Modern Mathematical Statistics With Applications Answers

Chapter 3: Discrete Random Variables and Their Probability Distributions

3.1 P(Y = 0) = P(no impurities) = .2, P(Y = 1) = P (exactly one impurity) = .7, P (Y = 2) = .1.

3.2 We know that P( HH) = P( TT) = P( HT) = P( TH) = 0.25. So, P(Y = -1) = .5, P(Y = 1) =

.25 = P (Y = 2).

3.3 p(2) = P( DD) = 1/6, p(3) = P( DGD) + P( GDD) = 2(2/4)(2/3)(1/2) = 2/6, p(4) =

P( GGDD) + P( DGGD) + P( GDGD) = 3(2/4)(1/3)(2/2) = 1/2.

3.4 Define the events: A: value 1 fails B: valve 2 fails C: valve 3 fails

)()2( CBAPYP ∩∩== = .83 = 0.512

)()())(()0( CBPAPCBAPYP ∪

∪∩== = .2(.2 + .2 - .22 ) = 0.072.

Thus, P ( Y = 1) = 1 - .512 - .072 = 0.416.

3.5 There are 3! = 6 possible ways to assign the words to the pictures. Of these, one is a

perfect match, three have one match, and two have zero matches. Thus,

p(0) = 2/6, p(1) = 3/6, p(3) = 1/6.

3.6 There are ⎟

⎟

⎠

⎞

⎜

⎝

⎛

5 = 10 sample points, and all are equally likely: (1,2), (1,3), (1,4), (1,5),

(2,3), (2,4), (2,5), (3,4), (3,5), (4,5).

a. p(2) = .1, p(3) = .2, p(4) = .3, p(5) = .4.

b. p(3) = .1, p(4) = .1, p(5) = .2, p(6) = .2, p(7) = .2, p(8) = .1, p(9) = .1.

3.7 There are 33 = 27 ways to place the three balls into the three bowls. Let Y = # of empty

bowls. Then:

p(0) = P(no bowls are empty) = 27

p(2) = P(2 bowls are empty) = 27

p(1) = P(1 bowl is empty) = 1 27

3.8 Note that the number of cells cannot be odd.

p(0) = P(no cells in the next generation) = P(the first cell dies or the first cell

splits and both die) = .1 + .9(.1)(.1) = 0.109

p(4) = P(four cells in the next generation) = P(the first cell splits and both created

cells split) = .9(.9)(.9) = 0.729.

p(2) = 1 – .109 – .729 = 0.162.

3.9 The random variable Y takes on vales 0, 1, 2, and 3.

a. Let E denote an error on a single entry and let N denote no error. There are 8 sample

points: EEE , EEN , ENE, NEE, ENN, NEN, NNE, NNN . With P (E ) = .05 and P (N ) = .95

and assuming independence:

P(Y = 3) = (.05)3 = 0.000125 P(Y = 2) = 3(.05)2(.95) = 0.007125

P(Y = 1) = 3(.05)2 (.95) = 0.135375 P(Y = 0) = (.95)3 = 0.857375.