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

#Combinatorics #HelpMe! #Help

Note by Gaurav Sharma
6 years, 6 months ago

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Comments

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

Take the polynomial f18(δ)=(5+δ)18f_{18}(\delta) = (5 + \delta)^{18} and find C18,2=∂2∂δ2f18(δ)∣δ=02!=(182)516C_{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 C18,2618\frac{C_{18,2}}{6^{18}}

In general, show that to find the probability of getting kk sixes when n≥kn \ge k die are thrown simultaneously is dkdδk(5+δ)nk!6n∣δ=0=(nk)5n−k6n \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 - 6 years, 6 months ago

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I think it is (182)×516618\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 - 6 years, 6 months ago

@Sanjeet Raria @Sandeep Bhardwaj

Gaurav Sharma - 6 years, 6 months ago

Let

p=p=Probability of getting success

q=q=Probability of getting failure

P(r)=P(r)=Probability of exactly rr successes out of total nn events.

P(r)=(nr)×pr×qn−rP(r)=\binom{n}{r} \times p^r \times q^{n-r}

where (nr)=n!r!×(n−r)!\binom{n}{r}=\dfrac{n!}{r! \times (n-r)!}

Sandeep Bhardwaj - 6 years, 6 months ago
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