I11-13.
(a)F(x) =∫x0αe−αxdx= 1−e−αx(c)F(x) =∫x−∞f(x)dxI11-14. (b)
(
n
n+ 1)
=n!
(m+ 1)!(n−(m+ 1))!=(n−m)n!
(m+ 1)m!(n−m)!=n−m
m+ 1(
n
m)I11-15. (c) (i) Area from−∞toz= 1is 0.8413 and area from−∞toz= 0is 0.5.
Therefore area between 0 and 1 is 0. 8413 − 0 .5 = 0. 3413 By symmetry, twice this is
0.6826 which representsP(− 1 < X≤1)I11-16. (b) P(Ace) = 4/ 52 and P(king) = 4/ 52 so that
P(Ace or King) =P(Ace) +P(King) = 8/52 = 2/ 13I11-17. (b) P(E 1 ) = 4/ 52 and P(E 2 ) = 13/ 52 and
P(E 1 ∪E 2 ) =P(E 1 ) +P(E 2 )−P(E 1 ∩E 2 ) = 4/52 + 13/ 52 − 1 /52 = 16/52 = 4/ 13Solutions Chapter 11