not able to understand a probability concept












-1












$begingroup$


Suppose there are patients suffering from a particular lung disease.
Either lung is diseased with a probability of 0.1.
How to find the probability of exactly n lungs being diseased ?
n = { 0,1,2 }
If I think of binomial distribution then what will be the probability
of success, p ? Will it be 0.1 ? If yes,Why ?






probability binomial-distribution







share|cite|improve this question









$endgroup$

















    -1












    $begingroup$


    Suppose there are patients suffering from a particular lung disease.
    Either lung is diseased with a probability of 0.1.
    How to find the probability of exactly n lungs being diseased ?
    n = { 0,1,2 }
    If I think of binomial distribution then what will be the probability
    of success, p ? Will it be 0.1 ? If yes,Why ?






    probability binomial-distribution







    share|cite|improve this question









    $endgroup$















      -1












      -1








      -1





      $begingroup$


      Suppose there are patients suffering from a particular lung disease.
      Either lung is diseased with a probability of 0.1.
      How to find the probability of exactly n lungs being diseased ?
      n = { 0,1,2 }
      If I think of binomial distribution then what will be the probability
      of success, p ? Will it be 0.1 ? If yes,Why ?






      probability binomial-distribution







      share|cite|improve this question









      $endgroup$




      Suppose there are patients suffering from a particular lung disease.
      Either lung is diseased with a probability of 0.1.
      How to find the probability of exactly n lungs being diseased ?
      n = { 0,1,2 }
      If I think of binomial distribution then what will be the probability
      of success, p ? Will it be 0.1 ? If yes,Why ?






      probability binomial-distribution




      probability binomial-distribution






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 6 '18 at 2:43









      shikharshikhar

      32




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          $begingroup$

          "Either lung" means. Either the left lung, the right, lung or both.



          So probability of either lung diseased is equal to $1 - P(text{neither lung diseased})$.



          And the probability of neither lung is $P(text{left lung healthy})P(text{right lung healthy}) = P(text{a specific lung healthy})^2$.



          And probability of $P(text{specific lung diseased}) = 1 -P(text{a specific lung healthy})$



          So we have $P(text{either lung diseased}) = 1-P(text{a specific lung healthy})^2=.1$ so $ P(text{a specific lung healthy})= sqrt{.9}approx 0.949$



          And $P(text{a specific lung diseased}) = 1 -sqrt{.9} approx 0.051$



          ====



          The probability of $0$ lungs diseased is $1- P(text{either lung diseased}) = 1 - .1 = .9$.



          Or we could say it is $P(text{specific lung healthy})^2 = sqrt{.9}^2 = .9$



          The probability of $1$ lung diseased i $P(text{left lung diseased and right healty}) + P(text{right lung diseased and left healthy}) = 2P(text{specific lung diseased})P(text{specific lung healthy}) = 2sqrt{.9}(1 - sqrt{.9})approx 0.097$



          The probability of $2$ lungs diseased is $P(text{specific lung diseased})^2 = (1 - sqrt{.9})^2 = 1 + .9 -2sqrt{.9} approx 0.003$



          We can generalize as exactly $n$ lungs diseased as



          $(text{# of ways to chose which lungs are diseased})P(text{specific n lungs are diseased})P(text{specific 2-n lungs are healthy})=$



          ${2 choose n}P(text {specific lung diseased})^nP(text{specific lung healthy})^{2-n}=$



          ${2 choose n} (1-sqrt{.9})^n sqrt{.9}^{2-n}$.



          I have no idea what you mean by "probability of success". Success of what?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            why is the probability of sucess equal to 0.1 ? Can you explain it more clearly please ?
            $endgroup$
            – shikhar
            Dec 6 '18 at 14:50






          • 1




            $begingroup$
            Um... because you said so. "Either lung is diseased with a probability of 0.1."
            $endgroup$
            – fleablood
            Dec 6 '18 at 16:23











          Your Answer





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          $begingroup$

          "Either lung" means. Either the left lung, the right, lung or both.



          So probability of either lung diseased is equal to $1 - P(text{neither lung diseased})$.



          And the probability of neither lung is $P(text{left lung healthy})P(text{right lung healthy}) = P(text{a specific lung healthy})^2$.



          And probability of $P(text{specific lung diseased}) = 1 -P(text{a specific lung healthy})$



          So we have $P(text{either lung diseased}) = 1-P(text{a specific lung healthy})^2=.1$ so $ P(text{a specific lung healthy})= sqrt{.9}approx 0.949$



          And $P(text{a specific lung diseased}) = 1 -sqrt{.9} approx 0.051$



          ====



          The probability of $0$ lungs diseased is $1- P(text{either lung diseased}) = 1 - .1 = .9$.



          Or we could say it is $P(text{specific lung healthy})^2 = sqrt{.9}^2 = .9$



          The probability of $1$ lung diseased i $P(text{left lung diseased and right healty}) + P(text{right lung diseased and left healthy}) = 2P(text{specific lung diseased})P(text{specific lung healthy}) = 2sqrt{.9}(1 - sqrt{.9})approx 0.097$



          The probability of $2$ lungs diseased is $P(text{specific lung diseased})^2 = (1 - sqrt{.9})^2 = 1 + .9 -2sqrt{.9} approx 0.003$



          We can generalize as exactly $n$ lungs diseased as



          $(text{# of ways to chose which lungs are diseased})P(text{specific n lungs are diseased})P(text{specific 2-n lungs are healthy})=$



          ${2 choose n}P(text {specific lung diseased})^nP(text{specific lung healthy})^{2-n}=$



          ${2 choose n} (1-sqrt{.9})^n sqrt{.9}^{2-n}$.



          I have no idea what you mean by "probability of success". Success of what?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            why is the probability of sucess equal to 0.1 ? Can you explain it more clearly please ?
            $endgroup$
            – shikhar
            Dec 6 '18 at 14:50






          • 1




            $begingroup$
            Um... because you said so. "Either lung is diseased with a probability of 0.1."
            $endgroup$
            – fleablood
            Dec 6 '18 at 16:23
















          0












          $begingroup$

          "Either lung" means. Either the left lung, the right, lung or both.



          So probability of either lung diseased is equal to $1 - P(text{neither lung diseased})$.



          And the probability of neither lung is $P(text{left lung healthy})P(text{right lung healthy}) = P(text{a specific lung healthy})^2$.



          And probability of $P(text{specific lung diseased}) = 1 -P(text{a specific lung healthy})$



          So we have $P(text{either lung diseased}) = 1-P(text{a specific lung healthy})^2=.1$ so $ P(text{a specific lung healthy})= sqrt{.9}approx 0.949$



          And $P(text{a specific lung diseased}) = 1 -sqrt{.9} approx 0.051$



          ====



          The probability of $0$ lungs diseased is $1- P(text{either lung diseased}) = 1 - .1 = .9$.



          Or we could say it is $P(text{specific lung healthy})^2 = sqrt{.9}^2 = .9$



          The probability of $1$ lung diseased i $P(text{left lung diseased and right healty}) + P(text{right lung diseased and left healthy}) = 2P(text{specific lung diseased})P(text{specific lung healthy}) = 2sqrt{.9}(1 - sqrt{.9})approx 0.097$



          The probability of $2$ lungs diseased is $P(text{specific lung diseased})^2 = (1 - sqrt{.9})^2 = 1 + .9 -2sqrt{.9} approx 0.003$



          We can generalize as exactly $n$ lungs diseased as



          $(text{# of ways to chose which lungs are diseased})P(text{specific n lungs are diseased})P(text{specific 2-n lungs are healthy})=$



          ${2 choose n}P(text {specific lung diseased})^nP(text{specific lung healthy})^{2-n}=$



          ${2 choose n} (1-sqrt{.9})^n sqrt{.9}^{2-n}$.



          I have no idea what you mean by "probability of success". Success of what?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            why is the probability of sucess equal to 0.1 ? Can you explain it more clearly please ?
            $endgroup$
            – shikhar
            Dec 6 '18 at 14:50






          • 1




            $begingroup$
            Um... because you said so. "Either lung is diseased with a probability of 0.1."
            $endgroup$
            – fleablood
            Dec 6 '18 at 16:23














          0












          0








          0





          $begingroup$

          "Either lung" means. Either the left lung, the right, lung or both.



          So probability of either lung diseased is equal to $1 - P(text{neither lung diseased})$.



          And the probability of neither lung is $P(text{left lung healthy})P(text{right lung healthy}) = P(text{a specific lung healthy})^2$.



          And probability of $P(text{specific lung diseased}) = 1 -P(text{a specific lung healthy})$



          So we have $P(text{either lung diseased}) = 1-P(text{a specific lung healthy})^2=.1$ so $ P(text{a specific lung healthy})= sqrt{.9}approx 0.949$



          And $P(text{a specific lung diseased}) = 1 -sqrt{.9} approx 0.051$



          ====



          The probability of $0$ lungs diseased is $1- P(text{either lung diseased}) = 1 - .1 = .9$.



          Or we could say it is $P(text{specific lung healthy})^2 = sqrt{.9}^2 = .9$



          The probability of $1$ lung diseased i $P(text{left lung diseased and right healty}) + P(text{right lung diseased and left healthy}) = 2P(text{specific lung diseased})P(text{specific lung healthy}) = 2sqrt{.9}(1 - sqrt{.9})approx 0.097$



          The probability of $2$ lungs diseased is $P(text{specific lung diseased})^2 = (1 - sqrt{.9})^2 = 1 + .9 -2sqrt{.9} approx 0.003$



          We can generalize as exactly $n$ lungs diseased as



          $(text{# of ways to chose which lungs are diseased})P(text{specific n lungs are diseased})P(text{specific 2-n lungs are healthy})=$



          ${2 choose n}P(text {specific lung diseased})^nP(text{specific lung healthy})^{2-n}=$



          ${2 choose n} (1-sqrt{.9})^n sqrt{.9}^{2-n}$.



          I have no idea what you mean by "probability of success". Success of what?






          share|cite|improve this answer











          $endgroup$



          "Either lung" means. Either the left lung, the right, lung or both.



          So probability of either lung diseased is equal to $1 - P(text{neither lung diseased})$.



          And the probability of neither lung is $P(text{left lung healthy})P(text{right lung healthy}) = P(text{a specific lung healthy})^2$.



          And probability of $P(text{specific lung diseased}) = 1 -P(text{a specific lung healthy})$



          So we have $P(text{either lung diseased}) = 1-P(text{a specific lung healthy})^2=.1$ so $ P(text{a specific lung healthy})= sqrt{.9}approx 0.949$



          And $P(text{a specific lung diseased}) = 1 -sqrt{.9} approx 0.051$



          ====



          The probability of $0$ lungs diseased is $1- P(text{either lung diseased}) = 1 - .1 = .9$.



          Or we could say it is $P(text{specific lung healthy})^2 = sqrt{.9}^2 = .9$



          The probability of $1$ lung diseased i $P(text{left lung diseased and right healty}) + P(text{right lung diseased and left healthy}) = 2P(text{specific lung diseased})P(text{specific lung healthy}) = 2sqrt{.9}(1 - sqrt{.9})approx 0.097$



          The probability of $2$ lungs diseased is $P(text{specific lung diseased})^2 = (1 - sqrt{.9})^2 = 1 + .9 -2sqrt{.9} approx 0.003$



          We can generalize as exactly $n$ lungs diseased as



          $(text{# of ways to chose which lungs are diseased})P(text{specific n lungs are diseased})P(text{specific 2-n lungs are healthy})=$



          ${2 choose n}P(text {specific lung diseased})^nP(text{specific lung healthy})^{2-n}=$



          ${2 choose n} (1-sqrt{.9})^n sqrt{.9}^{2-n}$.



          I have no idea what you mean by "probability of success". Success of what?







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 6 '18 at 16:49

























          answered Dec 6 '18 at 6:11









          fleabloodfleablood

          69.7k22685




          69.7k22685












          • $begingroup$
            why is the probability of sucess equal to 0.1 ? Can you explain it more clearly please ?
            $endgroup$
            – shikhar
            Dec 6 '18 at 14:50






          • 1




            $begingroup$
            Um... because you said so. "Either lung is diseased with a probability of 0.1."
            $endgroup$
            – fleablood
            Dec 6 '18 at 16:23


















          • $begingroup$
            why is the probability of sucess equal to 0.1 ? Can you explain it more clearly please ?
            $endgroup$
            – shikhar
            Dec 6 '18 at 14:50






          • 1




            $begingroup$
            Um... because you said so. "Either lung is diseased with a probability of 0.1."
            $endgroup$
            – fleablood
            Dec 6 '18 at 16:23
















          $begingroup$
          why is the probability of sucess equal to 0.1 ? Can you explain it more clearly please ?
          $endgroup$
          – shikhar
          Dec 6 '18 at 14:50




          $begingroup$
          why is the probability of sucess equal to 0.1 ? Can you explain it more clearly please ?
          $endgroup$
          – shikhar
          Dec 6 '18 at 14:50




          1




          1




          $begingroup$
          Um... because you said so. "Either lung is diseased with a probability of 0.1."
          $endgroup$
          – fleablood
          Dec 6 '18 at 16:23




          $begingroup$
          Um... because you said so. "Either lung is diseased with a probability of 0.1."
          $endgroup$
          – fleablood
          Dec 6 '18 at 16:23


















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