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A fair die is rolled 50 times. Find the probability of observing:

a) exactly 10 sixes
b) no more than 10 sixes
c) at least 10 sixes

I know how to do

a) $\frac{50!}{10!(40!)}$x$(\frac{1}{6})^{10}$x$(1-\frac{1}{6})^{50-10}$

Please help me out to do b) & c) I have tried same formula above and changing power to 9, 11 etc... But Can't get right answer.!

Appreciate your help!

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if you can compute the $a$ then just note that $b$ is just the sum of getting exaclty $1$ six plus exactly $2$ sixes$\ldots$ plus $10$ sixes. What about $c$ now? – clark Aug 21 '12 at 23:50
A related answer of mine:… It is related in the sense that it is another example of breaking an "at least" or "at most" problem into several "exactly" problems. – Austin Mohr Aug 22 '12 at 0:13
sorry! I have tried without changing power, changing 2 sixes,+ 3 sixes, 10 sixes...but can't get answer. – Sb Sangpi Aug 22 '12 at 1:19
(b) is 161379376595505164004862308502197265625 / 202070319366191015160784900114134073344. (c) should be easy to derive from this and (a). – Charles Aug 22 '12 at 19:15
up vote 4 down vote accepted

Hint: You will get no more than 10 sixes if you get no sixes or one six or two sixes or three sixes or ... or nine sixes. Since these possibilities are mutually exclusive, you can add the individual probabilities to get the total probability.

Once you've solved (b), think how you can use that answer to solve (c).

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does this mean $\frac{50!}{10!(40!)}$x$(\frac{1}{6})^{1}$x$(1-\frac{1}{6})^{50-1}$ – Sb Sangpi Aug 22 '12 at 11:30
Well, that is the chance of getting exactly 1 six. You need to work out something similar to this formula, and the one in question (a) for 10 sixes, but for each number of sixes from zero to nine... – Steve Kroon Aug 22 '12 at 17:11
@SbSangpi: That is neither the chance of getting exactly $1$ six nor the chance of getting exactly $10$ sixes. Formula $(1)$ in my answer gives the chance of getting exactly $k$ sixes. – robjohn Aug 22 '12 at 18:48
@robjohn is correct - sorry, I missed that Sb Sangpi hadn't changed the binomial coefficient in his comment. Remember that for 10 sixes, you used $\frac{50!}{10! (50-10)!} \times (\frac{1}{6})^{10} \times (1-\frac{1}{6})^{50-10}$. So when you want to move to exactly one six, you need to change the 10's occuring in the first factor to 1's. – Steve Kroon Aug 23 '12 at 5:57

Hint: If you get no six, you don't get exactly one six, or exactly two, or exactly three, and so on.

Similarly, if you get exactly one six, you don't get nonte, nor exactly two, and so on.

If you have several possibilities which exclude each other (i.e. if one of then occurs, none of the other has occurred), then the probability of getting any of those possibilities is just the sum of the probabilities of the individual probabilities.

"No more than 10" is "None, or one, or …, or 8, or 9, or 10".

"At least 10" is "more than 9", which is the opposite of "not more than none".

Since you already know how to calculate the probability of "Exactly $k$ sixes", those hints should enable you to solve the other questions.

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Hint: Your formula can be generalized to exactly $k$ sixes with $$ \binom{50}{k}\left(\frac16\right)^k\left(\frac56\right)^{50-k}\tag{1} $$ where $$ \binom{n}{k}=\frac{n!}{k!\,(n-k)!}\tag{2} $$

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can u provide more about question b) & c) ?. thx – Sb Sangpi Aug 22 '12 at 1:22
That was the intent of the hint. You can add the formula above for $k=0$ to $k=10$ to get b). Then note that the sum of the answers for b) and c) is $1$ plus the answer for a). – robjohn Aug 22 '12 at 12:40

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