probability - getting a red ball in thenext draw is












0












$begingroup$


An urn contains 5 red and 7 green balls. A ball is drawn at random and its color is noted. The ball is placed back into the urn along with another ball of same color. The probability of getting a red ball in the next draw is _____.



My approach:
I think both of these events are independent and hence probability of drawing a red ball random remains same which is equal to $frac{5}{12}$.



is it correct?






probability







share|cite|improve this question











$endgroup$












  • $begingroup$
    In mathematics, "I think" is the wrong way. "Why should it be?" is the right way.
    $endgroup$
    – user21820
    Dec 20 '18 at 5:03
















0












$begingroup$


An urn contains 5 red and 7 green balls. A ball is drawn at random and its color is noted. The ball is placed back into the urn along with another ball of same color. The probability of getting a red ball in the next draw is _____.



My approach:
I think both of these events are independent and hence probability of drawing a red ball random remains same which is equal to $frac{5}{12}$.



is it correct?






probability







share|cite|improve this question











$endgroup$












  • $begingroup$
    In mathematics, "I think" is the wrong way. "Why should it be?" is the right way.
    $endgroup$
    – user21820
    Dec 20 '18 at 5:03














0












0








0





$begingroup$


An urn contains 5 red and 7 green balls. A ball is drawn at random and its color is noted. The ball is placed back into the urn along with another ball of same color. The probability of getting a red ball in the next draw is _____.



My approach:
I think both of these events are independent and hence probability of drawing a red ball random remains same which is equal to $frac{5}{12}$.



is it correct?






probability







share|cite|improve this question











$endgroup$




An urn contains 5 red and 7 green balls. A ball is drawn at random and its color is noted. The ball is placed back into the urn along with another ball of same color. The probability of getting a red ball in the next draw is _____.



My approach:
I think both of these events are independent and hence probability of drawing a red ball random remains same which is equal to $frac{5}{12}$.



is it correct?






probability




probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 11:01







swapnil

















asked Dec 19 '18 at 10:56









swapnilswapnil

335




335












  • $begingroup$
    In mathematics, "I think" is the wrong way. "Why should it be?" is the right way.
    $endgroup$
    – user21820
    Dec 20 '18 at 5:03


















  • $begingroup$
    In mathematics, "I think" is the wrong way. "Why should it be?" is the right way.
    $endgroup$
    – user21820
    Dec 20 '18 at 5:03
















$begingroup$
In mathematics, "I think" is the wrong way. "Why should it be?" is the right way.
$endgroup$
– user21820
Dec 20 '18 at 5:03




$begingroup$
In mathematics, "I think" is the wrong way. "Why should it be?" is the right way.
$endgroup$
– user21820
Dec 20 '18 at 5:03










3 Answers
3






active

oldest

votes


















6












$begingroup$

begin{align}
& Bbb P(text{2nd draw red}) \
= & Bbb P(text{2nd draw red} ; | ; text{1st draw red}) cdot Bbb P(text{1st draw red}) + \
& Bbb P(text{2nd draw red} ; | ; text{1st draw green}) cdot Bbb P(text{1st draw green}) \
= & frac{6}{13} cdot frac{5}{12} + frac{5}{13} cdot frac{7}{12}
end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I am not getting it. we drew a ball we noted its color and we put it back in the urn with its neighbour having same color, but again we need to pick up how this becomes conditional probability?
    $endgroup$
    – swapnil
    Dec 19 '18 at 11:04










  • $begingroup$
    and also please tell me how you have come up with that formula.
    $endgroup$
    – swapnil
    Dec 19 '18 at 11:11






  • 2




    $begingroup$
    @swapnil "The ball is placed back into the urn with another ball of the same color..." This justifies the conditioning.
    $endgroup$
    – drhab
    Dec 19 '18 at 11:21



















3












$begingroup$

If the first draw is red, which happens with probability $frac{5}{12}$, the second draw is from an urn with $6$ reds (one extra) and $7$ greens, from which a red draw has probability $frac{6}{13}$.



If the first draw is green (happens with probability $frac{7}{12}$) we have a second draw from an urn with $5$ reds and $8$ greens, from which a red draw has probability $frac{5}{13}$.



So $$P(text{second red}) = P(text{second red}| text{first red})P(text{first red}) + P(text{second red}| text{first green})P(text{first green})$$



which equals $$frac{6}{13}frac{5}{12} + frac{5}{13}frac{7}{12}= frac{5}{12}$$



For $r$ red balls and $g$ green balls I get $$frac{r(r+1)+ rg}{(r+g+1)(r+g)}=frac{r(r+g+1)}{(r+g)(r+g+1)}=frac{r}{r+g}$$ as the answer, so the equal probability to the initial situation is no coincidence. Interesting.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    but why did you added that extra ball and also how do you come up with that formula?
    $endgroup$
    – swapnil
    Dec 19 '18 at 11:57










  • $begingroup$
    @swapnil the extra ball is in the description of the problem. The formula is just the general formula for the total probability based on conditional ones.
    $endgroup$
    – Henno Brandsma
    Dec 19 '18 at 11:59










  • $begingroup$
    so the ball is drawn is at random is not from the urn?
    $endgroup$
    – swapnil
    Dec 19 '18 at 12:07










  • $begingroup$
    I thought first we are picking a ball from urn then noticing its color and then putting it back again in urn.
    $endgroup$
    – swapnil
    Dec 19 '18 at 12:08










  • $begingroup$
    @swapnil we’re also putting in an extra ball of the same colour that was drawn. So two reds or two greens go back. “Along with another ball of the same color” it says.
    $endgroup$
    – Henno Brandsma
    Dec 19 '18 at 12:11





















2












$begingroup$

No. As another ball with a colour dependant on the first draw is placed in the urn before the second draw, the events are not independent.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    consider this example. I have a pack of card, I selected first card I noted it and again placed it back in the deck with another color of same card. Now the probability that I will get the ace won't it be 4/52.
    $endgroup$
    – swapnil
    Dec 19 '18 at 11:09










  • $begingroup$
    No, but the answer to what that probability will be also depend on the probability that the new card is an ace.
    $endgroup$
    – Henrik
    Dec 19 '18 at 11:11











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

begin{align}
& Bbb P(text{2nd draw red}) \
= & Bbb P(text{2nd draw red} ; | ; text{1st draw red}) cdot Bbb P(text{1st draw red}) + \
& Bbb P(text{2nd draw red} ; | ; text{1st draw green}) cdot Bbb P(text{1st draw green}) \
= & frac{6}{13} cdot frac{5}{12} + frac{5}{13} cdot frac{7}{12}
end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I am not getting it. we drew a ball we noted its color and we put it back in the urn with its neighbour having same color, but again we need to pick up how this becomes conditional probability?
    $endgroup$
    – swapnil
    Dec 19 '18 at 11:04










  • $begingroup$
    and also please tell me how you have come up with that formula.
    $endgroup$
    – swapnil
    Dec 19 '18 at 11:11






  • 2




    $begingroup$
    @swapnil "The ball is placed back into the urn with another ball of the same color..." This justifies the conditioning.
    $endgroup$
    – drhab
    Dec 19 '18 at 11:21
















6












$begingroup$

begin{align}
& Bbb P(text{2nd draw red}) \
= & Bbb P(text{2nd draw red} ; | ; text{1st draw red}) cdot Bbb P(text{1st draw red}) + \
& Bbb P(text{2nd draw red} ; | ; text{1st draw green}) cdot Bbb P(text{1st draw green}) \
= & frac{6}{13} cdot frac{5}{12} + frac{5}{13} cdot frac{7}{12}
end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I am not getting it. we drew a ball we noted its color and we put it back in the urn with its neighbour having same color, but again we need to pick up how this becomes conditional probability?
    $endgroup$
    – swapnil
    Dec 19 '18 at 11:04










  • $begingroup$
    and also please tell me how you have come up with that formula.
    $endgroup$
    – swapnil
    Dec 19 '18 at 11:11






  • 2




    $begingroup$
    @swapnil "The ball is placed back into the urn with another ball of the same color..." This justifies the conditioning.
    $endgroup$
    – drhab
    Dec 19 '18 at 11:21














6












6








6





$begingroup$

begin{align}
& Bbb P(text{2nd draw red}) \
= & Bbb P(text{2nd draw red} ; | ; text{1st draw red}) cdot Bbb P(text{1st draw red}) + \
& Bbb P(text{2nd draw red} ; | ; text{1st draw green}) cdot Bbb P(text{1st draw green}) \
= & frac{6}{13} cdot frac{5}{12} + frac{5}{13} cdot frac{7}{12}
end{align}






share|cite|improve this answer









$endgroup$



begin{align}
& Bbb P(text{2nd draw red}) \
= & Bbb P(text{2nd draw red} ; | ; text{1st draw red}) cdot Bbb P(text{1st draw red}) + \
& Bbb P(text{2nd draw red} ; | ; text{1st draw green}) cdot Bbb P(text{1st draw green}) \
= & frac{6}{13} cdot frac{5}{12} + frac{5}{13} cdot frac{7}{12}
end{align}







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 19 '18 at 11:01









glowstonetreesglowstonetrees

2,375418




2,375418












  • $begingroup$
    I am not getting it. we drew a ball we noted its color and we put it back in the urn with its neighbour having same color, but again we need to pick up how this becomes conditional probability?
    $endgroup$
    – swapnil
    Dec 19 '18 at 11:04










  • $begingroup$
    and also please tell me how you have come up with that formula.
    $endgroup$
    – swapnil
    Dec 19 '18 at 11:11






  • 2




    $begingroup$
    @swapnil "The ball is placed back into the urn with another ball of the same color..." This justifies the conditioning.
    $endgroup$
    – drhab
    Dec 19 '18 at 11:21


















  • $begingroup$
    I am not getting it. we drew a ball we noted its color and we put it back in the urn with its neighbour having same color, but again we need to pick up how this becomes conditional probability?
    $endgroup$
    – swapnil
    Dec 19 '18 at 11:04










  • $begingroup$
    and also please tell me how you have come up with that formula.
    $endgroup$
    – swapnil
    Dec 19 '18 at 11:11






  • 2




    $begingroup$
    @swapnil "The ball is placed back into the urn with another ball of the same color..." This justifies the conditioning.
    $endgroup$
    – drhab
    Dec 19 '18 at 11:21
















$begingroup$
I am not getting it. we drew a ball we noted its color and we put it back in the urn with its neighbour having same color, but again we need to pick up how this becomes conditional probability?
$endgroup$
– swapnil
Dec 19 '18 at 11:04




$begingroup$
I am not getting it. we drew a ball we noted its color and we put it back in the urn with its neighbour having same color, but again we need to pick up how this becomes conditional probability?
$endgroup$
– swapnil
Dec 19 '18 at 11:04












$begingroup$
and also please tell me how you have come up with that formula.
$endgroup$
– swapnil
Dec 19 '18 at 11:11




$begingroup$
and also please tell me how you have come up with that formula.
$endgroup$
– swapnil
Dec 19 '18 at 11:11




2




2




$begingroup$
@swapnil "The ball is placed back into the urn with another ball of the same color..." This justifies the conditioning.
$endgroup$
– drhab
Dec 19 '18 at 11:21




$begingroup$
@swapnil "The ball is placed back into the urn with another ball of the same color..." This justifies the conditioning.
$endgroup$
– drhab
Dec 19 '18 at 11:21











3












$begingroup$

If the first draw is red, which happens with probability $frac{5}{12}$, the second draw is from an urn with $6$ reds (one extra) and $7$ greens, from which a red draw has probability $frac{6}{13}$.



If the first draw is green (happens with probability $frac{7}{12}$) we have a second draw from an urn with $5$ reds and $8$ greens, from which a red draw has probability $frac{5}{13}$.



So $$P(text{second red}) = P(text{second red}| text{first red})P(text{first red}) + P(text{second red}| text{first green})P(text{first green})$$



which equals $$frac{6}{13}frac{5}{12} + frac{5}{13}frac{7}{12}= frac{5}{12}$$



For $r$ red balls and $g$ green balls I get $$frac{r(r+1)+ rg}{(r+g+1)(r+g)}=frac{r(r+g+1)}{(r+g)(r+g+1)}=frac{r}{r+g}$$ as the answer, so the equal probability to the initial situation is no coincidence. Interesting.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    but why did you added that extra ball and also how do you come up with that formula?
    $endgroup$
    – swapnil
    Dec 19 '18 at 11:57










  • $begingroup$
    @swapnil the extra ball is in the description of the problem. The formula is just the general formula for the total probability based on conditional ones.
    $endgroup$
    – Henno Brandsma
    Dec 19 '18 at 11:59










  • $begingroup$
    so the ball is drawn is at random is not from the urn?
    $endgroup$
    – swapnil
    Dec 19 '18 at 12:07










  • $begingroup$
    I thought first we are picking a ball from urn then noticing its color and then putting it back again in urn.
    $endgroup$
    – swapnil
    Dec 19 '18 at 12:08










  • $begingroup$
    @swapnil we’re also putting in an extra ball of the same colour that was drawn. So two reds or two greens go back. “Along with another ball of the same color” it says.
    $endgroup$
    – Henno Brandsma
    Dec 19 '18 at 12:11


















3












$begingroup$

If the first draw is red, which happens with probability $frac{5}{12}$, the second draw is from an urn with $6$ reds (one extra) and $7$ greens, from which a red draw has probability $frac{6}{13}$.



If the first draw is green (happens with probability $frac{7}{12}$) we have a second draw from an urn with $5$ reds and $8$ greens, from which a red draw has probability $frac{5}{13}$.



So $$P(text{second red}) = P(text{second red}| text{first red})P(text{first red}) + P(text{second red}| text{first green})P(text{first green})$$



which equals $$frac{6}{13}frac{5}{12} + frac{5}{13}frac{7}{12}= frac{5}{12}$$



For $r$ red balls and $g$ green balls I get $$frac{r(r+1)+ rg}{(r+g+1)(r+g)}=frac{r(r+g+1)}{(r+g)(r+g+1)}=frac{r}{r+g}$$ as the answer, so the equal probability to the initial situation is no coincidence. Interesting.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    but why did you added that extra ball and also how do you come up with that formula?
    $endgroup$
    – swapnil
    Dec 19 '18 at 11:57










  • $begingroup$
    @swapnil the extra ball is in the description of the problem. The formula is just the general formula for the total probability based on conditional ones.
    $endgroup$
    – Henno Brandsma
    Dec 19 '18 at 11:59










  • $begingroup$
    so the ball is drawn is at random is not from the urn?
    $endgroup$
    – swapnil
    Dec 19 '18 at 12:07










  • $begingroup$
    I thought first we are picking a ball from urn then noticing its color and then putting it back again in urn.
    $endgroup$
    – swapnil
    Dec 19 '18 at 12:08










  • $begingroup$
    @swapnil we’re also putting in an extra ball of the same colour that was drawn. So two reds or two greens go back. “Along with another ball of the same color” it says.
    $endgroup$
    – Henno Brandsma
    Dec 19 '18 at 12:11
















3












3








3





$begingroup$

If the first draw is red, which happens with probability $frac{5}{12}$, the second draw is from an urn with $6$ reds (one extra) and $7$ greens, from which a red draw has probability $frac{6}{13}$.



If the first draw is green (happens with probability $frac{7}{12}$) we have a second draw from an urn with $5$ reds and $8$ greens, from which a red draw has probability $frac{5}{13}$.



So $$P(text{second red}) = P(text{second red}| text{first red})P(text{first red}) + P(text{second red}| text{first green})P(text{first green})$$



which equals $$frac{6}{13}frac{5}{12} + frac{5}{13}frac{7}{12}= frac{5}{12}$$



For $r$ red balls and $g$ green balls I get $$frac{r(r+1)+ rg}{(r+g+1)(r+g)}=frac{r(r+g+1)}{(r+g)(r+g+1)}=frac{r}{r+g}$$ as the answer, so the equal probability to the initial situation is no coincidence. Interesting.






share|cite|improve this answer











$endgroup$



If the first draw is red, which happens with probability $frac{5}{12}$, the second draw is from an urn with $6$ reds (one extra) and $7$ greens, from which a red draw has probability $frac{6}{13}$.



If the first draw is green (happens with probability $frac{7}{12}$) we have a second draw from an urn with $5$ reds and $8$ greens, from which a red draw has probability $frac{5}{13}$.



So $$P(text{second red}) = P(text{second red}| text{first red})P(text{first red}) + P(text{second red}| text{first green})P(text{first green})$$



which equals $$frac{6}{13}frac{5}{12} + frac{5}{13}frac{7}{12}= frac{5}{12}$$



For $r$ red balls and $g$ green balls I get $$frac{r(r+1)+ rg}{(r+g+1)(r+g)}=frac{r(r+g+1)}{(r+g)(r+g+1)}=frac{r}{r+g}$$ as the answer, so the equal probability to the initial situation is no coincidence. Interesting.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 19 '18 at 17:31

























answered Dec 19 '18 at 11:17









Henno BrandsmaHenno Brandsma

111k348118




111k348118












  • $begingroup$
    but why did you added that extra ball and also how do you come up with that formula?
    $endgroup$
    – swapnil
    Dec 19 '18 at 11:57










  • $begingroup$
    @swapnil the extra ball is in the description of the problem. The formula is just the general formula for the total probability based on conditional ones.
    $endgroup$
    – Henno Brandsma
    Dec 19 '18 at 11:59










  • $begingroup$
    so the ball is drawn is at random is not from the urn?
    $endgroup$
    – swapnil
    Dec 19 '18 at 12:07










  • $begingroup$
    I thought first we are picking a ball from urn then noticing its color and then putting it back again in urn.
    $endgroup$
    – swapnil
    Dec 19 '18 at 12:08










  • $begingroup$
    @swapnil we’re also putting in an extra ball of the same colour that was drawn. So two reds or two greens go back. “Along with another ball of the same color” it says.
    $endgroup$
    – Henno Brandsma
    Dec 19 '18 at 12:11




















  • $begingroup$
    but why did you added that extra ball and also how do you come up with that formula?
    $endgroup$
    – swapnil
    Dec 19 '18 at 11:57










  • $begingroup$
    @swapnil the extra ball is in the description of the problem. The formula is just the general formula for the total probability based on conditional ones.
    $endgroup$
    – Henno Brandsma
    Dec 19 '18 at 11:59










  • $begingroup$
    so the ball is drawn is at random is not from the urn?
    $endgroup$
    – swapnil
    Dec 19 '18 at 12:07










  • $begingroup$
    I thought first we are picking a ball from urn then noticing its color and then putting it back again in urn.
    $endgroup$
    – swapnil
    Dec 19 '18 at 12:08










  • $begingroup$
    @swapnil we’re also putting in an extra ball of the same colour that was drawn. So two reds or two greens go back. “Along with another ball of the same color” it says.
    $endgroup$
    – Henno Brandsma
    Dec 19 '18 at 12:11


















$begingroup$
but why did you added that extra ball and also how do you come up with that formula?
$endgroup$
– swapnil
Dec 19 '18 at 11:57




$begingroup$
but why did you added that extra ball and also how do you come up with that formula?
$endgroup$
– swapnil
Dec 19 '18 at 11:57












$begingroup$
@swapnil the extra ball is in the description of the problem. The formula is just the general formula for the total probability based on conditional ones.
$endgroup$
– Henno Brandsma
Dec 19 '18 at 11:59




$begingroup$
@swapnil the extra ball is in the description of the problem. The formula is just the general formula for the total probability based on conditional ones.
$endgroup$
– Henno Brandsma
Dec 19 '18 at 11:59












$begingroup$
so the ball is drawn is at random is not from the urn?
$endgroup$
– swapnil
Dec 19 '18 at 12:07




$begingroup$
so the ball is drawn is at random is not from the urn?
$endgroup$
– swapnil
Dec 19 '18 at 12:07












$begingroup$
I thought first we are picking a ball from urn then noticing its color and then putting it back again in urn.
$endgroup$
– swapnil
Dec 19 '18 at 12:08




$begingroup$
I thought first we are picking a ball from urn then noticing its color and then putting it back again in urn.
$endgroup$
– swapnil
Dec 19 '18 at 12:08












$begingroup$
@swapnil we’re also putting in an extra ball of the same colour that was drawn. So two reds or two greens go back. “Along with another ball of the same color” it says.
$endgroup$
– Henno Brandsma
Dec 19 '18 at 12:11






$begingroup$
@swapnil we’re also putting in an extra ball of the same colour that was drawn. So two reds or two greens go back. “Along with another ball of the same color” it says.
$endgroup$
– Henno Brandsma
Dec 19 '18 at 12:11













2












$begingroup$

No. As another ball with a colour dependant on the first draw is placed in the urn before the second draw, the events are not independent.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    consider this example. I have a pack of card, I selected first card I noted it and again placed it back in the deck with another color of same card. Now the probability that I will get the ace won't it be 4/52.
    $endgroup$
    – swapnil
    Dec 19 '18 at 11:09










  • $begingroup$
    No, but the answer to what that probability will be also depend on the probability that the new card is an ace.
    $endgroup$
    – Henrik
    Dec 19 '18 at 11:11
















2












$begingroup$

No. As another ball with a colour dependant on the first draw is placed in the urn before the second draw, the events are not independent.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    consider this example. I have a pack of card, I selected first card I noted it and again placed it back in the deck with another color of same card. Now the probability that I will get the ace won't it be 4/52.
    $endgroup$
    – swapnil
    Dec 19 '18 at 11:09










  • $begingroup$
    No, but the answer to what that probability will be also depend on the probability that the new card is an ace.
    $endgroup$
    – Henrik
    Dec 19 '18 at 11:11














2












2








2





$begingroup$

No. As another ball with a colour dependant on the first draw is placed in the urn before the second draw, the events are not independent.






share|cite|improve this answer









$endgroup$



No. As another ball with a colour dependant on the first draw is placed in the urn before the second draw, the events are not independent.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 19 '18 at 11:01









HenrikHenrik

6,03592030




6,03592030












  • $begingroup$
    consider this example. I have a pack of card, I selected first card I noted it and again placed it back in the deck with another color of same card. Now the probability that I will get the ace won't it be 4/52.
    $endgroup$
    – swapnil
    Dec 19 '18 at 11:09










  • $begingroup$
    No, but the answer to what that probability will be also depend on the probability that the new card is an ace.
    $endgroup$
    – Henrik
    Dec 19 '18 at 11:11


















  • $begingroup$
    consider this example. I have a pack of card, I selected first card I noted it and again placed it back in the deck with another color of same card. Now the probability that I will get the ace won't it be 4/52.
    $endgroup$
    – swapnil
    Dec 19 '18 at 11:09










  • $begingroup$
    No, but the answer to what that probability will be also depend on the probability that the new card is an ace.
    $endgroup$
    – Henrik
    Dec 19 '18 at 11:11
















$begingroup$
consider this example. I have a pack of card, I selected first card I noted it and again placed it back in the deck with another color of same card. Now the probability that I will get the ace won't it be 4/52.
$endgroup$
– swapnil
Dec 19 '18 at 11:09




$begingroup$
consider this example. I have a pack of card, I selected first card I noted it and again placed it back in the deck with another color of same card. Now the probability that I will get the ace won't it be 4/52.
$endgroup$
– swapnil
Dec 19 '18 at 11:09












$begingroup$
No, but the answer to what that probability will be also depend on the probability that the new card is an ace.
$endgroup$
– Henrik
Dec 19 '18 at 11:11




$begingroup$
No, but the answer to what that probability will be also depend on the probability that the new card is an ace.
$endgroup$
– Henrik
Dec 19 '18 at 11:11


















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