Given $5$ white balls, $8$ green balls and $7$ red balls. Find the probability of drawing a white ball then a...
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
edited Nov 21 '18 at 20:13
N. F. Taussig
43.6k93355
asked Nov 21 '18 at 20:07
Sartr
949
add a comment |
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
edited Nov 21 '18 at 20:13
N. F. Taussig
43.6k93355
asked Nov 21 '18 at 20:07
Sartr
949
add a comment |
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
edited Nov 21 '18 at 20:13
N. F. Taussig
43.6k93355
asked Nov 21 '18 at 20:07
Sartr
949
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
probability combinatorics
edited Nov 21 '18 at 20:13
N. F. Taussig
43.6k93355
asked Nov 21 '18 at 20:07
Sartr
949
edited Nov 21 '18 at 20:13
N. F. Taussig
43.6k93355
asked Nov 21 '18 at 20:07
Sartr
949
edited Nov 21 '18 at 20:13
N. F. Taussig
43.6k93355
edited Nov 21 '18 at 20:13
N. F. Taussig
43.6k93355
edited Nov 21 '18 at 20:13
N. F. Taussig
43.6k93355
43.6k93355
asked Nov 21 '18 at 20:07
Sartr
949
asked Nov 21 '18 at 20:07
Sartr
949
asked Nov 21 '18 at 20:07
Sartr
949
949
add a comment |
add a comment |
2 Answers
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votes
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,
$$Pr(text{drawing white, then green}) = Pr(W)Pr(G mid W) = Pr(W)Pr(G) = frac{5}{20} cdot frac{8}{20} = frac{1}{10}$$
answered Nov 21 '18 at 20:18
N. F. Taussig
43.6k93355
add a comment |
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 $$P(White then Green) = P(White)*P_{White}(Green) = P(White)*P(Green) = frac{5*8}{20*20} = frac{1}{10}$$
answered Nov 21 '18 at 20:22
TheD0ubleT
39218
add a comment |
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2 Answers
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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,
$$Pr(text{drawing white, then green}) = Pr(W)Pr(G mid W) = Pr(W)Pr(G) = frac{5}{20} cdot frac{8}{20} = frac{1}{10}$$
answered Nov 21 '18 at 20:18
N. F. Taussig
43.6k93355
add a comment |
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,
$$Pr(text{drawing white, then green}) = Pr(W)Pr(G mid W) = Pr(W)Pr(G) = frac{5}{20} cdot frac{8}{20} = frac{1}{10}$$
answered Nov 21 '18 at 20:18
N. F. Taussig
43.6k93355
add a comment |
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,
$$Pr(text{drawing white, then green}) = Pr(W)Pr(G mid W) = Pr(W)Pr(G) = frac{5}{20} cdot frac{8}{20} = frac{1}{10}$$
answered Nov 21 '18 at 20:18
N. F. Taussig
43.6k93355
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,
$$Pr(text{drawing white, then green}) = Pr(W)Pr(G mid W) = Pr(W)Pr(G) = frac{5}{20} cdot frac{8}{20} = frac{1}{10}$$
answered Nov 21 '18 at 20:18
N. F. Taussig
43.6k93355
answered Nov 21 '18 at 20:18
N. F. Taussig
43.6k93355
answered Nov 21 '18 at 20:18
N. F. Taussig
43.6k93355
answered Nov 21 '18 at 20:18
N. F. Taussig
43.6k93355
43.6k93355
add a comment |
add a comment |
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 $$P(White then Green) = P(White)*P_{White}(Green) = P(White)*P(Green) = frac{5*8}{20*20} = frac{1}{10}$$
answered Nov 21 '18 at 20:22
TheD0ubleT
39218
add a comment |
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 $$P(White then Green) = P(White)*P_{White}(Green) = P(White)*P(Green) = frac{5*8}{20*20} = frac{1}{10}$$
answered Nov 21 '18 at 20:22
TheD0ubleT
39218
add a comment |
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 $$P(White then Green) = P(White)*P_{White}(Green) = P(White)*P(Green) = frac{5*8}{20*20} = frac{1}{10}$$
answered Nov 21 '18 at 20:22
TheD0ubleT
39218
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 $$P(White then Green) = P(White)*P_{White}(Green) = P(White)*P(Green) = frac{5*8}{20*20} = frac{1}{10}$$
answered Nov 21 '18 at 20:22
TheD0ubleT
39218
answered Nov 21 '18 at 20:22
TheD0ubleT
39218
answered Nov 21 '18 at 20:22
TheD0ubleT
39218
answered Nov 21 '18 at 20:22
TheD0ubleT
39218
39218
add a comment |
add a comment |
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