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# help

What is the probability of getting exactly 2 sixes when 18 die are thrown simultaneously?

Note by Gaurav Sharma
2 years, 3 months ago

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Since this seems like a homework question, I'll leave you with a short puzzle to ponder with.

Take the polynomial $f_{18}(\delta) = (5 + \delta)^{18}$ and find $C_{18,2} = \frac{\left.\dfrac{\partial^2}{\partial \delta^2} f_{18}(\delta) \right|_{\delta = 0}}{2!} = {18\choose 2} 5^{16}$. Prove that the probability of getting exact 2 sixes when 18 die are thrown simultaneously is $\frac{C_{18,2}}{6^{18}}$

In general, show that to find the probability of getting $$k$$ sixes when $$n \ge k$$ die are thrown simultaneously is $\left.\dfrac{d^k}{d\delta^k}\frac{(5+\delta)^n}{k!6^n}\right|_{\delta = 0} = \frac{{n\choose k} 5^{n-k}}{6^n}$ · 2 years, 3 months ago

Let

$$p=$$Probability of getting success

$$q=$$Probability of getting failure

$$P(r)=$$Probability of exactly $$r$$ successes out of total $$n$$ events.

$P(r)=\binom{n}{r} \times p^r \times q^{n-r}$

where $$\binom{n}{r}=\dfrac{n!}{r! \times (n-r)!}$$ · 2 years, 3 months ago

@Sanjeet Raria @Sandeep Bhardwaj · 2 years, 3 months ago

I think it is $\frac{{18 \choose 2} \times 5^{16}}{6^{18}}$

The denominator is the size of the sample space.

The numerator is the size of the required event space. To get what you like, you choose 2 die out of the 18 in (18 choose 2) ways and make them 6, and then have 5 options for each of the 16 dies which you can now configure in 5^16 ways · 2 years, 3 months ago