0
$\begingroup$

I am working my way through an example problem from Goldberg's "Probability: An Introduction".

There are x red balls and x green balls in an urn. Total number of balls in the urn is 5. You must guess the colour of the ball being drawn by Mr Y using a number of strategies. I am having problems with strategies 4 and 5:


Strategy 4:

Draw a ball from the urn and replace it. Then draw another ball and replace it. If both balls are red, then guess Y will draw a red ball. If both balls are green then guess Y will draw a green ball. If you draw one red and one green ball, then draw one more ball from the urn. If this ball is red, then guess Y will draw a red ball. If it is green, then guess Y will draw a green ball.

Strategy 5:

Same as strategy 4, except that the first ball is not replaced before the second is drawn. Also, if a third ball is required, it is done without replacing the first two balls.


The example suggests drawing a table with the strategies as columns and the number of red balls as rows. The cells contain the probability of guessing a red ball given the number of red balls in the urn. My problem is with strategy 5 where you have just one red ball in the urn.

You have 1/5 probability of obtaining the red ball in the first draw. Let's say it is obtained, then, because it is not replaced, there is 0 probability of obtaining the red ball in the second draw. This means that the probability of obtaining a red ball is 0 for all cases where there is only one red ball and you draw it first up. If that is right, then it also means that there is 0 probability of guessing the red ball if you guess green then red, or red then green. Am I on the right track? The answer is a probability of 0.8 for 1 red ball in the urn for strategy 5, but this is not what I am getting.

Thanks in advance for any tips or help.

$\endgroup$
  • $\begingroup$ "There are x red balls and x green balls in an urn. Total number of balls in the urn is 5. " ... So then $x= 2.5$ ? Or are we to assume there are other coloured balls? Perhaps one of those two $x$ was intended to be a $y$ ? $\endgroup$ – Graham Kemp May 26 '16 at 4:07
  • $\begingroup$ Yeah, I went with the wording of the problem. If this makes it easier there are $a$ red balls and $b$ green balls. $a+b=5$. $a$ and $b$ can only be positive integers. Does that make more sense? $\endgroup$ – user183974 May 26 '16 at 4:12
1
$\begingroup$

You have to follow the strategy given.

The strategy is: take two draws, if they are both red then guess "red", else if they are both green then guess "green", otherwise they are different colours and you ask to take a third draw then guess the fourth draw will be the same colour as that.

If there is one red ball then using the outlined strategy you will either:

  • With $\tfrac 2 5$ probability obtain red on the first or second draw, ask for another draw (which will be green) and your guess of green on the fourth draw will then certainly be okay.
  • With $\tfrac 3 5$ probability obtain green on both first and second draw, then guess green on the third draw with $\tfrac 2 3$ certainty.

If there is just one red ball, this strategy will yield a correct guess with probability: $\tfrac {2}{5}+\tfrac 3 5\tfrac 2 3 = \tfrac {4}{5}$


So if there are $x$ red balls among the $5$, what is the probability of:

  • Obtaining two red balls then being correct by guessing "red" for the third draw.
  • Obtaining two green balls then being correct by guessing "green" for the third draw.
  • Obtaining one of each colour then either
    • obtaining red on the third draw and being correct by guessing "red" for the fourth.
    • obtaining green on the third draw and being correct by guessing "green" for the fourth.

Complete the table $$\begin{array}{l|l}x & \mathsf P(\text{Correct}) \\ \hline 0 & 1 \\ \hdashline 1 & 0.8 \\ \hdashline 2\\ \hdashline 3\\ \hdashline 4\\ \hdashline 5\end{array}$$

Note: symmetry

$\endgroup$
  • $\begingroup$ Thanks. I got .3 for x=2. But that can't be right. The answer given is .54. I added the probability of drawing two greens and then guessing green (1/10), then the possibility of getting a green,red,green followed by a green (1/10), and finally the possibility of getting a red then three greens (1/10). The other options are not possible given that you would have to draw more than two reds. I am assuming that the last draw by Mr Y does not involve any replacement. $\endgroup$ – user183974 May 26 '16 at 8:55
  • $\begingroup$ I concur. Under that assumption it is $0.3$. $\endgroup$ – Graham Kemp May 26 '16 at 9:18
  • $\begingroup$ Thanks. So I assume the textbook answer of .54 occurs if you replace all balls for the final draw. But if that is the assumption, then we cant get .8 for the first draw. Is that right? $\endgroup$ – user183974 May 27 '16 at 11:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.