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I had a discussion with my students today on simulations and gave them the problem below. After some time talking about it we came to the conclusion that there might be two solutions. So can someone please answer the following question and maybe list any alternative solutions:

45% of donors have type O blood type. Using a random number table (5 runs) estimate the probability of getting 4 type O donors before 12 donors arrive at the blood bank?

Given random numbers:

6841 35013  15529 72765 85089 57067 50211   47487   82739   57890   20807

81676 55300 94383 14893 60940 72024 17868   24943   61790   90656   87964

73311 12190 06628 71683 12285 39814 29103   81733   73035   57446   99209

(I found this question online and can't find the solution).

My estimate was approx. 47% and my students estimate was 80%. Are there two ways to interpret this question?

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    Is there an entry missing in the beginning of the random number table? It is supposed to be a set 5, isn't it? – 2017-01-12
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    Thank you for noticing. Yes it is supposed to be 68417 – 2017-01-12
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    An analytical solution yields a figure of $\approx 81\%$ – 2017-01-12
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    @trueblueanil can you please explain how you calculated this approximation? – 2017-01-13
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    Analytical solution: X ~ BINOM(11,.45). You want $P(X\ge4).$ That is, four or more Type O's among 11 visitors. In R statistical software `sum(dbinom(4:11, 11, .45))` returns 0.8088773. Normal approx. might also work. @trueblueanil must have done something like that. (Difficult to see how you can use random numbers like yours to get a reasonably accurate result.) – 2017-01-13
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    Sorry, I was away. It's exactly what BruceET has described. – 2017-01-13
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    Thank you all for your replies. @BruceET your solution makes sense. – 2017-01-13

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To calculate the analytical solution suppose there is a sequence of independent Bernoulli trials with probability of success given by $\pi$. The negative binomial distribution, also known as the Pascal distribution or Pólya distribution, gives the probability of $k-1$ "successes" and $y$ "failures" in $y + k - 1$ trials with a success following on the $(y + k)$th trial. The probability density function is given by

$$ P[Y = y] = {{y + k - 1}\choose{k - 1}} \pi^k (1 - \pi)^y $$

Let $A$ be the event of getting k = 4 O type donors before 12 independent donors arrive to the blood bank. With $\pi = 0.45$

$$ P[A] = \sum_{y = 0}^{7} {{y + 4 - 1}\choose{4 - 1}} 0.45^4 (1 - 0.45)^y = 0.808877 $$

To simulate, assume that random values between 00 and 44 are O type donors. If the fourth time a value falls within this range occurs before a run of 12, then count this as a success and begin a new run, otherwise it will be a failure.

Although the value presented by @BruceET is the same as above, the binomial distribution conceptually does not work since it calculates the probability of $x$ successes in $n$ independent trials, so $P(X \ge 4)$ is the probability that in a group of 11 donors, 4 or more are of type O. This clearly is not the event of getting 4 type O donors before 12 donors arrive at the blood bank.