The correct value is 3 e y 1215320 91 2527 36 36 24

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Unformatted text preview: 2.2 Let X be the number showing for a roll of a fair die. Find E [X ]. Solution Since pX (i) = 1/6 for 1 ≤ i ≤ 6, E [X ] = 1+2+3+4+5+6 6 = 7. 2 Example 2.2.3 Let Y be the number of distinct numbers showing when three fair dice are rolled. Find the pmf and mean of Y. Solution The underlying sample space is Ω = {i1 i2 i3 : 1 ≤ i1 ≤ 6, 1 ≤ i2 ≤ 6, 1 ≤ i3 ≤ 6}. There are six ways to choose i1 , six ways to choose i2 , and six ways to choose i3 , so that |Ω| = 6 · 6 · 6 = 216. What are the possible values of Y ? Yes, Y takes values 1, 2, 3. The outcomes in Ω that give rise to each possible value of Y are: {Y = 1} = {111, 222, 333, 444, 555, 666} {Y = 2} = {112, 121, 211, 113, 131, 311, 114, 141, 411, · · · , 665, 656, 566} {Y = 3} = {i1 i2 i3 ∈ Ω : i1 , i2 , i3 are distinct}. Obviously, |{Y = 1}| = 6. The outcomes of {Y = 2} are listed above in groups of three, such as 112,121,211. There are thirty such groups of three, because there are six ways to choose which number appears twice, and then five ways to choose which number appears once, in the outcomes in a group. Therefore, |{Y = 2}| = 6 · 5 · 3 = 90. There a...
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This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Tech.

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