HW#2 Solutions
AMS 311 Probability Theory (Spring 2011)
Read: Ross, sections 3.1-3.4 of Ch3. You’d better go through the following examples before doing the
assignments: Ch3: 2a-2g; 3a-3g; 3i-3n; 4a-4f
Part 1 (10 points):
PRINT
your name and StonyBrook ID (
Last, First, #ID
) on the top on your work
sheet. Staple your HW if more than 1 pages.
Part 2: Problems (90 points)
Write down your work step by step to get full score.
(1) (5 points) To find P(at least one 6 | sum at least 15):
{Sum at least 15}={(3,6,6); (4,5,6);(4,6,6);(5,5,5);(5,5,6);(5,6,6);(6,6,6)}, notice (4,5,6) has 3!=6
different arrangements, (3,6,6), (5,5,6) and (5,6,6) have 3 different arrangements, so #{Sum at
least 15}=6+1+3+3+3+3+1=20;
P(at least one 6 | sum at least 15)=P(at least one 6 and sum at least 15)/P(sum at least 15)=19/20
(2). (10 points)
P(Male)=0.72, P(Female)=0.28;
P(Success|Male)=0.17, P(Success|Female)=0.41
(a) To find P(Female|Success)
Applying Bayes’s Theorem, P(Female|Success)=
P(Female)*P(Success|Female)/[P(Female)*P(Success|Female)+P(Male)*P(Success|Male)]
=0.28*0.41/(0.28*0.41+0.72*0.17)=0.4840
(b) P(attending party)=P(Success)=P(Female)*P(Success|Female)+P(Male)*P(Success|Male)
=0.28*0.41+0.72*0.17=0.2372
(3). (12 points)
(a) P(1
st
=3 and 2
nd
>4)=0.1*6/9=1/15;
(b) P(both <3)=;
(c) P(both>3)=
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