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Dr. Hackney STA Solutions pg 7

Dr. Hackney STA Solutions pg 7 - 1-4Solutions Manual for...

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Unformatted text preview: 1-4Solutions Manual for Statistical Inferencethe remainingktasks. Thus for each of thenk+1we haven1×n2× ··· ×nkways of com-pleting the job by the induction hypothesis. Thus, the number of ways we can do the job is(1×(n1×n2× ··· ×nk)) +···+ (1×(n1×n2× ··· ×nk))|{z}nk+1terms=n1×n2× ··· ×nk×nk+1.1.16 a) 263.b) 263+ 262.c) 264+ 263+ 262.1.17 There are(n2)=n(n-1)/2 pieces on which the two numbers do not match. (Choose 2 out ofnnumbers without replacement.) There arenpieces on which the two numbers match. So thetotal number of different pieces isn+n(n-1)/2 =n(n+ 1)/2.1.18 The probability is(n2)n!nn=(n-1)(n-1)!2nn-2. There are many ways to obtain this. Here is one. Thedenominator isnnbecause this is the number of ways to placenballs inncells. The numeratoris the number of ways of placing the balls such that exactly one cell is empty. There arenwaysto specify the empty cell. There aren-1 ways of choosing the cell with two balls. There are(n2)ways of picking the 2 balls to go into this cell. And there are (ways of picking the 2 balls to go into this cell....
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