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A fair die is rolled 10 times. What is the probability of the event of getting a 3 exactly 4 times?

This is what I have so far...

$(\frac{1}{6})^4 \cdot (\frac{5}{6})^6 \cdot \binom{10}{4} \approx.049$

My teacher said my answer is wrong.

Part 2:

What is the probability of a 3 at least once?

This is what I had but my teacher said to keep going.

$P(\text{No 3s})= (\frac{5}{6})^{10}$

$1-(\frac{5}{6})^{10} \approx .162$

She said to keep going

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    $\begingroup$ Your exact calculation is correct on the left of $(\frac{1}{6})^4(\frac{5}{6})^6\binom{10}{4}$ however it is approximately $0.078143...$, not $0.049$. You made an arithmetic mistake somewhere. The second problem, the number $0.162...$ represents $(\frac{5}{6})^{10}$. You hadn't yet subtracted it away from one. It should be $0.83849...$ $\endgroup$
    – JMoravitz
    Feb 17 '15 at 4:34
  • $\begingroup$ $\left(\frac{1}{6}\right)^4 \cdot \left(\frac{5}{6}\right)^6 \cdot \binom{10}{4} \approx 0.054$ $\endgroup$
    – clay
    Feb 17 '15 at 4:59
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You have the formulae and theory right, however, you aren't evaluating them correctly.

For the first: $\dfrac{5^6}{6^{10}}\dfrac{10!}{4!\; 6!} = \dfrac{15625\times 210}{60466176} = \dfrac{546875}{10077696}\quad \approx 0.0543\ldots$

For the second: $(\tfrac 5 6)^{10} \approx 0.162\ldots\;$ so $\;1-(\tfrac 5 6)^{10} = \Box\;$?

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