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An urn has $4$ black, $5$ white balls. A ball is drawn, the color is noted, and then the ball is returned back to the urn.

I would like to verify my answers:


a) What is the probability that 5 black balls are observed in $15$ draws?

Since it is replaced, we can think of it as: I want $5$ black balls, $10$ white balls, in any order:

$$\binom{15}{5}\left(\frac{4}{9}\right)^5\left(\frac{5}{9}\right)^{10}$$

b) What is the probability that $15$ draws are required until the first black ball comes.

This wording confuses me. Am I to assume there are a total of $16$ draws then, which the $16$th one being the first black?

$$\left(\dfrac{5}{9}\right)^{15}\left(\dfrac{4}{9}\right)$

In this, there is only one way this can happen (first $15$ are white, last is black)

c) What is the probability that $15$ draws are required until the fifth black ball comes?

Once again, am I to assume that on the $16$th draw, the fifth black ball will come?

In that sense I can have $15$ spots to store $4$ blacks, and $11$ whites.

Thus, a possible sequence is $BBBBWWWWWWWWWWWB$

$$\binom{15}{4}\left(\frac{4}{9}\right)^4\left(\frac{5}{9}\right)^{11}\left(\frac{4}{9}\right)$$

Is this correct? Also, I am learning about distributions right now, and was wondering if this problem can be done using a negative binomial distribution because it follows similar logic about $x$ amount of failures before the $r^{th}$ success?

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    $\begingroup$ I would read this as on the 15th draw (not the 16th), the first or fifth black ball is drawn. $\endgroup$ – aschepler Sep 28 '18 at 0:04
  • $\begingroup$ (b) Refers to negative binomial dist'n. There are two formulations: (1) Random variable X is the number of the trial on which the 5th black ball is drawn, (2) X is the number of white balls drawn before the 5th black ball appears. Wikipedia article shows formulas for both versions. Make sure to find out which version is used in your textbook or course. // I agree with @aschepler that your statement sounds like (1), but check pgs before problem appears for the full context. // Geometric dist'n in which you're waiting for first black ball also has two versions. $\endgroup$ – BruceET Sep 28 '18 at 6:29
  • $\begingroup$ Formatting tips: Type $\dbinom{n}{k}\left(\dfrac{m}{n}\right)$ to obtain $\dbinom{n}{k}\left(\frac{m}{n}\right)$ within a line and $$\binom{n}{k}\left(\frac{m}{n}\right)$$ to display $$\binom{n}{k}\left(\frac{m}{n}\right)$$ on its own line. $\endgroup$ – N. F. Taussig Sep 28 '18 at 10:41
  • $\begingroup$ Thank you. I will use this next time. @BruceET, I was thinking this too, but does the fact that we add the ball back in play any role in this? i.e, the ball is replaced. $\endgroup$ – K Split X Sep 28 '18 at 22:20
  • $\begingroup$ Yes, the distinction between sampling with replacement (as here) and without replacement can make an important difference in the distribution of the number of black balls drawn (or or the number of trials to get a particular number of black balls). The distinction is especially important if the total number of balls in the urn is small, so that the ratio of black and white balls changes greatly is balls are drawn without replacement. // If you have 4 black and 5 white balls, it makes a huge difference whether sampling is with or without replacement; if 40000 black and 50000 white, not so much. $\endgroup$ – BruceET Sep 28 '18 at 22:29

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