chapter 1

Chapter 3: Discrete Random Variables and Probability Distributions71.a. With S = a female child and F = a male child, let X = the number of F’s before the 2 nd S.Then P(X = x) = nb(x;2, .5)b. P(exactly 4 children) = P(exactly 2 males)= nb(2;2,.5) = (3)(.0625) = .188c. P(at most 4 children) = P(X ≤ 2)(2)(.5)2= ∑x=0nb ( x;2,.5)= .25+2(.25)(.5) + 3(.0625) = .688d. E(X) = = 2 , so the expected number of children = E(X + 2).5= E(X) + 2 = 472. The only possible values of X are 3, 4, and 5.p(3) = P(X = 3) = P(first 3 are B’s or first 3 are G’s) = 2(.5) 3 = .250p(4) = P(two among the 1 st three are B’s and the 4th is a B) + P(two among the 1 st three are⎛3⎞⎝2⎠G’s and the 4th is a G) = 2 ⋅ ⎜ ⎟(.5)4 = . 375p(5) = 1 – p(3) – p(4) = .37573. This is identical to an experiment in which a single family has children until exactly 6 femaleshave been born( since p = .5 for each of the three families), so p(x) = nb(x;6,.5) and E(X) = 6( = 2+2+2, the sum of the expected number of males born to each one.)74. The interpretation of “roll” here is a pair of tosses of a single player’s die(two tosses by A ortwo by B). With S = doubles on a particular roll, p = 6 1 . Furthermore, A and B are reallyidentical (each die is fair), so we can equivalently imagine A rolling until 10 doubles appear.The P(x rolls) = P(9 doubles among the first x – 1 rolls and a double on the x th roll =⎛ x −1⎞⎛5 ⎞ ⎛ 1 ⎞ ⎛ 1⎞⎛ x −1⎞⎛5⎞⎜ ⎟⎜⎟ ⎜ ⎟ ⋅⎜⎟ = ⎜ ⎟⎜⎟⎝ 9 ⎠⎝6 ⎠ ⎝ 6 ⎠ ⎝ 6⎠⎝ 9 ⎠⎝6⎠5r(1− p)10(6)E(X) = = = 10(5) = 501px−109x−106V(X) =10⎛ 1⎞⎜ ⎟⎝ 6⎠5r(1− p)10(6)= = 10(5)(6) = 3002 1 2p( )6117