Given $5$ white balls, $8$ green balls and $7$ red balls. Find the probability of drawing a white ball then a...

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Given $5$ white balls, $8$ green balls and $7$ red balls in an urn. Find out the probability to draw a white ball and then a green one if the drawing is done consecutively and after drawing the ball is returned into the urn.



What I ended up with as an answer is $1/20$ by taking the chance for drawing a white ball and multiplying it by the chance to get a green ball and then dividing by two, since I only want the case where the white ball is first, which I assume is half the cases.










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    Given $5$ white balls, $8$ green balls and $7$ red balls in an urn. Find out the probability to draw a white ball and then a green one if the drawing is done consecutively and after drawing the ball is returned into the urn.



    What I ended up with as an answer is $1/20$ by taking the chance for drawing a white ball and multiplying it by the chance to get a green ball and then dividing by two, since I only want the case where the white ball is first, which I assume is half the cases.










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      Given $5$ white balls, $8$ green balls and $7$ red balls in an urn. Find out the probability to draw a white ball and then a green one if the drawing is done consecutively and after drawing the ball is returned into the urn.



      What I ended up with as an answer is $1/20$ by taking the chance for drawing a white ball and multiplying it by the chance to get a green ball and then dividing by two, since I only want the case where the white ball is first, which I assume is half the cases.










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      Given $5$ white balls, $8$ green balls and $7$ red balls in an urn. Find out the probability to draw a white ball and then a green one if the drawing is done consecutively and after drawing the ball is returned into the urn.



      What I ended up with as an answer is $1/20$ by taking the chance for drawing a white ball and multiplying it by the chance to get a green ball and then dividing by two, since I only want the case where the white ball is first, which I assume is half the cases.







      probability combinatorics






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      edited Nov 21 '18 at 20:13









      N. F. Taussig

      43.6k93355




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      asked Nov 21 '18 at 20:07









      Sartr

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          There is no need to divide by $2$. You need to multiply the probability of drawing a white ball on the first draw by the probability of drawing a green ball on the second draw. Since the draws are independent and the first ball is replaced, the probability of drawing a green ball on the second draw given that a white ball was drawn on the first draw is just the probability of drawing a green ball from the urn. Hence,
          $$Pr(text{drawing white, then green}) = Pr(W)Pr(G mid W) = Pr(W)Pr(G) = frac{5}{20} cdot frac{8}{20} = frac{1}{10}$$






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            First you calculate the probability of getting a white ball. $$P(white)=frac{Number of white balls}{Total number of balls} = frac{5}{20}$$

            Then the probability of getting a green ball is $$P(green)=frac{Number of green balls}{Total number of balls} = frac{8}{20}$$

            Because you put the ball back in the urn all the probabilities are independant

            Thus $$P(White then Green) = P(White)*P_{White}(Green) = P(White)*P(Green) = frac{5*8}{20*20} = frac{1}{10}$$






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              2 Answers
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              There is no need to divide by $2$. You need to multiply the probability of drawing a white ball on the first draw by the probability of drawing a green ball on the second draw. Since the draws are independent and the first ball is replaced, the probability of drawing a green ball on the second draw given that a white ball was drawn on the first draw is just the probability of drawing a green ball from the urn. Hence,
              $$Pr(text{drawing white, then green}) = Pr(W)Pr(G mid W) = Pr(W)Pr(G) = frac{5}{20} cdot frac{8}{20} = frac{1}{10}$$






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                There is no need to divide by $2$. You need to multiply the probability of drawing a white ball on the first draw by the probability of drawing a green ball on the second draw. Since the draws are independent and the first ball is replaced, the probability of drawing a green ball on the second draw given that a white ball was drawn on the first draw is just the probability of drawing a green ball from the urn. Hence,
                $$Pr(text{drawing white, then green}) = Pr(W)Pr(G mid W) = Pr(W)Pr(G) = frac{5}{20} cdot frac{8}{20} = frac{1}{10}$$






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                  There is no need to divide by $2$. You need to multiply the probability of drawing a white ball on the first draw by the probability of drawing a green ball on the second draw. Since the draws are independent and the first ball is replaced, the probability of drawing a green ball on the second draw given that a white ball was drawn on the first draw is just the probability of drawing a green ball from the urn. Hence,
                  $$Pr(text{drawing white, then green}) = Pr(W)Pr(G mid W) = Pr(W)Pr(G) = frac{5}{20} cdot frac{8}{20} = frac{1}{10}$$






                  share|cite|improve this answer












                  There is no need to divide by $2$. You need to multiply the probability of drawing a white ball on the first draw by the probability of drawing a green ball on the second draw. Since the draws are independent and the first ball is replaced, the probability of drawing a green ball on the second draw given that a white ball was drawn on the first draw is just the probability of drawing a green ball from the urn. Hence,
                  $$Pr(text{drawing white, then green}) = Pr(W)Pr(G mid W) = Pr(W)Pr(G) = frac{5}{20} cdot frac{8}{20} = frac{1}{10}$$







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                  answered Nov 21 '18 at 20:18









                  N. F. Taussig

                  43.6k93355




                  43.6k93355























                      1














                      First you calculate the probability of getting a white ball. $$P(white)=frac{Number of white balls}{Total number of balls} = frac{5}{20}$$

                      Then the probability of getting a green ball is $$P(green)=frac{Number of green balls}{Total number of balls} = frac{8}{20}$$

                      Because you put the ball back in the urn all the probabilities are independant

                      Thus $$P(White then Green) = P(White)*P_{White}(Green) = P(White)*P(Green) = frac{5*8}{20*20} = frac{1}{10}$$






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                        1














                        First you calculate the probability of getting a white ball. $$P(white)=frac{Number of white balls}{Total number of balls} = frac{5}{20}$$

                        Then the probability of getting a green ball is $$P(green)=frac{Number of green balls}{Total number of balls} = frac{8}{20}$$

                        Because you put the ball back in the urn all the probabilities are independant

                        Thus $$P(White then Green) = P(White)*P_{White}(Green) = P(White)*P(Green) = frac{5*8}{20*20} = frac{1}{10}$$






                        share|cite|improve this answer
























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                          First you calculate the probability of getting a white ball. $$P(white)=frac{Number of white balls}{Total number of balls} = frac{5}{20}$$

                          Then the probability of getting a green ball is $$P(green)=frac{Number of green balls}{Total number of balls} = frac{8}{20}$$

                          Because you put the ball back in the urn all the probabilities are independant

                          Thus $$P(White then Green) = P(White)*P_{White}(Green) = P(White)*P(Green) = frac{5*8}{20*20} = frac{1}{10}$$






                          share|cite|improve this answer












                          First you calculate the probability of getting a white ball. $$P(white)=frac{Number of white balls}{Total number of balls} = frac{5}{20}$$

                          Then the probability of getting a green ball is $$P(green)=frac{Number of green balls}{Total number of balls} = frac{8}{20}$$

                          Because you put the ball back in the urn all the probabilities are independant

                          Thus $$P(White then Green) = P(White)*P_{White}(Green) = P(White)*P(Green) = frac{5*8}{20*20} = frac{1}{10}$$







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                          answered Nov 21 '18 at 20:22









                          TheD0ubleT

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