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Chapter 1: Q5E (page 45)

Suppose thatnbalanced dice are rolled. Determine the probability that the numberjwill appear exactlynjtimes(j=1, . . . ,6), wheren1+n2+. . .+n6=n.

Short Answer

Expert verified

The probability that if\({n_1} + {n_2} + ... + {n_6} = n\)then no. j will appear exactly \({n_j}\) times\(\left( {j = 1,2,...,6} \right)\) is \(\frac{1}{{{6^n}}} \times \frac{{n!}}{{{n_1}!{n_2}!{n_3}!{n_4}!{n_5}!{n_6}!}}\).

Step by step solution

01

Given information

No. of balanced dice which are rolled is n.

02

Finding the possible arrangement

The Multinomial coefficient is

\(\left( {\begin{array}{*{20}{c}}n\\{{n_1},{n_2},...,{n_k}}\end{array}} \right) = \frac{{n!}}{{{n_1}!{n_2}!...{n_k}!}}\)

So, among n rolls, we can arrange\({n_j};j = 1,2,...,6\)as,\(\left( {\begin{array}{*{20}{c}}n\\{{n_1},{n_2},{n_3},{n_4},{n_5},{n_6}}\end{array}} \right) = \frac{{n!}}{{{n_1}!{n_2}!{n_3}!{n_4}!{n_5}!{n_6}!}}\)

Total no. of possible equally likely rolls are \({6^n}\).

03

Computing the probability

Referring to the equation 1.9.3 for the following probability.

The required probability is given by,

\({n_1} + {n_2} + ... + {n_6} = n\)

The required probability that number j will appear exactly\({n_j}\)times

\(\frac{{\frac{{n!}}{{{n_1}!{n_2}!{n_3}!{n_4}!{n_5}!{n_6}!}}}}{{\frac{1}{{{6^n}}}}} = \frac{1}{{{6^n}}} \times \frac{{n!}}{{{n_1}!{n_2}!{n_3}!{n_4}!{n_5}!{n_6}!}}\).

Thus required probability is:

\(\frac{1}{{{6^n}}} \times \frac{{n!}}{{{n_1}!{n_2}!{n_3}!{n_4}!{n_5}!{n_6}!}}\)

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