IIT JEE , AIEEE Forum - Mathematics , Algebra

shubham_sachdeva

the probability of a missile hitting a target is 50 %.

two missiles are required to completely destroy a target. how many missiles should be fired to have 99% or more satisfaction that target has been destroyed...???

akhil_o

^n-1C₁(1/2)ⁿ+ ^n-2C₁(1/2)^n-1+ ^n-3C₁(1/2^)n-2.+...^.1C₁1/2²=0.99

(n-1)(1/2)ⁿ+(n-2)(1/2)^n-1+...1/4=0.99

This gives the sum of each case...such that the nth missile hits and one of the n-1 missiles has hit before it

So to get the value of n

^{add this series and find the sum...}

^{the first term for which sum=0.99 gives the value for n}

akhil_o

The lowest value of n is coming out to be

n=11,

where the probability becomes 0.9941

shubham_sachdeva

11 is not in the options...

Ronnieryan

we need P(hit twice or more)>0.99
So P(hit never)+P(hit once)<0.01

=> 0.5ⁿ + n(0.5)ⁿ < 0.01

We get n=10(min. value)

Hope u get me.

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