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
\[
\left.\dfrac{d^k}{d\delta^k}\frac{(5+\delta)^n}{k!6^n}\right|_{\delta = 0} = \frac{{n\choose k} 5^{n-k}}{6^n}
\]
–
Lee Gao
·
2 years, 5 months ago

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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 \(\binom{n}{r}=\dfrac{n!}{r! \times (n-r)!}\)
–
Sandeep Bhardwaj
·
2 years, 5 months ago

@Gaurav Sharma
–
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
–
Agnishom Chattopadhyay
·
2 years, 5 months ago

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TopNewestSince 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 \[ \left.\dfrac{d^k}{d\delta^k}\frac{(5+\delta)^n}{k!6^n}\right|_{\delta = 0} = \frac{{n\choose k} 5^{n-k}}{6^n} \] – Lee Gao · 2 years, 5 months ago

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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 \(\binom{n}{r}=\dfrac{n!}{r! \times (n-r)!}\) – Sandeep Bhardwaj · 2 years, 5 months ago

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@Sanjeet Raria @Sandeep Bhardwaj – Gaurav Sharma · 2 years, 5 months ago

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@Calvin Lin @Krishna Ar @Agnishom Chattopadhyay @ – Gaurav Sharma · 2 years, 5 months ago

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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 – Agnishom Chattopadhyay · 2 years, 5 months ago

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