Question

. Lightning Strikes It has been said that the probability of being struck by lightning is about 1 in 750,000, but under what circumstances? Below are listed the numbers of deaths from lightning since 1996. Golf/ball field Boating/ in water Outside/camping Construction Under a tree Phone Other 1996–2000 16 23 117 9 40 0 30 2001–2005 17 16 112 3 35 0 23 2006–2010 15 17 91 0 42 1 16 Choose one fatality at random and find each probability. a. Given that the death was after 2000, what is the probability that it occurred under a tree? b. Find the probability that the death was from camping or being outside and was before 2001. c. Find the probability that the death was from camping or being outside given that it was before 2001

Answer 1

Complete Question

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Answer:

a

      $P(U |D ) = 0.198$

b

   $P(O\ n \ B) = 0.188$

c

  $P(O | B) = 0.498$

Step-by-step explanation:

The total number of deaths is mathematically represented as

      $T = 16 + 23 + \cdots + 16$

        $T =623$

The total number of deaths in 1996 - 2000 is mathematically represented as

     $T_a = 16 + 23+ \cdots + 30$

      $T_a = 235$

The total number of deaths in 2001 - 2005 is mathematically represented as

     $T_b = 17 + 16 + \cdots + 23$

      $T_b = 206$

The total number of deaths in 2006 - 2010 is mathematically represented as

     $T_c = 15 + 17 + \cdots + 16$

      $T_d = 182$

Generally the the probability that it would occur under the tree given that the death was  after  2000 is mathematically represented as

     $P(U |D ) = \frac{P(A \ n\ U )}{P(A)}$

Here  $P(A \ n\ U )$ represents the probability that it was after 2000 and it was under the tree and this is mathematically represented as

         $P(A \ n\ U ) = \frac{Z}{ T}$

Here Z is the total number of death under the tree after 2000 and it is mathematically represented as

         $Z = 35 + 42$

=>       $Z = 77$

=>       $P(A \ n\ U ) = \frac{77}{ 623}$

=>      

Also

     $P(A)$ is the probability of the death occurring after 2000  and this is mathematically represented as

        $P(A) = \frac{T_b + T_c}{ T}$

=>      $P(A) = \frac{ 206+ 182}{623}$

=>  

So

         $P(U |D ) = \frac{\frac{77}{ 623} }{ \frac{ 206+ 182}{623}}$

=>      $P(U |D ) = 0.198$

Generally the probability that the death was from camping or being outside and was before 2001 is mathematically represented as

      $P(O | B) = \frac{T_z}{ T}$

Here $T_z$ is the total number of death outside / camping before 2001  and the value is  117  

So

            $P(O \ n \ B) = \frac{117}{623}$

=>          $P(O\ n \ B) = 0.188$

Generally the probability that the death was from camping or being outside given that it was before 2001 is mathematically represented as

       $P(O | B) = \frac{ P( O \ n \ B)}{ P(B)}$

Here $P(B)$ is the probability that it was before 2001 , this is mathematically represented as  

          $P(B ) = \frac{T_a}{T}$

=>       $P(B ) = \frac{235}{623}$

So

          $P(O | B) = \frac{ \frac{117}{623}}{ \frac{235}{623}}$

=>       $P(O | B) = 0.498$