all for your reply.
The correct answer is (D) 4. Thank Rajesh, I got it now.

Thanks.
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highhope
Thank you all for your reply.
The correct answer is (D) 4. Thank Rajesh, I got it now.

Thanks.

Pleasure that i could help you :)

Post your doubts and do not forget to help others too.
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Hey thanks for the solution. My miss, set up a wrong equation.
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highhope
A certain type of missile hits target with probability p=0.4. Find the number of missile that should be fired so that there is at least an 80% probability of hitting the target?

Ans:
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Nice problem. I did not get the solution when I solved the problem. I am somehow now able to understand the solution.



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