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STAT 408 Spring 2009
Homework
(due Friday, January 30, by 3:00 p.m.)
1. Suppose that P(A) = 0.40, P(B) = 0.30, P(A ∩ B) = 0.20.
What is the probability that …
a) either A occurs or B occurs (or both); b) B does not occur;
c) B occurs and A does not occur; d) neither A nor B occurs;
a) P ( A ∪ B ) = 0..
b) P ( B ' ) = 0..
c) P ( B ∩ A ' ) = 0..
d) P ( A ' ∩ B ' ) = 0..
2. Let events A and B and sample space S be defined as the following intervals:
S = { x : 0 ≤ x ≤ 10 }, A = { x : 0 < x < 5 }, B = { x : 3 ≤ x ≤ 7 }.
Characterize the following events:
a) A ' b) A ∩ B c) A ∪ B
{ 0 } ∪ [ 5, 10 ] [ 3, 5 ) ( 0, 7 ]
d) A ∩ B ' e) A ' ∪ B f) A ' ∩ B '
( 0, 3 ) { 0 } ∪ [ 3, 10 ] { 0 } ∪ ( 7, 10 ]
3. An urn contains six chips numbered 1 through 6. Three are drawn out. What
outcomes are in the event “Second smallest chip is a 3”? Assume that the order of the chips is irrelevant.
4. Suppose a baseball player steps to the plate with the intention of trying to
“coax” a base on balls by never swinging at a pitch. The umpire, of course, will necessarily call each pitch either a ball (B) or a strike (S). What outcomes make up the event A, that a batter walks on the sixth pitch? Note: A batter “walks” if the fourth ball is called before the third strike.
S S B B B B B S S B B B B B S S B B S B S B B B B S B S B B B B S B S B S B B S B B B S B B S B B B B S S B S B B B S B
5. Suppose S = { 0, 1, 2, 3, … } and
P( 0 ) = p , P( k ) = (^) k 5
1 , k = 1, 2, 3, ….
a) Find the value of p that would make this a valid probability model.
Must have P( 0 ) + P( 1 ) + P( 2 ) + P( 3 ) + P( 4 ) + … = 1.
1 = p + (^) ∑
k k^
= p + (^) ∑
k
k
= p +
= p +
1. p =
b) Find P( odd ).
P( 1 ) + P( 3 ) + P( 5 ) + P( 7 ) + … = ... 5
1 +^3 + 5 + 7 + =^ ∑
=
0
2 1 5
k
k
= (^) ∑
=
k
k
25
OR
a) ( 1, 2 ) ( 1, 3 ) ( 1, 4 ) ( 1, 5 ) ( 2, 3 )
- 1.2-
- a) P ( A ∪ B ) = P ( A ) + P ( B ) – P ( A ∩ B ) = 0.4 + 0.5 – 0.3 = 0.
- b) P ( A ∩ B ' ) = P ( A ) – P ( A ∩ B ) = 0.4 – 0.3 = 0.
- c) P ( A ' ∪ B ' ) = 1 – P ( A ∩ B ) = 1 – 0.3 = 0. - a) P ( A ∪ B ) = 0. - b) P ( A ∩ B ' ) = 0. - c) P ( A ' ∪ B ' ) = 0.
- 1.2-
- 0.7 = 0.4 + 0.5 – P ( A ∩ B ) P ( A ∩ B ) = 0. a) P ( A ∪ B ) = P ( A ) + P ( B ) – P ( A ∩ B )
- b) P ( A ' ∪ B ' ) = 1 – P ( A ∩ B ) = 1 – 0.2 = 0.
- 1.2-
- a)^1 / 3 b)^2 / 3 c) 0 d)^1 /
- 1.2-
- b) (i)^1 / 10 (ii)^5 / ( 2, 4 ) ( 2, 5 ) ( 3, 4 ) ( 3, 5 ) ( 4, 5 )
