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The Martians are trying to destroy Earth by launching an anti-matter rocket (it will be disastrous it will hit earth). In response, Earth launches 500 missiles towards the rocket. Each missile will hit the rocket with probability 0.01, independently of other missiles. What is the probability that Earth will be saved, if at least 3 missiles must hit the rocket in order to destroy it before it hits Earth

Answer :

okpalawalter8

Answer:

The probability the earth is saved is [tex]P(X >  3) = 0.8753 [/tex]

Step-by-step explanation:

From the question we are told that  

   The number of missiles lunched by earth is n =  500

   The probability that each missile will hit the rocket is  p(x) =  0.01

   

Generally the expected number of  missiles that will hit the rocket is  

    [tex] \lambda =  n  *  p(x)[/tex]

    [tex]\lambda =  500  *  0.01 [/tex]

    [tex]\lambda =  5[/tex]

Gnerally the probability that Earth will be saved is mathematically represented as

      [tex]P(X >  3) =  1 - P(X \le 2)[/tex]

=>    [tex]P(X >  3) =  1 -  P( X = 0 ) + P(X =1 ) + P( X = 2) [/tex]

Gnerally the probability distribution function of a Poisson distribution is

      [tex]P(K) =  \frac{e^{- \lambda t } *  (\lambda *  t)^k}{k!}[/tex]

Here t =  1

Hence

      [tex]P( X = 0 ) =  \frac{e^{-5} *  5^0 }{0!}[/tex]

      [tex]P(X =1 ) =\frac{e^{-5} *  5^1 }{1!}[/tex]

     [tex]P( X = 2) = \frac{e^{-5} *  5^2 }{2!}[/tex]

 [tex]P(X >  3) =  1 - [ \frac{e^{-5} *  5^0 }{0!} + \frac{e^{-5} *  5^1 }{1!} +  \frac{e^{-5} *  5^2 }{2!}][/tex]

[tex]P(X >  3)= 1 - [ \frac{0.00673 }{1} + \frac{0.033689}{1} +  \frac{0.168449 }{2}][/tex]

[tex]P(X >  3) = 0.8753 [/tex]

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