A fair die is tossed 360 times. The probability that a six comes up on 70 or more of the tosses is

A. greater than 0.50 B. between 0.16 and 0.50 C. between 0.02 and 0.16 D. between 0.01 and 0.02 E. less than 0.01

the answer is C.

Why is that? I know the probabily = P(a six comes up exactly 70 times) + P(a six comes up exactly 71 times) + ... + P(a six comes up exactly 360 times), which is a sum of binomial probabilities, but how do we approximate it?

Thank you very much.

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The probability of rolling a $6$ is $1/6$, and the complementary probability is $5/6$, so the expected number of $6$’s is $(1/6)\cdot 360=60$, with standard deviation $\sqrt{360(1/6)(5/6)}=\sqrt{50}\approx 7$. The distribution is approximately normal, so you can estimate the desired probability as the probability that a normally distributed random variable is at least $(70-60)/7=10/7$ standard deviations above the mean. The $68-95-99.7$ rule of thumb tells you that this is somewhere between $\frac12(1-0.68)=0.16$ and $\frac12(1-0.95)=0.025$, and C is the only answer that fits, even without using any accurate calculations.

By the way, dice is plural: the singular is die.

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You've got 60 where you need 360. And your way of explaining how you got $10/7$ makes no sense at all; it looks as if the $10$ came from the claim that that's the expected number of 6s rather than from the fact that 70 exceeds 60 by 10. And you're neglecting the continuity correction. Maybe the continuity correction isn't so important if you're only trying to show the answer is in a certain interval, but it's not such a great idea to fail to make it habitual. – Michael Hardy Nov 10 '11 at 16:46
@Michael: The first is a genuine typo, which I’ve now fixed. I don’t see any basis for your statement that ‘it looks as if the $10$ came from ...’, but I’ve gone ahead and clarified it anyway. And of course I neglected the continuity correction: my purpose was to show that even a crude estimate using only mental arithmetic was sufficient to answer the question as it was asked. I thought that that would be evident from all the other crude approximations that I used, especially $\sqrt{50}\approx 7$. – Brian M. Scott Nov 10 '11 at 16:53
You haven't entirely fixed the typo, since you now say $(1/6)\cdot360=10$. Even now, there might be confusion as to where the $10$ came from, because of that. – Michael Hardy Nov 10 '11 at 18:33
@Michael: Thanks; fixed. – Brian M. Scott Nov 10 '11 at 18:38
Thank you very much, I just look up the 68-95-99.7 rule. This is very useful for me. I don't know any of what you guys discuss in here before, maybe this is not in the standard syllabus in the country I took my undergrad. Thanks a lot for your help. – lnth Nov 10 '11 at 19:52

Use the central limit theorem and a continuity correction.

The expected number of $6$s is $360/6=60$. The standard deviation of the number of $6$s is $\sqrt{360\cdot(1/6)\cdot(5/6)}= \sqrt{50} = 5\sqrt{2}$.

"70 or more" is the same as "more than 69". That's where the "continuity correction" tells you to use $69.5$ when you use a continuous approximation to this discrete distribution.

So $$\frac{69.5-\text{expected value}}{\text{standard deviation}} = \frac{69.5-60}{5\sqrt{2}} \approx 1.3435.$$ You're asking: what's the probability of being more than that many standard deviations above the mean. Plug that number into the appropriate table or software for the normal distribution.

BTW, I'd have said "a fair die", and used "dice" only as the plural.

Later edit: I find it being pointed out that calculators aren't allowed, as if that means this is all of no use. But the answer was phrased as "between" something and something. So you shouldn't need a calculator to apply this; you don't need a precise number, but only that it's between something and something else. You don't need a calculator to know immediately that $\sqrt{50}$ is slightly more than $7$, so that $9.5/\sqrt{50}$ is somewhat more than $1$. Even before doing any of this you know that the probability is less than $0.5$, since $70$ is well above the expected value. And "somewhat more than $1$ is certainly less than $2$. In this case, all you really need is that it's between $1$ and $2$ standard deviations above the mean. So what you see above should do it even if you don't have a calculator.

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It's the GRE exam in the US (like SAT, but for grads) so there's no calculator.. but thanks a lot for the formula and the English, I appreciate it. – lnth Nov 10 '11 at 16:51
@user19242: I think you are expected to figure out that 70 is between 1 and 2 standard deviations high (as Michael Hardy has done), then to know that B would represent between 0 and 1 standard deviation high and D would represent more than 3. – Ross Millikan Nov 10 '11 at 17:23