Conditional Probability of multiple independent events.












0














This problem is from my teacher and I think their answer is wrong.



The problem is in the context of trapping rats.
problem



I agree with my teacher for the answer in part A as that is just $$P(R=1)=3(0.3cdot0.7cdot0.7)=0.441$$



I do not agree with my teacher for the answer in part B. The answer my teacher gave me was equal to 0.63 using the probability tree below.
enter image description here



I found an alternate answer using this equality given to me.



$$P(X|Y)=frac{P(Xcap{Y})}{P(Y)}$$



The question wants



$$P(text{2 rats, given rat is caught in first trap})=frac{{P(text{2 rats }cap{text{ rat is caught in first trap}})}}{P(text{rat is caught in first trap})}$$



So by using substitution



$P(text{2 rats }cap{text{ rat is caught in first trap}})=2cdot0.63=0.126$



$P(text{rat is caught in first trap})=0.3$



$P(text{2 rats, given rat is caught in first trap})=frac{0.126}{0.3}=0.42$



I belive that my teacher incorrectly added the third possibility of 2 rats being caught which does not satisfy the fact that a rat was caught in the first trap. This leads to $P(text{2 rats} cap text{rat is caught in first trap})=0.189$ instead of $P(text{2 rats} cap text{rat is caught in first trap})=0.126$



After I asked about the answer to this question, my teacher checked the answer with another teacher and they insist that they are correct.



Am I missing something obvious and if so what? or is the teacher's answer incorrect?






conditional-probability







share|cite|improve this question
























  • I think you are right & the teachers wrong.
    – kimchi lover
    Nov 26 at 1:31
















0














This problem is from my teacher and I think their answer is wrong.



The problem is in the context of trapping rats.
problem



I agree with my teacher for the answer in part A as that is just $$P(R=1)=3(0.3cdot0.7cdot0.7)=0.441$$



I do not agree with my teacher for the answer in part B. The answer my teacher gave me was equal to 0.63 using the probability tree below.
enter image description here



I found an alternate answer using this equality given to me.



$$P(X|Y)=frac{P(Xcap{Y})}{P(Y)}$$



The question wants



$$P(text{2 rats, given rat is caught in first trap})=frac{{P(text{2 rats }cap{text{ rat is caught in first trap}})}}{P(text{rat is caught in first trap})}$$



So by using substitution



$P(text{2 rats }cap{text{ rat is caught in first trap}})=2cdot0.63=0.126$



$P(text{rat is caught in first trap})=0.3$



$P(text{2 rats, given rat is caught in first trap})=frac{0.126}{0.3}=0.42$



I belive that my teacher incorrectly added the third possibility of 2 rats being caught which does not satisfy the fact that a rat was caught in the first trap. This leads to $P(text{2 rats} cap text{rat is caught in first trap})=0.189$ instead of $P(text{2 rats} cap text{rat is caught in first trap})=0.126$



After I asked about the answer to this question, my teacher checked the answer with another teacher and they insist that they are correct.



Am I missing something obvious and if so what? or is the teacher's answer incorrect?






conditional-probability







share|cite|improve this question
























  • I think you are right & the teachers wrong.
    – kimchi lover
    Nov 26 at 1:31














0












0








0







This problem is from my teacher and I think their answer is wrong.



The problem is in the context of trapping rats.
problem



I agree with my teacher for the answer in part A as that is just $$P(R=1)=3(0.3cdot0.7cdot0.7)=0.441$$



I do not agree with my teacher for the answer in part B. The answer my teacher gave me was equal to 0.63 using the probability tree below.
enter image description here



I found an alternate answer using this equality given to me.



$$P(X|Y)=frac{P(Xcap{Y})}{P(Y)}$$



The question wants



$$P(text{2 rats, given rat is caught in first trap})=frac{{P(text{2 rats }cap{text{ rat is caught in first trap}})}}{P(text{rat is caught in first trap})}$$



So by using substitution



$P(text{2 rats }cap{text{ rat is caught in first trap}})=2cdot0.63=0.126$



$P(text{rat is caught in first trap})=0.3$



$P(text{2 rats, given rat is caught in first trap})=frac{0.126}{0.3}=0.42$



I belive that my teacher incorrectly added the third possibility of 2 rats being caught which does not satisfy the fact that a rat was caught in the first trap. This leads to $P(text{2 rats} cap text{rat is caught in first trap})=0.189$ instead of $P(text{2 rats} cap text{rat is caught in first trap})=0.126$



After I asked about the answer to this question, my teacher checked the answer with another teacher and they insist that they are correct.



Am I missing something obvious and if so what? or is the teacher's answer incorrect?






conditional-probability







share|cite|improve this question















This problem is from my teacher and I think their answer is wrong.



The problem is in the context of trapping rats.
problem



I agree with my teacher for the answer in part A as that is just $$P(R=1)=3(0.3cdot0.7cdot0.7)=0.441$$



I do not agree with my teacher for the answer in part B. The answer my teacher gave me was equal to 0.63 using the probability tree below.
enter image description here



I found an alternate answer using this equality given to me.



$$P(X|Y)=frac{P(Xcap{Y})}{P(Y)}$$



The question wants



$$P(text{2 rats, given rat is caught in first trap})=frac{{P(text{2 rats }cap{text{ rat is caught in first trap}})}}{P(text{rat is caught in first trap})}$$



So by using substitution



$P(text{2 rats }cap{text{ rat is caught in first trap}})=2cdot0.63=0.126$



$P(text{rat is caught in first trap})=0.3$



$P(text{2 rats, given rat is caught in first trap})=frac{0.126}{0.3}=0.42$



I belive that my teacher incorrectly added the third possibility of 2 rats being caught which does not satisfy the fact that a rat was caught in the first trap. This leads to $P(text{2 rats} cap text{rat is caught in first trap})=0.189$ instead of $P(text{2 rats} cap text{rat is caught in first trap})=0.126$



After I asked about the answer to this question, my teacher checked the answer with another teacher and they insist that they are correct.



Am I missing something obvious and if so what? or is the teacher's answer incorrect?






conditional-probability




conditional-probability






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share|cite|improve this question













share|cite|improve this question




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edited Nov 26 at 1:30

























asked Nov 26 at 1:15









Zavier

185




185












  • I think you are right & the teachers wrong.
    – kimchi lover
    Nov 26 at 1:31


















  • I think you are right & the teachers wrong.
    – kimchi lover
    Nov 26 at 1:31
















I think you are right & the teachers wrong.
– kimchi lover
Nov 26 at 1:31




I think you are right & the teachers wrong.
– kimchi lover
Nov 26 at 1:31










1 Answer
1






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You identified correctly the mistake your teacher made.



You can derive your result using the values of your teacher by correctly restricting the consideration to the upper branch of the depicted tree:




  • $P(R=2) = 0.189$


  • $color{red}{P(R'RR) = 0.063}$ does not belong to the event $1st, R$
    $$P(R=2, | , 1st, R)= frac{(R=2) cap (1st, R)}{P(1st ,R)} = frac{0.189 color{red}{- 0.063}}{0.3} = 0.42$$






share|cite|improve this answer





















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    1 Answer
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    1 Answer
    1






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes









    1














    You identified correctly the mistake your teacher made.



    You can derive your result using the values of your teacher by correctly restricting the consideration to the upper branch of the depicted tree:




    • $P(R=2) = 0.189$


    • $color{red}{P(R'RR) = 0.063}$ does not belong to the event $1st, R$
      $$P(R=2, | , 1st, R)= frac{(R=2) cap (1st, R)}{P(1st ,R)} = frac{0.189 color{red}{- 0.063}}{0.3} = 0.42$$






    share|cite|improve this answer


























      1














      You identified correctly the mistake your teacher made.



      You can derive your result using the values of your teacher by correctly restricting the consideration to the upper branch of the depicted tree:




      • $P(R=2) = 0.189$


      • $color{red}{P(R'RR) = 0.063}$ does not belong to the event $1st, R$
        $$P(R=2, | , 1st, R)= frac{(R=2) cap (1st, R)}{P(1st ,R)} = frac{0.189 color{red}{- 0.063}}{0.3} = 0.42$$






      share|cite|improve this answer
























        1












        1








        1






        You identified correctly the mistake your teacher made.



        You can derive your result using the values of your teacher by correctly restricting the consideration to the upper branch of the depicted tree:




        • $P(R=2) = 0.189$


        • $color{red}{P(R'RR) = 0.063}$ does not belong to the event $1st, R$
          $$P(R=2, | , 1st, R)= frac{(R=2) cap (1st, R)}{P(1st ,R)} = frac{0.189 color{red}{- 0.063}}{0.3} = 0.42$$






        share|cite|improve this answer












        You identified correctly the mistake your teacher made.



        You can derive your result using the values of your teacher by correctly restricting the consideration to the upper branch of the depicted tree:




        • $P(R=2) = 0.189$


        • $color{red}{P(R'RR) = 0.063}$ does not belong to the event $1st, R$
          $$P(R=2, | , 1st, R)= frac{(R=2) cap (1st, R)}{P(1st ,R)} = frac{0.189 color{red}{- 0.063}}{0.3} = 0.42$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 26 at 3:48









        trancelocation

        9,0751521




        9,0751521






























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