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, 3 months ago

Log in to reply

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, 3 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, 3 months ago

## Comments

Sort by:

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, 3 months ago

Log in to reply

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, 3 months ago

Log in to reply

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

Log in to reply

@Calvin Lin @Krishna Ar @Agnishom Chattopadhyay @ – Gaurav Sharma · 2 years, 3 months ago

Log in to reply

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, 3 months ago

Log in to reply