4)a) Let Si denote the score obtained by the players who throw i,i=1,2,…,6 , and let Xi be the number of people who throw i, and 1ij be the indicator function which is 1 only if the jth person throws i. So, we have Xi=∑nj=11ij. E[Si|Xi]=Xi(Xi−1)/2 and so E[Si]=(n2−n)/72 and therefore E[S]=(n2−n)/12, where S=∑6i=1Si is the total score.
b) The indicator variables Yij that are 1 if i and j throw the same number are pairwise independent and can thus be used to find both the expectation and the variance. The expectation of Yij is 1/6 and the variance is "\\frac16\\cdot\\left(1-\\frac16\\right)=\\frac5{36}" ,so the expectation of the score is "\\binom n2\\cdot\\frac16=\\frac{n(n-1)}{12}" and the variance of the score is "\\binom n2\\cdot\\frac5{36}=\\frac{5n(n-1)}{72}"
5)suppose there are 6 types of objects
the probabilities of getting a new type go on decreasing from 1 to 5/6, to 4/6....as we get more & more types, and using E[x] = 1/p, the expected # of days to get the next type goes on increasing
E[days to get 1st type] = 6/6 ...... 6/(7-1)
E[more days to get 2nd type] = 6/5 ..... 6/(7-2)
E[more days to get 3rd type] = 6/4 ..... 6/(7-3)
a. E[more days to get (j+1)th new object] = c/(c-j)
b. E[days to get full set] = Σ(c-j+1)
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