Question

Part I - Throwing a Dice with Six Faces (32 points) Consider a dice with six faces, i.e. a standard dice. The possible outcomes after throwing the dice once are 1, 2,3, 4,5,6 (a) Assume that the dice is thrown once. Let w represents the outcome. Explain why P(w-i) for i = i..6, when the dice is fair (b) The fair dice is thrown twice. Find the probability of occurance of at least one 6 The fair dice is thrown until a 6 shows up. Let X be the random variable representing the number of throws until the first 6 appears. (c) Determine the mass function of X, i.e. P(x-k) for k21 (d) Find the expected value of X and its variance. The mean value of X is the average number of throws to make a 6 appear for the first time. c) Determ

Answer 1

$(a)~Since~dice~is~unbiased~so~each~outcome~is~equally~likely~so~the~total~no.\\ of~equally~likely,~mutually~exclusive~and~exhaustive~cases~is~6~and~so\\ P(w=i)=Probability~of~getting~no.~i=\frac{1}{6},~i=1,2,3,4,5,6.$

$(b)~Probability~of~occurrence~of~at~least~one~6\\ =P(\{(1,6),(6,1),(2,6),(6,2),(3,6),(6,3),(4,6)(6,4),(5,6),(6,5),(6,6)\})\\ =\frac{11}{36}.$$(c)~P(X=k)=Probability~of~getting~6~at~kth~throw\\ =(1/6)(5/6)^{k-1},~k=1,2,3,....\\ since~in~first~k-1~throws~we~get~any~number~other~than~6~and~at~kth\\ trial~we~get~6~and~trials~are~independent.\\$

$(d)~Let~p=1/6,~q=5/6\\ then~E(X)=\sum_{k=1}^\infty kpq^{k-1}=p\sum_{k=1}^\infty\frac{\mathrm{d} q^k}{\mathrm{d} q}=p\frac{\mathrm{d} }{\mathrm{d} q}\sum_{k=1}^\infty q^k=p\frac{\mathrm{d} }{\mathrm{d} q}\left( \frac{q}{1-q}\right )=\frac{1}{p}\\ =6\\ E(X(X-1))=\sum_{k=1}^\infty k(k-1)pq^{k-1}=pq\sum_{k=1}^\infty\frac{\mathrm{d}^2 q^k}{\mathrm{d} q^2}\\ =pq\frac{\mathrm{d}^2 }{\mathrm{d} q^2}\sum_{k=1}^\infty q^k=pq\frac{\mathrm{d}^2 }{\mathrm{d} q^2}\left( \frac{q}{1-q}\right )=\frac{2q}{p^2}\\$

$Var(X)=E(X^2)-E^2(X)=E(X(X-1))+E(X)-E^2(X)=\frac{q}{p^2}\\ =30$