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?

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

\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}

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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 supports the community Dec 19 '18 at 11:11

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